Trigonometry for Electrical Engineering | Graham Van Brunt | Skillshare

Trigonometry for Electrical Engineering

Graham Van Brunt, Professional Electrical Engineer

Trigonometry for Electrical Engineering

Graham Van Brunt, Professional Electrical Engineer

Play Speed
  • 0.5x
  • 1x (Normal)
  • 1.25x
  • 1.5x
  • 2x
12 Lessons (5h 4m)
    • 1. Ch 00 Intro to Trig Course

      8:15
    • 2. Ch 01 Point Lines & Angles

      12:55
    • 3. Ch 02 Intersecting & Parallel Lines Triangles

      11:26
    • 4. Ch 03 Triangles

      21:46
    • 5. Ch 04 The Pythagorean Theorem

      9:37
    • 6. Ch 05 Unique Triangles & Ratios

      13:01
    • 7. Ch 06 Trig functions & Tangents

      33:40
    • 8. Ch 07 The Unit Circle

      80:23
    • 9. Ch 08 Trig Identities

      42:12
    • 10. Ch 09 Trig Identities

      22:53
    • 11. Ch 10 Trig Identities Double & Half Angle Formulas

      31:33
    • 12. Ch 11 Trig Functions in the Time Domain

      16:39
  • --
  • Beginner level
  • Intermediate level
  • Advanced level
  • All levels
  • Beg/Int level
  • Int/Adv level

Community Generated

The level is determined by a majority opinion of students who have reviewed this class. The teacher's recommendation is shown until at least 5 student responses are collected.

69

Students

--

Projects

About This Class

This course takes you from the fundamentals of trigonometry to the more sophisticated required in electrical engineering. As you work and study in electrical engineering you are going to run int proofs and equations that are based on trigonometry. Over my engineering experience, I have noted and gathered what I believe are the most significant.

Meet Your Teacher

Teacher Profile Image

Graham Van Brunt

Professional Electrical Engineer

Teacher

Hello, I'm Graham Van Brunt B.Sc; P.Eng.I have spent an entire career in the power sector of electrical engineering, I graduated from Queen's University in Kingston Ontario, Canada. Coupled with subsequent studies with Wilfrid Laurier University I have traveled the globe and applied my skills to garner my protection and control experience internationally.

I have a passion for staying in touch with my profession as an electrical engineer and have a kinship for mentoring that has kept me in front of an audience of learners.

See full profile

Related Skills

Technology Data Science

Class Ratings

Expectations Met?
  • Exceeded!
    0%
  • Yes
    0%
  • Somewhat
    0%
  • Not really
    0%
Reviews Archive

In October 2018, we updated our review system to improve the way we collect feedback. Below are the reviews written before that update.

Your creative journey starts here.

  • Unlimited access to every class
  • Supportive online creative community
  • Learn offline with Skillshare’s app

Why Join Skillshare?

Take award-winning Skillshare Original Classes

Each class has short lessons, hands-on projects

Your membership supports Skillshare teachers

Learn From Anywhere

Take classes on the go with the Skillshare app. Stream or download to watch on the plane, the subway, or wherever you learn best.

phone

Transcripts

1. Ch 00 Intro to Trig Course: trigonometry is a branch of mathematics that focuses on the relationship between the sides and angles of triangles. The word trigonometry comes from the Latin derivative of Greek words for Triangle Trigon on and measure met Ron Trigonometry or in its shortened form, Trig is an intricate piece of other branches of mathematics such as geometry, algebra and calculus as well as it is a fundamental piece to the proof for electrical engineering equations. The course starts, Ah, by describing a point in the universe and how a second point in universe, when connected, describes a straight line. The placing a the third point in the universe would uniquely define a plane. And you can further define that plane by setting a set of rectangular coordinates on that plane, and it would be now described by the X Y coordinates of the rectangular playing. And if those three points are then joined by two lines they would uniquely describe on angle on the X Y plane from there, the course then goes on to describe how that angle is measured, and it is split up into a circular diagram, with the various angles measuring as degrees from zero degrees through this 360 degrees. The other way of measuring angles is described, and that is radiance. And the radiance are explained, as well as the conversion factors of how we can transpose radiance into degrees or degrees into radiance. From here we go into the triangles and we start out by defining what is meant by an acute angle, triangle or right angle triangle, or in a two single triangle and the characteristics of the various triangles. Next, the course gets into the various serums, starting with some of the basic theorems that are out there, such as triangles that have the same interior angles or the same sides are equally proportional as faras their sides are concerned. From here we get into one of the best known theorems. That air up there in the in the world of trigonometry or engineering is the Pythagorean theorem. And not only do we look at the results of the path agree in fume, but we actually prove the PPA, thuggery and serum. In other words, the fact that you take the square of the two sides of a right angle triangle on, add them up and they will equal the square that's on the high pot in use or in other words , a squared plus B squared is equal to C squared. This gets us to where the course starts to get very interesting as we start to look at special or unique triangles that have set angles such as the 30 60 90 trained colony and the Sauce Elise Triangle, which is having 45 degrees in the Cy and the angles next to the equal sites. We look at these triangles because they crop up from time to time quite often in the analysis of some of the electrical systems, as you will see later on in other courses. For example, when looking at the connection of ah three face Transformer Bank, you will see that the 30 60 90 triangle crops up and helps us do the analysis off that transformer. In describing the fees relationships between the primaries and the secondaries, this brings us to the core of trigonometry and that is the trigger, the metric functions. This is where we define what the signs co signs tensions, ko sexy cans, the sea canes and coo tangents are and how we represent them. We also have placed them inside what we call the unit circle, which will be described and tell us the usefulness of the unit circle. We then start to study what we conduce with East regular metric functions in the way of describing what is known as the law, the signs and the law of the co signs, which are nothing more than special equations that fall out of a combination of the Trigana metric functions. These laws, if as they're called, are actually proven in the course, but also some of the usefulness of it. You will see as we start using them in the various courses that are on my site. Further work with the Trigana metric functions will lead us into describing the reciprocal identities, quotient, identities, negative angle identities and Pythagorean identities, all of which are proven and again pointing out the usefulness of these equations. More useful identities are developed, such as the summon difference formulas for signs and co signs. The product to some formulas and the sum to product formulas and further identities that are also developed will be done during this course. The tangent is given a large consideration on this course, and we spend a lot of time developing what is meant by a tangent and proving what a tangent is capable of, as well as its relationship to a radius line in a circle where it meets the change in at point A. And before finishing up on tangents, the student will learn how to draw a tangent from an external point to a point A on the circle. The final thing that we will be looking at is these functions in the time domain, which is very significant in regard to how it's used in electrical power systems. So the time domain is described very extensively, and we're looking at the various things that change with time. In this particular facet of Trigana metric functions, this is the end of the introduction. 2. Ch 01 Point Lines & Angles: trigonometry for electrical engineering. Trigonometry is, ah, huge feel that you could spend a lot of time in. But I've condensed a lot of the trigonometry that you'll run across in electrical formulas , and we'll go through them from the very basics up to the more intricate. And we'll have a look at the proofs of some of these serums and equations. We're going to start off in Chapter one, calling it Points, Lines and Angles. Trigonometry is a branch of mathematics that focuses on relationships between the sides and angles of triangles. The word trigonometry comes from the Latin derivative of Greek words for Triangle Trigon on and Measure Met Ron Trigonometry or Trig for short is an intricate piece of other branches of mathematics such as geometry, algebra and calculus, as well as fundamental proofs for electrical engineering equations. A point is symbolized by a dot and represents a specific location somewhere in the universe . It has no size and no shape. A lying is made up of a series of points that extends infinitely in two directions. A line segment is defined by two points on the line and has only one dimension. It is part of a line that has to defined endpoints and represents a collection of points inside the in points and is named by the end points quite often no for expedience purposes . Sometimes a lying is given one designation rather than designation of the two endpoints, but it is usually very intuitive, and you can tell whether we're talking about ah line or a point when it's being used. 1/3 point will define a plane. A plane is a flat surface without edges, and it extends infinitely in two dimensions and is determined by three points in the plane that are not on the same line, and in this example it is points a, B and C. An example of a plane would be the rectangular coordinate plain. The three points could be determined by measuring perpendicular distances from any point along the X axis and any point along the Y axis an angle is formed by two line segments. Had that have the same endpoint? The end points of the two segments are either lettered or numbered. In this case, it is a B C with a single letter designating the common point, and in this case it's the letter a. So the three letters can designate an angle, and that angle can either be designated in this case C A, B or B, A. C. They're both the same thing, as long as the Vertex or the common point of the two lines or lying segments are in the middle of the designation. And this symbol here is usually used to indicate we're talking about an angle, and it can be followed by the three letters. So it be an angle C A B and an angle B A C. So you could. You use either one of those to designate this angle and sometimes for expedience purposes. Again, the three letters can be dropped in favour of a single letter or a single number that's place between the two lines inside what is known as the angle and even if numbers or letters aren't used, sometimes Greek letters are used, and quite often you'll see data being used as a zey designation of a particular angle. Angles have an initial side and a terminal side. The vertex of the angle is the common point at which initial side and terminal side meat the initial side is said to be in reference position when it's Vertex is located at the origin of the rectangular coordinate plain, and the initial side goes along the positive access. Measuring an angle is similar to cutting up a pie, which is basically a circle, which is divided into 360 major wedges called degrees, each of which may be subdivided into smaller way to save tents or even hundreds etcetera. The bigger the wedge of pie, the more degrees it contains. The number of degrees is written with a superscript zero following the number, indicating that that number represents degrees. Another way to measure angles or other than using degrees is by radiance, which seems to be favored by mathematicians. The name is derived from the radius of a circle. One radiant is the distance measured along the circumference of the circle, which is the exact same length. As of the circle's radius is, this forms a pie wedge measuring one radius along one side of the wedge, one radius alone, the circumference in one radius along the other side of the wedge. As I said, this is a method preferred by mathematicians because it works very nicely and a lot of calculations because if the wedge we see here is one radiant and we add a 2nd 1 making it to radiance. And if we add 1/3 1 making it three radiance and at a fraction of a radiant and that fraction is 0.14159 making it all totaled 3.14159 radiance. We get exactly 100 and 80 degrees, 1/2 of the circle. In other words, if we measure Pi radiance or 3.14159 radiance, we get 180 degrees. Or if we measure to pie Iranians, it is the whole circle sold. The measurement of angles can be done either by doing it with degrees for with radiance. Measuring angles because we do it in two ways is similar to measuring links, which weaken due in in more than one way. In fact, we can measure them in kilometers, or we can measure them in miles. We can measure and feet, or you can measure in meters. It doesn't matter what form it takes in the measurement. It's still the same distance. If you're measuring volume, whether was gallons or leaders, it would still be the same amount. It would just be different, whether you would be measuring in gallons foreign leaders, and you can convert from one to the other with just one simple mathematical manipulation. And so it is with angles. You can chop up a or divide off a pie. In this case, it's a pumpkin pie. And if we divided the pie up this way and measured in radiance, the big piece would be because it's half the pie and it's 180 degrees is pie radiance. We just found that out. If we were to look at 1/4 of the pie, it would be a pi over two radiance. And if we look at the eight of the Pyrgos to small pieces, which are a need to the pie, they would measure pi over four radiance. Now it's the same pie, and we're dividing it up the same way. But we could use degrees, and in the case of the big piece, it would be 100 and 80 degrees. In the case of the quarter pie, it would be 90 degrees and in case of a NATO, the pie, it would be 45 degrees for each one of those small pieces. So if we're looking at the markings in degrees around a circle, we could see that if we had a 45 degree angle, we could convert that 45 degree angle to pi over four, which we saw in the case of the pumpkin pie. We could convert the 90 degrees to pi over two, and we could go right around a circle, converting all of the degrees to radiant measurements. And there is a conversion of formula that you can remember in this one sticks in my mind that pie radiance is equal to 180 degrees. From that formula, you can convert from radiance two degrees or degrees to radiance. One radiant would be 180 degrees all over pie, and that would give you degrees. Or, if you had degrees and you wanted radiance, you would multiply by pi over 180 degrees, and not would give you the number of radiance that you would have. A counterclockwise rotation of the terminal side oven angle with respect to the initial side, is considered a positive angle, regardless of whether you're measuring the angle in degrees or in radiance. A counterclockwise rotation of the terminal side, with respect to the initial side is considered a negative angle again, regardless of whether you are measuring the angle in degrees or in radiance, angles have a type name, depending on the size of the angle on an angle that measures between zero and 90 degrees is called acute angle. Right angles have a measure of 90 degrees and usually designated by a little box symbol. At the vertex of the angle, an angle that measures between 90 and 180 is called an obtuse angle. And finally, where a straight angle measures on 180 degrees, it is called a straight angle, and this ends the chapter. 3. Ch 02 Intersecting & Parallel Lines Triangles: trigonometry for electrical engineering, Chapter two intersecting and parallel lines. Two lines that meet at a single point are called intersecting lines. Two lines that are in the same plane and that will never meet or never intersect are called parallel lines, and they're usually indicated by one to or three Chevron's lines that are perpendicular our lines that intersect at right angles. The symbol for perpendicular is the same as that for a right angle. Because they are perpendicular, they are 90 degrees. We're now going to look at the first in the simplest of trigonometry identities, and that is the that opposite angles of intersecting lines are equal. In this case, we have designated the angles of these intersecting lines to be A, B, C and D. What we're about to prove is that angle A is equal toe ankle. See an angle B is equal to Angle D. If we look at angle a an angle B, we see that they are making up a straight line, so if we add them together, they will equal 100 90 degrees. If we look at angles, see and angle B, they also add up to 100 degrees because they form another straight angle or a straight line . If we look at those two equations, we see that the left hand sides are both equal to 190 degrees. Solder to left hand sides must be equal to each other. In other words, a plus B is equal to C plus B. Well, we can cancel out the be on both sides of the equation, and we're left with a is equal to see. In other words, angle A is equal to angle, see, And if we look at a angle a plus angle D that is also adding up to another straight line. So it is. They are, when added together, equal to 180 degrees. And if we look at angle a plus angle B, which we already have looked at, it is already equal. We know that it is equal to 100 90 degrees. So we have these two equations now that left outside Zehr both equal to 180 degrees. So the left hand side have to be equal to each other and on both sides of this equal sign, we have the angle A so we can cancel that out, and it says that Angle B is equal to angle D. So we have just proven that angle A is equal to angle. See an angle. B is equal to Angle D, and that is what we set out to prove in the first place. For the moment, let's consider an angle and let's call it fate a one formed by drawing two straight line sectors joined at one point, which is the vertex of the angle. Now let's extend the initial side and duplicate another terminal line at the same angle again joined at the one point, which is the vertex of the new angle. Because we are have just duplicated the terminal line, it will be at the same angle. Fate a one. It is intuitive that the two terminal lines are parallel, but let's go through a little exercise just to make that definitely clear. And let's rotate clockwise slightly the second angle. Clearly, the angle will be larger than data one, and because it is larger, if the two lines are extended at some point, they will intersect because of the difference in the angle. And the more we rotate clockwise, the larger the angle becomes and the sooner the intersection will take place. This means that lines with angles larger than data one cannot be parallel to the first terminal line, going back to where we started with the terminal line at Angle one Earl Angle theta one. Now let's rotate the terminal line counterclockwise. Clearly, that angle will be smaller than Taito one, and because it is smaller, if the two lines are extended at some point, they will intersect because of the difference of the angles. And the more we rotate counterclockwise, a smaller, the angle becomes, and the sooner the intersectional take place. This means that lines with angle smaller than data one cannot be parallel to the first terminal line on this plane, since there must be at least one lying that passes through the initial line at the vert tax of the angle that is parallel to the first terminal line and all lines passing through the Vertex that form angles and her greater or less than theater. One are not parallel to the first room on the line, then the angle fado one must be the one and only angle that forms a parallel line, which means that two terminal lines must be parallel. Now let's look at this situation further. By extending all of the lines, we're now going to go through a series of definitions or terminologies that we're going to use when looking at trigonometry. For the rest of this course, we have already defined the term parallel lines, and any line that cuts parallel lines is No. One as a trans verceles to help us with the the discussion of terminology, I'm going to number the angles formed by the parallel lines and the trans versatile as 123 and four and 567 and eight. The trans versatile same side exterior angles are 164 and seven that Trans Verceles same side interior angles are 238 and five. Corresponding angles are on the same side of the trans versatile and in the same position with respect to the parallel lines. In our example here, the corresponding angles are one in five two and six, four and eight, three and seven can grow in. Angles are angles that are equal and are usually designated with unequal sign and sometimes with a squiggly line on top of it. or just a plain equal sign. Supplemental angles add to 180 degrees. Complementary angles add to 90 degrees, using our example of two parallel lines being trans first by 1/3 line. We already know that because we have intersecting lines, the opposite angles of those intersecting lines are equal or congruent, and they are one and three two and four, five and seven, six and eight. Furthermore, looking at the Trans Verceles of Parallel Lines, the following angles are also congruent one in five, two and six for an eight, three and seven, and that angle one is congruent with 35 and seven and angle to is congratulating with 46 and eight alternate angles. One in seven are equal three and five to an eight four in six supplemental angles. These air angles that add to 180 degrees. One plus two, five plus six, four plus three and eight plus seven two plus three, six plus seven, one plus four and five plus eight. And this ends the chapter 4. Ch 03 Triangles: Chapter three triangles. If we look at an angle, it has two sides, and it's open on one side. We designate the angle with three letters. In this case, you can see it. It's a B and C. If we close that angle, it forms what we call a triangle, and the triangle is designated with the three letters that you see there. The sides of the triangle are designated by two letter combinations. We have side A, B side B, C and side. See a. The angles of the triangle are designated by three letters. In some cases, C A B, or in some cases, we'll call that angle. Just angle A. And ABC will describe what you see in green there, and sometimes you will call that angle just angle B and similarly angle that you see there in red is B C. A. And sometimes we just designated angle See, and sometimes the interiors of the triangle are numbered or lettered to describe the angle . In this case, angle C A B could be having a one in it, and we had call it angle one. Instead of angle a angle, ABC could be called angle to instead of angle B and angle B C A. Could be called angle three instead of angle. See the area of a triangle and we're gonna call a triangle ABC. The area is given by multiplying 1/2 times the base times the height. Now the base could be any one of the three sides, but whichever side you pick, the height of the triangle would be the perpendicular vertical distance to the next vertex of the angle. So BC If that is chosen as the base, then it would be the vertical height to Vertex. A. Now we could have picked the side a B as the base, in which case we would have had to pick the height to be the vertical distance from the base to Vertex. See, And if we would have chose Side A C as the base, we would have had to choose the height of the vertical distance to the Vertex Be. We have names for different types of triangles, and the triangle is classified by its angles and its sights. A triangle that has three acute angles is called an acute triangle. A triangle that has one right angle is called a right triangle or a right angle triangle, A triangle that has one of two single is called an obtuse triangle, when a triangle has three equal sides, hence three equal angles. We call the triangle and equal lateral triangle, and we mark the equal sides by a slash mark. As you see there, the angles in an equal lateral triangle are always 60 degrees. When a triangle has to equal size, it's called an ex sauce Elise triangle. The angles opposite to two equal sides of the same length are congrats. A triangle without any equal sides or angles is called a scale lean triangle. When two triangles are congruent, it means that they have the same size and shape. This means that they have the same angles and their sides are equal. The red slash marks show us which sides are equal, and the angles are marked as shown in green, indicating which angles are congruent. Mathematically they have. They are shown with an equal sign that has a squiggly line on top of it, triangles that have congruent angles but not the same sides. Same size sides are called similar triangles, and similarity is shown with just a little squiggly symbol. As you see here, we're now going to look at the angle some property of a triangle. This serum states that the some of the interior angles of a triangle must add to 180 degrees. What you see in front of you is a typical triangle. We are going to call it a B C. On the sides are a BBC and see A and the angles are contained inside those three sides. And I'm going to start by drawing a parallel line to side BC through the vergis E A. And I'm going to call that line P a que. Now, because P a que is a straight line or a straight angle, it has to be ah, 100 needy degrees. Therefore, angle P A B plus angle B, A C plus angle Q A C have to add to 180 degrees. The two sides A B and a C of that triangle are trans verceles to the two parallel lines. And because they are trans verceles, the alternate angles have to be equal. That means angle. Q. A C must equal angle a CB because their alternate angles and on the other side, we have alternate angles. Angle P A. B must be equal to angle C B A. So I can substitute the interior angles of least two of them of the triangle in that equation so I can plug in angle C B A for P A B and I can plug in angle a CB for Q A. C in the equation, an angle B A C is already on interior angle of the triangle, so what we have just proved is that angle C B A plus angle B, A C plus angle a CB are equal to 180 degrees, which means since we arbitrarily chose any triangle to start with, then the some of the interior angles of a triangle is 180 degrees. We're now going to look at the triangle proportionality, the're, um This involves two triangles that are similar Now. Remember that similar triangles air triangles that are different sizes but have the same interior angles, the theorem states. If two triangles are similar, their sides are proportional, mathematically stated the ratio of the length of Syed a B of the first triangle divided by the length of the side X y of the second triangle is equal to the ratio of the length of side. A sea of the first triangle divided by the length of the side X zed of the second triangle , which is also equal to the ratio of the length of side BC of the first triangle, divided by the length of the side wise ed of the second triangle. We're going to start out the proof. But considering two parallel lines, I'm going to construct a random green triangle with one side, the base along the bottom of the two parallel lines and the vertex of the other two sides touching the top parallel lying. Next, I'm going to construct a second random red triangle whose base is the same as a green triangle, and its vertex of the other two sides is also touching the top parallel line, but at a different location than that of the green trey angle. Next, I want to look at the areas of these two triangles. Now, remember that the area of any triangle is given by multiplying 1/2 the base of the top triangle times its height. The hike H of these two triangles, given by the distance between the two parallel lines and the base of bowl triangles is the same. Therefore, the area of the green triangle is given by 1/2 B times H and the area. The rant triangle is given by the same 1/2 be times H, which means that the two areas are equal and generally speaking. Then, when triangles air drawn between two parallel lines with the same base, they will have equal areas keeping the same two parallel lines separated by the same distance, H. I'm again going to construct a green triangle similar to the way I did before. This time, however, I'm going to construct a red triangle between the two parallel lines, but with a different size base. As a reminder. The area of a triangle is given by 1/2 the base times the heights, which means the area of the Red Triangle is 1/2 the base, which I've indicated with a red B times. The height, which is each. The area of the green terrain triangle is 1/2 the base, which is I've indicated, with the green B times the height. Now I'm going to divide the area of the Red Triangle by the area of the Green Triangle, which I conduce with both sides of the equation. The common elements of the right side, numerator and denominator cancel out. Leaving the ratio of the areas of the triangles with equal heights is equal to the ratio of their basis. We need to look at one more theory and before proceeding to the proof of the triangle proportionality through. It's called the basic proportionality Fear, um, which essentially is a subset of the triangle proportionality. Theroux. The basic proportionality theorem states that a line parallel to one side of a triangle divides the other two sides in equal proportions. What we have to prove is that a D divided by D B is equal to a E divided by E. C. When we're talking about lengths of the sides now, I'm going to start to proof by constructing a red and green triangle. As you see in the diagram, we will again be using the equation for calculating the area of a triangle, which I've put up there for reminder purposes. So let's move the triangles to the left, rotating them so that the bases a d e r on the bottom as shown, and they both have the same I, which I'm gonna call h The Green Triangle area is 1/2 times a d Times H. The area of the Red Triangle is 1/2 D B times h. Previously, we discovered that the ratio of the areas of the triangle with equal heights are equal to the ratio of their basis. That means the ratio of the area of the green triangle to the area of the Red Triangle is equal to the ratio a d. Two d b Going back to our original diagram, I am leaving the green triangle but replacing the red triangle with a blue triangle. As sure remember that I'm still using the same two parallel lines. I will move the green and blue triangles to the right, rotating them such that the triangle's base is on the bottom. C. E. A. The area of the green triangle is given by 1/2 e times. The H. The area of the Blue Triangle is 1/2 times E. C. Times H, and previously we discovered that the ratio of the areas of triangles with equal heights are equal to the ratio of their basis. That means the ratio of the area of the green Trang go to the area of the Blue Triangle is equal to the ratio off a E all over E. C. I want to go back to the center set of triangles, replacing the red Triangle, and then I'm going to remove the green triangle. And for the moment, I'm going to also remove the labelling cause I want to rotate the blue and red triangle 280 degrees without turning the labeling upside down. I will now replace the labelling right side up so they can be easily read. We now have two triangles, a red one and a blue one between the same two parallel lines and whose bases are equal. Hence their areas must be equal, which means the two ratio equations are equal and that a D all over D B is equal to a E all over E. C, which is what we originally set out to prove. I'm returning to the beginning now because we did a lot of manipulation with triangles and this is what we started out with one triangle, which is being cut by a parallel line, and we can see that the line drawn parallel to one side of the triangle cuts the other two sides in equal proportions, which is a D divided by D B, which is mathematically equal to E all over E. C. And now the last step proving the triangle proportionality therapy, which states if two triangles air similar, their sides are proportional. Here we have two triangles of blue one and a red one whose angles are equal but whose sides are not the Blue Triangle. ABC is larger than the Red Triangle, X Y said. What we're looking to prove is a be all over X Y is equal to a C all over. X ed is equal to B C all over. Why is it Let's move the red triangle over the Blue Triangle. Such that diverted sees A and X overlap. And since angle A is equal to Angle X, the two sides of these angles coincide. And since angle why is equal to Angle, B, side, Wise said, is parallel to B C. The basic proportionality, the're um, tells us at the first part of the equation is true that a be all over. X Y is equal to a C all over, X said. So we're part way there now. Let's go back to the beginning, And this time let's move the red triangle over the Blue Triangle. Such the diverted sees C and zed overlap, and since angle C is equal to angle zed, the two sides of these angles coincide. And since angle why is equal to angle B Side X wise, parallel to side A B. The basic proportionality theorem tells us again that the last part of the equation that we're setting up to prove is true. A C all over ex Zach is equal to B C all over. Why's that? So the bottom line is a be divided by X Y is equal to a C, divided by X said and BC. All over wise, it is also equal to a C divided by X zed. Therefore, a be divided by X Y is equal to a C divided by X set is equal to B C, divided by y zed, meaning if two triangles are similar, ther sides are proportional and we're done. This ends the chapter 5. Ch 04 The Pythagorean Theorem: Chapter four, The Pythagorean theorem. Now this is a short chapter, and we're only going to deal with the Pythagorean theory. The Pythagorean theorem states that for any right angle triangle, the square of the length of the triangles high pot muse is the sum of the squares of the lengths of the triangles. Other two sides. This serum is represented by the formula. C Square is equal to a squared plus B squared, which can be rewritten. A squared plus B squared is equal to C squared. Another way of saying this is if squares are drawn on all sides of the right angle triangle the area of the squares on the two sides of the right angle triangle when added equal to the area of the squares on the iPod News. The formula for the Pythagorean theorem is often used but seldom proved. So let's look at that proof now. And to do that, we're going to start with a right angle triangle and let's designate the sides to be a B C . And the right angle contained is contained by the sides A and B, and I'm going to designate the other two angles of this right angle Triangle one and two, as you can see on the side. No, here that the 90 degree angle plus angle one and angle to must add to 180 degrees because they're the interior angles of a triangle. Then that means that angle one plus angle to must equal 90 degrees. What we are looking to prove is the formula A square plus B squared is equal to see square . We start by placing this right angle triangle or its equivalent with this high pot in use on the horizontal. And they've done that. And I've colored the triangle red just and you'll see, cause we wanna be able to tell it from the other triangles and I'm gonna be putting on the diagram and I'm going to place the side designations on the inside of the triangle. Just allow me room on the outside of the triangle, and I'm also gonna place the angles numbers one and two on the inside of the triangle. Now I'm going to place another triangle which is equivalent to our original triangle. And this time I colored it blue, and I'm going to place the apex of angle one touching the apex of the angle to of the Red Triangle. As shown with these two, Apex is touching. I'm going to rotate the Blue Triangle. Or I have rotated the blue triangle about the apex of angle one until side B of the blue triangle overlaps and coincides with side A of the red triangle, as you see in the in the slide. Now, this is also unidentifiable triangle, so the sides are going to be labeled a B C. And the angles are going to be one and two. Now I've colored the labelling Azaz blue and the Red Triangle as the labeling is red, but they're still the same lengths. Now I'm going to place another triangle which is identical to our original triangle. And I'm gonna color this one green and I'm gonna place it. So the apex of the green Triangle one is going to be on the apex of angle to of the Blue Triangle, and then I'm going to rotate it or line it up so that the side B aligns with side A. That is a lying the green triangle side. Be alliance with the blue triangle side A and I'm going to add one more. Try angle to this group and I'm coloring it yellow. But it's identical to our original triangle, and I'm gonna place it so the apex of angle one is going to be rate over the apex of angle to of the green triangle, and I'm gonna label the sides and the angles, as I did on the previous ones. And the labels are going to be yellow, however, the same lengths as the other four triangles. Of course, now, since the high Pontin use of all four triangles form four angles at 90 degrees and the four triangle High Pontin uses are equal, that is of length C, they will form a perfect square. Looking at the little square formed on the insight of the four triangles. The length of each side is given by a minus B. For example, the bottom side of the little square is given by read a minus blue be the right side of the little square is given by blue. A minus green be on the top side of the little square is given by green, a yellow be, and the left side of the little square is given by yellow a minus red B. Now, regardless of the color, the size of this square are equal to a minus B, and they are perpendicular angles to each other. Therefore, they also form a perfect square. The area of the little square, plus the area of the four triangles, is equal to the area of the big square. You can see that from the diagram. Now. Mathematically, we can come up with this equation, which is the quantity A minus B squared plus four times 1/2 a B is equal to C squared. The A minus B squared is the area of the little square. The 1/2 a times B is 1/2 the base times the height of each little triangle. Eso we have four of them, so we have four times 1/2 a times B is equal to C squared, which is the area of the big square. We can now rewrite that equation, expanding out the binomial a minus B squared, which gives us a squared minus two a B plus B squared. And that's added to a to A B because the four in the 1/2 result in the number two and that's equal to C squared. Well, the minus two a b in the plus two a b cancel out the brackets don't mean anything in this case because it's all additional or add signs or minus signs, and that's going to leave us with a squared. Plus B squared is equal to see square, which is what we set out to prove in the first place, which is the path agree, and the're, um, this ends the chapter. 6. Ch 05 Unique Triangles & Ratios: Chapter five unique triangles and ratios. We're gonna look at some unique triangles now and ah, there. Ratios of the sides will depend on the angles of the various triangles that we're looking at. And I'm going to start with the equal lateral triangle. And we saw earlier that the equal lateral triangle, besides of this unique triangle are equal. Which means the interior angles must also be equal and because they have to add to 100 needing degrees if we divide 100 and 80 into by three. Ah, we're going to come up with 60 degrees for each one of the angles. Now, in studying this triangle, I'm going to drop a perpendicular from the top, angle down, toothy if you would call it the base of the triangle, and it's going to split the top angle of 60 degrees into two angles of 30 degrees, and it's going to split the base of the triangle from a side of X to to to, uh, links of ex all over to. So, in actual fact, we have two triangles, identical triangles that air back to back that have 30 60 90 degrees as their interior angles and the two back to back sides. I'm going to designate as why now if we look at these triangles with the Perth agree and the're, um in mind X squared or the high pot in use of each one of those two identical triangles that air back to back the high pot in use is X. So if we square it, it's going to be equal to the square of the other two sides, which is X over two squared plus y squared. I'm now gonna rearrange this equation just mathematically, and I'm gonna take the y squared over the left hand side and the X squared over to the right hand side. And that's going to mean that the equation will now read why squared is equal to minus the quantity X over two squared plus X squared. I'm going to rewrite the equation, getting rid of the bracket in the square outside the bracket that's going to leave us with on the right hand side, minus X squared over four plus X squared. And if I find a common denominator so that I can add the two quantities on the right hand side, I'll multiply the last term by 4/4, which is the same as multiplying it by one. And that's going to give me minus X squared over four plus four x squared all over four. And the common denominator is being four. That means the numerator will be minus X squared plus four x squared. And if we carry out that mathematical operation, we will be left with three x squared all over four. And if we solve for why we have to find a square root of three x squared all over four, which reduces to the square root of three times X all over, too. Now, if I want to deal with the ratios of these triangles of 30 60 90 train goes, I'm gonna pull one out of there and just so that I won't get all mixed up with the rest of the stuff that's on that diagram and I'm gonna deal with only one, 30 60 90 triangle, and I'm going to say let X equal to in the equation that we just developed for this 30 60 90 triangle. And if X is equal to two than the high Pontin use, which is X is going to be too. And the X over two, if X is two, is just one. And why will be to all over route three times to over two, Which, of course, just reduces to root three. So the last side of that 30 60 90 triangle is Route three, so the ratios of the lengths of the sides of a 30 60 90 triangle are going to be always in the ratio of one to Route three. Now this ratio is going to be cropping up all the time in in science and electrical equations just to give you an example. And you don't have to commit this to memory because I have it in another another course. But in this other course, we start studying the phase relationship of a three phase system on and a transformer bank in that three phase system. We're looking at the phase relationship, and they form a unequal lateral triangle, which has contained in that equal lateral triangle the 30 60 90 triangle. So that helps us in analyzing the phase relationship of the voltages and occurrence in this three phase power bank because they are in the form of a 30 60 90 triangle which are in the ratio of 12 and Route three. So we then condense side. Or he can decipher the phase relationship between the face to neutral voltage and the phase two phase voltage, as well as the angles that are which those voltages are. The next unique triangle that I want to look at is going to be the SA Seles triangle, which has two sides equal, and the two sides are gonna have a contained angle of 90 degrees. I'm just going to flip the triangle over here to work with it, because I can work with a little bit better in this form, and you can see that the to side or two contained angles on each side of the equal sides are 45 degrees. Because the sides are equal, the interior angles have to be equal. And because the other angle of this triangle is 90 degrees than the other two, angles have to add up to 90 degrees. And because they're equal, we'll split 90 degrees and two and I will give us 45 degrees. So a nice awesome these triangle, with the contained angle being 90 degrees. The other two angles air 45 degrees and I'm going to designate the last side of that triangle to be our And if we were again, look at the Pythagorean theorem. For this right angle, Triangle R squared would be equal to X squared plus X squared. And of course, we can rewrite that equation to R squared is equal to two x squared. And if we find the square root of both sides of that equation, then we have our is equal to root two of X and this time I'm gonna let x equal one. So the sides the equal sides of this I saw Seles triangle are going to be one each and r Times X, which is root Osoria R is equal to root two times x So ex being one means that in this particular example, where X is equal to one our would equal route to. So this unique triangle of being in a sauce lee strangle with the contained angle of 90 degrees means the sides of this special triangle are in the ratio of 11 route to the next unique triangle That I want to have a look at is a right angle triangle with one side being three and the other side being four and the contained angle being 90 degrees. Now that three and four can be three inches, three feet, three miles, it doesn't matter. And we're just going to say, Let let one side B three on the other side before. If that is the case, then, according to the path agree. In theory, Lipan, news of that right angle triangle would have to be the square root of three squared Plus four squared, which is the square root of nine plus 16 which is the square root of 25 which is five. So this special right angle triangle is what we call a 345 triangle, and that crops up from time to time. So if you see a right angle triangle anywhere in physics or in electrical or in mathematics , where the right angle has se sides three miles and four miles, then you know automatically know that the high pine use of that right angle triangle is gonna be five miles. And if we made the triangle bigger, the ratios of aside still have to remain the same ratios. We've already proven that so in the case where we multiply the sides of that 345 triangle by SE X, where X is greater than one. Then you got a bigger 345 triangle, but the sides air still in the ratio of 34 and five. And if we let X be less than one, then you'd have a smaller triangle. However, the ratios would still be 345 for that triangle and the X in the 345 combination that we just showed. There is one, of course, and just out of for knowledge, I guess, if you would the angles, the other angles in the 345 triangle are 53 degrees and 37 degrees. But that doesn't usually crop up in the calculations, although it might from time to time and you never know. But it's easy to remember 345 as the sides of the triangle, and that ends the discussion of unique triangles and their ratios. So this ends the chapter 7. Ch 06 Trig functions & Tangents: Chapter six triggered a metric functions and a tangent lines drinking a metric functions comes from a special area of the triangle. Proportionality, fear, um, which states if two triangles air similar, their sides are proportional and characterized by this equation that you see here and in moving there, we're going to concentrate on three specific parts of this equation. Namely the first part of the equation. A be all over X Y is equal to a C all over ex zed. The first and last part of the equation, which is a be all over X y, is equal to B. C all over. Why, I said, And the last part of the equation, which is a C all over ex zed, is equal to B. C all over. Why is that now I want to rearrange the first equation, and I'm going to do that by multiplying both sides of the equation by ex Zed over a B. Now I can do that in an equation. In other words, I can operate on both size of the equation as long as I do it equally. And since I'm multiplying by the same thing on both sides of the equation. This is a legitimate operation for a mathematical equation. And you can see here that a B cancels out on the left side and X said cancels out on the right side, leaving us with X ed all over Ex, Why a c all over a b and we don't have to really do that. Weaken just would look at the results of what happens there and that is actually called cross multiplication, which we could do just like this, leaving us with the same results x x ed all over ex Why physical day, See all over a B and we can cross multiply the second equation as well, as well as the third equation. This keeps the blue triangle side ratios either on one side or the other of the three equations, as well as keeping the red trey angle side ratios on one side or the other of the equations . So now I would like to do another change, and I'm going to change the triangles into right angled triangles. The triangles remain similar. That is, their angles are equal. We have just made one of the angles 90 degrees. Therefore, the ratios remain unchanged now I'm going to concentrate on one angle, which is the same angle in each of the two triangles and that iss the angle that is immediately to the left of the right angle. And let's designate that angle as theater in establishing triggered a metric function, functions the size of a right angle triangle are named relative to this angle theta, which is immediately to the left of the right angle. This goes for any right angle triangle where we're looking at the sides of the triangle relative to the angle that is immediately to the left of the right angle. In this case, that angle for both red and blue triangles is data the side of the triangle that is opposite Thena. We call it the opposite side, which stands to reason and the side of the triangle that is immediately below Fada or in Jason to fado. We call it the adjacent side, which is obvious. And, of course, the side of the triangle that is opposite the right angle is called the high pot noose. Now, if we take the ratio of the opposite side over the high pot in use of any right angle triangle, including our blue and are red triangles here that is called the sign of the angle. And it is written like this. It is shortened to S i N. And it is equal to the ratios designated in the first equation. You'll notice the ratios in the red box is the opposite. Over the high pot in use of the red triangle in the ratio in the blue box is the opposite over the high pot news of the Blue triangle. Now you can see why I manipulated the equations the way I did. Now, if we take the ratios of the adjacent side over the high pot in use of any right angle triangle, including the ones year that is called the co sine of the angle and it is written like this usually shortened to just three letters Khost fada and again you'll notice it's equal to the ratios in the red and the blue box, which are the ratios of the ah, adjacent over the high pot in use of the red and the blue triangles, respectively. Now, if we take the ratio of the opposite over the adjacent of any right angle triangle that is called the tangent of the angle, and it is written like this shortened, usually to tan, which is using the data angle is called tan data, and it also is equal to the ratios designated in the third equation. Again, you'll notice the ratios in the red box are the opposite over the adjacent of the Red Triangle, and the ratio in the blue box is the opposite over the adjacent of the blue train angle. These are probably the most important equations, or trig functions that you will come across in working with geometry or trigonometry as you relate to the various science fields, for example, physics, electrical, even chemistry. So as you work in these various fields, these functions will become second nature to you, and you won't even have to think about, um, very hard to remember what sign CO sign. And the tangent ratios are, needless to say, if these are new to you, and it will take you a while to memorize them in a way of remembering him, there is a simple tool for bringing to mind these ratios, and that tool is so cata. The phrase, of course, is meaningless in any dictionary. However, it does help with the memorization. The first part. So, uh, S O. H s stands for sign and the O. C. Stands for the opposite and the h stand for the high pot news. So that means that the sign is the opposite. Over the high pot news. The middle part car stands for the co sign of the adjacent over the high pot news and the Torah T o A. Stands for the tangent of the opposite. Over the adjacent there are three more trig ratios that are important. However, they are secondary importance to the 1st 3 We don't need a meaningless phrase to memorize them because they are just the reciprocal off the 1st 3 trig functions. In other words, there the reciprocal of sign co sine and tangent. The 1st 1 is inverted, giving us a reciprocal of the opposite over the iPod in use, and it is given the name Cho Si Cand, and it is sometimes shortened to C. S. C. As you can see in the brackets, the next one is just the reciprocal of the co sign, and that is the C can't of the angle, and it is usually shortened to S E. C. The 3rd 1 is the reciprocal of the tangent, and that is called the Co Tangent, and it is usually shortened to see O. T for caught data. When we started out, we placed data to the left of the 90 degree angle. We could have placed data just above the 90 degree angles, as you see here. In that case, of course, data, the angle would have been different, and the opposite side and the adjacent side would also have to be moved around in order to be the opposite in the adjacent of the angle Sita. However, the proportionality, the're, um of triangle still holds true, So the ratios of the sides haven't changed. But what has changed is which side is now the opposite and which side is now the adjacent and they are different lengths. So are sign and our co sign and are tangent, of course, would change. Those ratios would change because of the size of the sides which have changed. However, we're still talking about one angle, and that has an opposite side and an adjacent site. So it doesn't matter which angle of the triangle you're talking about as long as you know what the adjacent and the opposite sides off that angle are. Here is a simple of example of how you might use thes trig functions. In this example of Surveyor is observing a high rise building from a distance of 500 feet. The angle of elevation to the top of the building, as seen through the transit, is 42 degrees. Our task is to find how high the building is, and for the moment, we'll neglect the height of the transit. I'm gonna let the height of the building equal H The surveyor finds the angle of elevation to the top of the building to be 42 degrees. The surveyor now has all of the information needed to calculate the height of the building . And, as I said, we're gonna neglect the height of the of the transit for now. So we're going to let the height of the building less the transit height B h primed. We have a right angle triangle. Who's angle? 42 degrees is immediately to the left of the right angle or the 90 degree angle, making the height of the building the opposite side of the right angle triangle and the side. On the bottom of this triangle is the adjacent side of the 42 degrees taking the opposite or the adjacent our trig function for the opposite. Over the adjacent is the tangent of the angle, which is the tangent of 42 degrees, and we can look up 42 degrees, either in a smartphone calculator or a series of tables. But we can find out what tangent of 42 degrees is, and we know that the uncorrected height of the building h primed or we don't know what it is, what we're gonna let it represent. H prime. And we know the distance off The Surveyor from the building is 500 feet so we can find out what age prime is by multiplying the tangent of 42 degrees by 500. And we know the tangent after looking it up 0.900 So we can immediately find out what each prime is by multiplying 500 times 0.900 and that would give us 450 feet. Now we can look at the transit height. Let's say the transit height was exactly five feet, then all we have to do is add five feet to the 450 feet to get the exact height of the building, which is a church and that is equal to 455 feet, I would like for a moment to consider a circle with a C can't line cutting the circumference at two locations. Notice the C can't line that I've drawn it parallel to the horizontal. It doesn't have to be, but it just makes it easier to look at and quicker to analyze. Next, I'm going to draw to radius lines from the center of the circle to the point where the sea Kant's A C can't line constant circumference, and I'm going to designate the angle formed by the two radius lines as X. And because these are radius lines, they are equal in length. The triangle formed by the radius lines and the sea can't forms, and I saw silly strangle, which means the angle formed by the two equal sides and the Sea Kent line are equal, and I'm going to designate those angles to be. Why now, the interior angles of a triangle have to add to 180 degrees. So if we add up the interior angles of Theus softly strangle that we have here, we will have the equation. X plus y plus y is equal to 180 degrees. We can simplify the equation by collecting like terms of why, and we have X Plus two y is equal to 180 degrees and weaken solve for why we'll bring why, to the left hand side and everything over bring everything else over to the right hand side . And because we had to, I we're going to divide everything on the right hand side by two, leaving us with the fact that why is equal to 180 degrees divided by two minus X all over too, which can be further reduced to the fact that why is equal to 90 degrees minus ex all over , too. And I'm going to write those on the triangle that we have. They're replacing. Why, with the term 90 degrees minus X all over, too. Now I'm going to move the C can't line a little bit away from the center of the circle, and if I do that, you can see that angle X will be reduced. It'll get smaller. And the radius lines, of course, are still the same. However, the triangle is now kind of shortening from left to right, which means the angle X is getting smaller and the two equal angles of the A sausage please triangle are getting larger. If I further reduce the angle, you can see that the triangle is getting a little bit skinnier and angle. X, of course, is getting smaller, and the two angles of the equal sides of the triangle are still getting bigger. And I can reduce it one more time, and you can see it getting even the X angle getting even smaller. And if we to reduce a smaller and I'm going toe, have to zoom in on it so I can see what's happening on that skinny triangle. So as the triangle gets skinnier and skinnier, X is getting smaller and smaller, and the two inside equal angles that you see in yellow 90 degrees minus X over two are getting larger. In fact, the second term X over two is getting smaller and smaller, so the angles are really approaching 90 degrees. I'd like to look at the outside angles now, and that is the angles that are formed by the radius. Lines with the C can't lying outside the triangle, and I'm coloring those angles. Purple, the angle of the purple angle and the yellow angle must add to 180 degrees because the C can't line is a straight line and those two angles add up to a straight line on both the left hand side in the right hand side of the triangle that you see there. So if you're looking at just the purple angle, it's going to be 100 and 80 degrees minus yellow angle. What you see inside the bracket, 90 degrees minus X all over, too. Now, let's just go to the final step and let X equal to zero. Now a lot of things are happening when you let X equal zero. The first thing we have to understand is the two radius lines are now overlapping, which means they are one on top of the other, so they are no longer cutting a circumference at two locations. They're cutting the circumference at only one location, which means that the radius lines are meeting the Seacat line at one location, and that is on the circumference of the circle, which is really means the Seacat line has now become a tangent line. So we have a radius line or lines, both going from the center of the circle to the tangent line, touching it at only one location. The equation for the angle that is formed by the Radius Lines. Meeting that tangent line is still given by 180 degrees, minus the quantity inside the brackets, 90 degrees minus X all over, too. Simplify that because we don't need the brackets. We have, ah, 180 degrees, minus 90 degrees plus X divided by two. And if we subtract 90 degrees from 180 degrees, we're left with just 90 degrees. So that quantity now becomes 90 degrees plus X over two. And because we've made X zero, we will have 90 degrees plus 0/2, which is just 90 degrees. This means that if X is equal to zero, a radius line meeting a tangent line on the circumference of the circle is meeting the tangent line at 90 degrees. At the risk of repetition. If the radius lying meets a tangent line on the circumference of the circle, it can only meet the tangent line and the circle circumference at one point. And at that point, the radius line forms and angle of 90 degrees with the tangent line. We're not gonna look at something called feels if your um, which will have helping the analysis of tangent lines as the relate two circles and tail serum, states that if a B and C are distinct points on a circle such that a line segment A C, is a diameter of the circle, then the angle C B A or A B C is a right angle. In other words, Triangle ABC is a right angle triangle. Furthermore, no matter where Point B is selected on the circle, the angle ABC will always be a right angle or 90 degrees. Another way of looking at it is the diameter. AOC will always support a right angle triangle whose right angle Vertex is on the circumference of the circle. Now the proof is easy. First, we'll draw a radius line from the center of the circle to Point B, where wherever that is on the circumference and we just stopped it there so that we can concentrate on one point anyway. But it doesn't matter where B is on the circumference, because the line is a radius line of the circle. It is equal to the diameter are half of the diameter, which is also a radius line. Oh, a as well as O. C. And since these lines far equal, they form to mawr triangles inside the right angle triangle. And those triangles are a sauce. Elise triangles, which means the angles contained on the equal sides of each of those triangles are equal. And I'm gonna let the there I'm gonna name the angle of one triangle X, and I'm gonna name the other two equal angles as why. And I'm gonna look at this right angle triangle and we know that the interior angles of a triangle have to add to 180 degrees, which means if we add up those individual angles, they have to equal 180 degrees, which is X plus X plus. Why plus why is equal to 180 degrees or two x plus two? I is 180 degrees. And if we divide both sides, the equation by two x plus why is 90 degrees, which is the angle? A. B C, which we stated at the beginning, could be anywhere on the circumference, but it still contains those two angles, ex and why? And adding those two angles together will always come out to 90 degrees. There is another way to simply prove they'll serum, and that is working with the fact that the that the diameter AOC is a straight line. We know that the interior angles of Triangle A O. B. Must add 280 Greece, so that angle a albi, which I've colored in blue, must be equal to 180 degrees minus two x and when added to angle, B O. C, which for the same reason is equal to 180 degrees minus two. Why I must add up to 180 degrees because diameter. As I said, AOC is a straight line rearranging that equation and collecting like terms and gives us this equation 360 minus two X minus two y is equal to 180 and then we can divide the equation both sides by two and leaves us with 180 degrees, minus X minus. Why is equal to 90 degrees and can further simplify the equation and we're left with X plus . Why is equal to 90 degrees, which is equal to angle a B. See, I'm now going to go through uninterested exercise using tail serum, and that is, I want to draw a tangent line to a circle from a point or to a point on the circumference of that circle. And I'm going to use a compass, which is something that you used to draw circles or arcs with pencil on one end and a point in the other. And the point is usually at the center of any of the circles or arcs that you're gonna draw . And I'm going to start out by drawing just a random circle, and I'm coloring it read. Just discern it from the other arcs and circles that I might be drawing later and the center of that circle I'm going to designate as old and I'm picking a point on the circumference of the circle, and I'm going to call it a And that is the point to which I want to draw a tangent. And of course, I could do it by just I in a ruler. But we want to do it accurately. So we're gonna use our compass to do it accurately. And I'm gonna place the point of the compass or the center part of the compass on my point A and I'm gonna open up the compass. Ah, little bit more than the radius of the circle that I just drew. And I'm gonna draw an arc approximately where I think the tangent might be from point A. Then I'm going to place the compass with the same diameter that I used for drawing the first dark. But I'm gonna place the center now on the center of my original circle, and I'm going to draw another arc to cut the first arc. And there is one point where the two arcs intersect and I'm going to place my compass there . And I want to draw a circle with that point being a center using the same, uh, radius that we used to draw the arcs. But this time the center is gonna be at where the arcs intersected. So the circumference of the circle, because that radius is the same, is gonna pass through the center of the circle as well as this. The point A on the circumference of the circle, and I'm going to draw that circle will colored it black so we can differentiate it from the other marks on this slide. I'm now going to draw a diameter line. Uh, O B. And then I'm going to draw a line from O to a and then from a to be. And we know from Thales serum that the angle of those two lines because they are on the circumference of the circle and will be as a diameter of the circle. Then they have to support a 90 degree angle. And if a radius line is supporting a 90 degree angle on the circumference of a circle, that line has to be lying along the tangent line. So we have successfully drawn a tangent line accurately through Point A on the circumference of the original circle that I just drew and we're done. This ends the chapter 8. Ch 07 The Unit Circle: Chapter seven, the unit circle and trade functions. When looking at Trigana Metric functions, we start with an angle and we give that angle a name. It could be you a number or could be a letter. In this case, it's a Greek letter and we call it data and ah, the trig function is made by dropping a perpendicular from the terminal side of the angle to the initial side. And I've colored that perpendicular red, and I'm coloring the ah Jason side blue and the sign is given by putting the opposite side over the height pot news that is the length of the opposite side over the length of the high pot in use. And that ratio is known as Sign Fada, The coast Seda is made up of putting the length of the adjacent side over the high pot news , and that gives us the co sign of Fada. Now I'm gonna look at one of the simplest trig identities, and I'm going to take the sign of Fada and divided by the co sign of Feta. And if I've done that with what is the left hand side of these equations I have to do it with the right hand side of the equation, which is the opposite side over thy pot news all over the adjacent side, all over the high pot. Use now. Remember, those are ratio. So they are numbers, so we have just a number over a number. However, I'm not gonna, of course, put numbers in there right now. I'm just going to look at this fraction over a fraction, and that can be rewritten in that manner. And you can see that the two high pot news will cancel Obed. And that will leave us with just the opposite side over the adjacent side. And we know from our trig functions that the opposite side over the adjacent side is just the tangent of the angle. So this trigger identity, which is very simple, tells us that if we want to know what the tangent of the angle is, all we have to do is divide the sine of the angle. By the co sign of that angle, I'm going to name the side opposite the data as Oh, and I'm gonna call the side of the adjacent side. A. It's just a little bit easier than writing opposite side, high pot news and and Jason Side Night Pot News. So I'm going to call the iPod news H and I've colored them so we can follow them along a little bit easier. And that means that the sine of the angle is going to be oh, all over H. The co sign is gonna be a all over H now if I was to take the square of the sine of the angle In other words, I'm going to square both sides of this equation. I will have sine squared data is equal to oh squared all over h squared. And if I'm going to square both sides of the co sign equation, I'm gonna have co squared. Data is equal to a squared all over H squared. Now, if I'm going to add sine squared plus coast squared, I have to add the right hand side of those equations which gives us oh, squared over a squared plus a squared all over each squared. Now each squared is a common denominator, So I only have to write at once as long as I have the fraction site underneath both of the numerator. Now the Pythagorean theorem tells us that a squared plus O squared Oro squared plus a squared is equal to H squared. So I'm going to be able to replace the numerator of that fraction. Oh, squared, plus a squared with H squared. And that gives us eight squared over H squared, which is equal to one. And that gives us a second trig identity, which is used quite often, and that is sine squared. Theta plus co squared data is equal to one now. It doesn't matter whether the triangles are huge or small, as long as the angles are equal are triangle proportionality. Theorem tells us that the ratios, in other words, the sign co sine and tangent of the angle theta, remains the same regardless of the size of the training. So at the risk of being repetitions here, it doesn't matter whether we have a large triangle or a small triangle. Whether there was sides are measured in inches or miles or in feet or centimeters, the ratios remain the same as long as Theodora remains the same. Now what does change the ratios is, of course, if that angle changes, and if that angle changes, of course, the opposite side as you see here gets larger and the adjacent side gets smaller. So the ratio certainly will change for sign co sine and tangent. And as long as that angle is less than 90 degrees but greater than zero, it's pretty easy to come up with the sign co sine and tangent of the angle. But now what happens when data becomes greater than 90 degrees? Then we have to set some rules, and we're gonna use the rectangular coordinate system to help us out here. The right angle triangle that we're going to analyze is sitting with the adjacent side along the X axis parallel to the X axis, and the opposite side is parallel to the Y axis and the coordinates for the tip of the high pot. News where it meets the opposite side is given by the co ordinates X one y one. Now, in this case, X one and why one are positive. But they could be depending on where we are in the Cartesian or their rectangular coordinates system, they could be positive or negative. The high pot news, however, is going to be given the length of the high pot. News is going to be given by finding the square of X one and adding it to the square of why one then finding the square root of the some of those two that will give us the length of the high pot in use. Now, even if X one is negative and why one is negative, the square of those terms produce a positive number. So what we're saying is in this system that we're analyzing this right angled triangle. The high Pontin use is always positive. So that's one thing that we have to keep in mind. The other thing that we will keep in mind is because the adjacent side is on the X axis than the beginning of the opposite side is zero and the beginning of the adjacent side is also zero than the length of the opposite side is going to be given by why one and the length of the adjacent side is going to be given by X one. So if we're to find the sign of Fada of this particular right angled triangle, as you see here on the rectangular coordinates system, why one is positive, which is the side size of the opposite side and the high pot news is always positive. So the sign of Fada of this right angle triangle is going to be a positive number, and the adjacent side is going to be given by plus X one, which is a positive number. And the high pot news is positive. So the co sign of this triangle is going to be positive. And similarly, the tangent. The opposite side is Ah, positive number in the adjacent side is a positive number. So Tan data or the tangent Athena is going to be a positive number if we move a little bit further. In other words, increase Fada but still staying less than 90 degrees x one and why one are still positive. So sign co sine and tangent of this angle are all going to be positive. And if we increase it even further, but we're still less than 90 degrees signed co sign intention of this angle is still positive. Now, if we rotate that angle such that the angle is greater than 90 degrees, but less than 180 degrees, then X one is going to be a minus number because it's to the left of the origin. Why one is still going to be a positive number. So that sign data. The opposite side is given by why one is a positive number. The high potting use we've already discussed is always positive. So sign of this angle is a positive number. The co sign. On the other hand, the adjacent side is given by minus X one. So the adjacent side is a minus number, and the high pot news is a positive number. So the co sign off this angle is going to be a negative number, and a tangent is the opposite side over the adjacent side. We have a positive number over a negative number, and that will generate a negative number as well. If we rotate the angle through 180 degrees, but less than 270 degrees, we's find that both x one and why one are negative numbers. So the opposite side is negative. The high pot news is still positive, so sign Fada is going to be a negative number. The adjacent side is a negative number, and I pot news is a positive number. So the co sign of this angle is going to be a negative over a positive generating and negative number. Now the tangent is the opposite side is a negative number at over the adjacent site, which is a negative number, and that will generate a positive number. And we're gonna look at one more example. And that is, if our angle is greater than 270 degrees, but less than zero, then our X one will be a positive number, but the why one will be a negative number. So sign Fada is going to be positive. Sorry, Negative. Why one which is a negative number over high pot news, which is a positive number, and that's going to generate a negative number. The co sign is going toe have, plus X one all over, plus high pot news, which will generate a positive number and 10 the tangent of this angle. The opposite side is going to be a negative number, and the adjacent side is going to be a positive number, so a negative over a positive is going to generate a negative number. So we've gone around 360 degrees on, I guess, in the rectangular coordinate plain and if you remember back when we did when you studied the rectangular coordinate plane, the X and Y axis, divide this plane up into four quadrants. The first quadrant is you see here is 0 to 90 degrees, and the second quadrant is from 90 degrees to 180 degrees. The third quadrant is from 180 degrees to 270 degrees on the last and fourth quadrant is from 270 degrees back to zero degrees. And in the first quadrant, what we discovered is sign Fada Coast data and tan theta are all positive in the second quadrant. Sine theta is positive, Coast data is negative, and tangent of the Fada is negative. In the third quadrant, scientist Ida's negative Coast data is negative and tan theta is positive. And in the fourth quadrant, sine theta is negative. Coast data is positive and tan theta is negative. I want to look at something now that will bring out some more unique properties of the trig functions on. I'm going to do that with the assistance of a circle that has its center at the origin and the radius. Length is going to be one could be one inch one centimetre when mile. It doesn't matter. We don't We don't care about exactly what the denomination of the oneness where he's going to call it one because we're dealing with ratios anyway. Now, as it at the point where the radius line cuts the circumference of the circle, I'm going to call that X one and why one? In order to look at the trig functions, I'm going to drop a perpendicular, which is the red line you see there. And if I drop that red line to the horizontal, it's gonna be exactly why one in length. Next, I'm going to draw a horizontal line, joining the origin of the rectangular plain to where the Why, uh, lying? Or the red line meets the X axis, and I'm going to call that length x one, or it will be X one in length. And given these dimensions now, the sign of this angle Fada is going to be why one all over one, which will reduce to simply sign Fada is equal to y one. And if that's the case, then I can replace why one with sign data now remember, science data is just a number, and it depends on what the size of the angle is. However, whatever the angle is, the length of the red line is going to be signed data. And if that's the case, then co sign of Fada is going to be X 1/1, which will reduce to co sign is equal to X one, which means the blue line. The adjacent line of that triangle is going to be exactly Khost Fada in length now for clarity. Before moving on, I'm gonna erase most of the stuff from the diagram just leaving a circle in its radius line of one. And I'm going to draw a tangent line at the point where the radius line meets the circumference. And we learned that the radius line is going to meet the tangent line at 90 degrees or at right angles. Next, I'm going to join our draw a horizontal from the center of the circle to where it will meet this tangent line. And lastly, I'm going to drop a perpendicular from the tangent point to that horizontal line, and I'm gonna concentrate for the moment on the little train. The let's formed by that perpendicular, and I've shaded one of the angles in yellow on gonna call that angle one and I sheeted the other another angle of that little triangle in blue. And I'm going to call that angle number two. And the third angle I'm going to look at is of the bigger triangle. And I'm gonna shape that in green, and I'm gonna call it angle three. Now we know that the intern interior angles of a triangle have to add 280 degrees. So in the little triangle with angles 12 and the right angle, those angles have toe add up to 100 meaty degrees, so angle one plus angle to must also equal 90 degrees, or the some of those two have to add to 90 degrees. And looking at angle number three and angle number one, we know that that is a 90 degree angle because it's a tent. It's meeting a tangent line, so angle number one plus angle number three also has to be 90 degrees. And if that is the case, we could subtract one, uh, equation from the other, or just saying that those two equations are true, then that means angle three must equal angle to and I'm going to now remove the numbers on . I'm gonna rename angles three and two angle theta. Now, when I have a look at these two triangles independently or a little bit independently. So I'm going to split them apart and rotate the little, uh, triangle. So it kind of looks like the Big triangle, or it does only in miniature form. And those two angle are triangles are similar because they have the same angles. They just are different sizes in the I'm gonna mark the place where I split them apart because I want to keep track of that lying there. And we know because they were one line at one time before I split day triangles apart. Now, in the way of recalling the triangular proportionality fear room where you have to similar triangles that two triangles of different sizes but with the same angles their sides are proportional. So this equation a B over X Y is equal day See, all over ex Zet is equal to B C all over wise that I'm going to transpose the letters of the similar triangles on the bottom to the right angles that we have just manipulated and are comparing right now so that we can reuse the same proportionality forum equation, which tells us that the anger they'll side a B over X y of the right angle triangles is equal to a C all over. Ex zed is equal to B. C all over. Why said, I'm going to look at just the last part of that equation. A C all over ex Zed is equal to B C over why Zed and I'm going to invert both the left side and the right side of that equation, which you're allowed to do mathematically and that will leave us with X zed all over A C is equal to y zed all over BC, and I'm going to cross multiply. And I'm going to manipulate the equation, keep just accept on the left hand side. So that leaves us with wise that all over bc times a c on the right hand side of the equation. Now you remember we put the green markings on the two right angle triangles to keep us or help us remember that those two sides used to be coinciding with with each other. In other words, they were equal in mathematical terms. BC was equal to X y so I can replace BC with X Y, and the equation now will read. X ed is equal to y zed all over x y times a c Looking at the larger triangle, we can see that Wise said all over X Y is equal to 10 data or the tangent of theater so we can rewrite the equation now so that X zed is equal to a C Times tan data. Now where we're going with this, I'm going to go back to where we were before we split the two triangles apart and put our circle back in the picture. And we can see that a C is equal toe one because this was a unit circle. So the equation x zed equal that a sea panther data is just x zed is equal to tan theta and x zed. It is just the portion of the tangent line that you see there. So let's remove some of the surplus superfluous data that's in front of our face and put our rectangular coordinates back into position. And remember before we started to look at for the answer of how long that tangent line Waas and we found out that it was tan. Data are red line in the red are in the unit circle was equal to sign data. The blue line was equal to CO Seda. Now we got a measurement for the tangent of Fada. Next, I want to extend the tangent line backwards and upwards until it intersects the Y axis of our rectangular coordinates and you can see I've colored the extension of the tangent line . I'm gonna I've colored it orange. Now I'm going to where that orange line starts, I'm going to draw a parallel line to the X axis. I'm coloring in blue and then I'm gonna draw a vertical line along the Y axis and I've got a very small little right angle triangle at the top of the circle. The red line that we designated was of length Sign Fada because it meets the X axis at 90 degrees and the Y axis meets the X axis at 90 degrees. We can say that those two lines are parallel lines and if you look at the blue line that we said was of Khost Fada in length because it meets the two parallel lines at 90 degrees and up in the new little triangle that we just drew the blue line. There also meets the parallel lines at 90 degrees. So we've formed a rectangle with the two parallel lines and the two blue lines. So the length of the upper blue line that we just drew must be of length coast data as well . I'm going to highlight the little triangle and yellow just so that we can keep track of it . And if you look at the tangent line, I'm gonna color, agreeing for now so that it stands out as well. It is a trans versatile of the two parallel lines. So if that is a trans versatile, then the angle contained by the left parallel line and the Green Line has to be equal to Fada because these are congruent angles of a trans versatile of two parallel lines. Now let's call the high pot news of our little Triangle H age all over coast Fada, which is h over the length of the blue line of our little triangle, is equal to one all over scientific data because the two right angle triangles that I've highlighted in yellow Arkan grew. And so the sides this are proportional and the sides were talking about are the high pot in use and the opposite side? I can rewrite that equation and solve for H and H is equal to coast data all over science data which is equal to the co tangent of Fada. So that means that h the length of the high pot news of our little triangle is equal to co tangent beta. Now I want to look at this large right angle triangle that we have or I've colored it in Burgundy and I want to call the the High Ponte news of that right angle triangle each and the sign of data of that large right angle triangle is tan theta all over H. Because tan data is the opposite side to our angle. Theta and H is the high pontin use. So signed data is equal to tan data all over each. That means H is equal to tan data all over Sign fada, which is weaken substitute signed data over co Seda for tan data in the numerator and I can put signed data just over one on. And now I have a fraction over a fraction which can be rewritten as two fractions sine theta all over coast data of the numerator bringing the denominator up beside it. I have to invert it that is equal to one over Sign Keita, the sign Data's cancel out and I'm left with one over coast data which is equal to see can't Fada. So that means I can replace the length of ah, hi pot news which I let equal to H with C can't beta. So now I've got another length described as a Trigana metric function The hi pot news of that right angle triangle is C can't data carrying this observation to the next step I've shaded in a right angle triangle here. This time it is a large right angle triangle, and I'm going to let the length of that right angle triangle but lies along the y axis. I'm gonna say, let that be represented by a now that fada in at the top of the big right angle triangle that I just shaded in. It is data which is the same angle as our little right angle triangle. But we're talking about the big right angle triangle here. But the angled is the same, and it is stada. So if I take tangent of Fada that is equal to the length of the line of the right angle triangle, it lies along the X axis. It is of length C camp data. And if I want take the tangent of feed, I would have to put that all over a so tan theta is equal to see Can't data over a And if I solve for a I have C camp data all over the tangent of Fada and that is equal to for a moment. I'm going to just remind everybody that the seek and of data is one over the coast of Fada and one over The tangent of Fada is equal to Coast data over sign Fada. So I'm going to replace C camp data with one over Coast data, and I'm gonna replace one over Tan Theta with co signed data over scientific data like this and like this. So I just rewritten the equation. But I want to show you how I got there. Now it's clear that the two Coast Data's will cancel out and that will leave us with one over sign data. And you remember what the inverse of sine theta is? It is just cosi can't feta. So now a can be replaced with CO c can't CDA So this is rather interesting. We started out with a unit circle and the unit circle had the radius line of one and we proceeded to draw right angle triangles along this unit circle And what we found Waas. We could describe the lengths of the right angle triangles that we drew here in terms of Trigon a metric functions. Now remember, these trig functions are just numbers. They will change and they will change as long as the angle changes and the angle that we're changing is data. And I want you to look at the radius line of our unit, circle the one and it forms an angle with the horizontal axis, the X axis off data. Now, if I rotated that radius lying about the origin in other words, it would rotate around the unit circle. Fada would change, of course, and it would go from zero all the way through 360 degrees and back to zero again. However, the lines of the triangle that would be formed those triangles you see there they would change. But they would still be equal to scientific Tako se to see campaign, etcetera. And let me demonstrate that now you can see that the right angles are changing. But the sides of those right angles or the triangles are changing in proportion to Scient ate a casita tan theta, etcetera. And you can see that the sign Data and CO. Seda never get larger than one. They go from 0 to 1 to negative zero r sorry, zero and negative one. But they are always less than one or one. The lines for tangent and co tension they spent off to infinity from time to time. That's when the cose ada is zero because scientific data over coast data if Costa to become zero then, uh, the tan theta will go off to infinity. But you can see that all happening here with the unit circle. Now there's a couple of other lines that are on this diagram and there are ever sign and X c can't and CVS don't worry about those those are other things that are related, but we haven't covered those, and we won't be in in this course. But I thought it be interesting. Just tow. Watch this, uh, this unit circle with the radius line rotating about 360 degrees and watching how the right angles change with that angle, I'm gonna go back to have a look at our unique trey angles. And we looked at them in regard to the sides of the various angles of these unique triangles and what the ratios were, if you would. But now I want to have another look at them in regard to the triggered a metric functions Signs, co sines, tangents, etcetera. And our first unique triangle that I want to look at is a 30 60 90 degree triangle. And I gonna look at the 30 degree angle to start with. And if I look at the sign of 30 degrees, I'm gonna look at the opposite over the high pot in use. Now, the high Pontin use is always going to be too. But the opposite side of a 30 60 90 triangle is one. So sign 30 degrees is going to be 1/2 or 1/2 co sign of 30 degrees is the adjacent side all over the iPod in use, and we said the hypothesis was, too. The adjacent side is a square root of three, so the co sign of 30 degrees is Route three. All over, too. Tangent of 30 degrees is the opposite. Over the adjacent. The opposite side is one, and the adjacent side is the square root of three, so it's one over Route three. Now the C can't Cosi canned in co tension are just the inverse of the sign. Co sign intention so he can zip through those pretty fast. See Can't 30 is to over one. Cosi can't 30 degrees is to all over Route three, and the co tension of 30 degrees is Route three all over one. Now let's look at the 60 degree angle of a 30 60 90 degree triangle, and if we were to look at the sign of 60 it's the opposite over the high pot news. So that would give you root 3/2, and we could go through all the rest of them the same way. The Coast co. Sign of 60 greasy adjacent side is one, and the iPod uses to the tangent is the opposite over the adjacent, which is Route three all over one and you just inverse are invert all of the 1st 3 that we just looked at there to get the sea Can Cosi Canon Co. Tangent. The next unique triangle is the I saw Slowly strangle, which has 32 equal angles 45 degrees. So we're gonna look at the right angle triangle of a 45 degree angle, and the sides air in the ratio of 11 and the root of two. We've already seen that. So if we take the Trigana metric functions of 45 degrees, the opposite over the high Pontin use is one all over the root of two, and the adjacent is also one. So co sign of 45 would be the adjacent or the iPod news, which is also equal to one over the root of two. Now, if we're taking the tangent of 45 it's going to be the opposite over the adjacent, which is 1/1. Or it could be just one if you want. The inverse of signed 45 is route to over one or just route to that is C can't 45 degrees. The Cosi can't of 45 degrees is also route to or route to over one. The co tangent is the inverse of 1/1, which is just one all over one the next triangles or the next special angles that we want to look at. I'm gonna use the rectangular coordinate system in orderto Look at that because I'm going to start with our angle theta and our unit circle, where the the radius lying of a unit circle is one and the, uh triggered a metric functions are going to be given by the ratios of X one. And why one and the of course hi pot in use of one. And the first angle I want to look at is when Fada is equal to zero. When data is equal to zero. X one is one. Why one is zero and the high pot news is one. So sign data is going to be. Why one all over one, which is 0/1, which is zero. So the sign of zero degrees is zero. Co sign Fada is x one all over one and x one is one which is 1/1. So the co sign of zero is one. The tangent of Athena is why one over X one, which is 0/1, which is zero. Now, let's rotate the angle so that it is 180 degrees x one is gonna be minus one. Why one is gonna be zero. So sign is being why one all over one is 0/1, which is zero co sign data is X one all over one, which is minus 1/1, which is minus one tan. Data is why one all over X one, which is zero over minus one, which is still zero. Now let's let beta B 90 degrees. And if data is 90 degrees X one is zero. Why one is one. So science data being why 1/1 is equal to 1/1 or just one co sign Theatre co sign of 90 degrees is X over one. Our sorry x one all over one which is equal to 0/1, which is equal to zero and tan data is why one all over X one, which is 1/0, which is infinity. So as the angle changes and gets larger and larger or sorry gets closer and closer to 90 degrees, Tank Veda is going to get larger and larger until it becomes infinity. And if you remember the rotating right angles that we had with our unit circle, you could see that at various times. In other words, if you watched it close enough, whenever the angle Fada was 90 degrees. And this is kind of a hint of things to come, when data is equal to 270 degrees. That's when tangent will go to infinity or minus infinity. But for now, just remember that scientific co Seda are one and zero, respectively, and tan theta of 90 degrees being they being 90 degrees. Tanta is infinity. Now let's rotate the angle to 270 degrees. The coordinates on the unit circle when we're at 270 degrees is zero minus one, so x one is zero. Why one is minus one? So scientific data is why 1/1, which is minus 1/1 or just minus one. Coast data is X one all over one, which is 0/1, which is zero tan uh, data. If data is 270 degrees, why one is minus one and x one is zero. So again, the detain Geant will go to infinity. But we'll call it negative infinity this time because it gets to be a larger and larger negative number until X one becomes zero and then it's goes to infinity. I'm going to go through a exercise that I called Easy Trig, and it's more fun than something that you have to remember. But it's worth while going through the exercise, and it may become useful. Teoh. You never know. It's a way of not having to memorize the ratios of special or unique triangles or unique angles. And I'm going to start out by listing the angles that we are going to observe. And I'm going to observe untangle that is zero than 30 degrees than 45 degrees than 60 degrees, then 90 degrees, and I'm going to wanna find out what the sign co sign tangent Cosi can't see can't and co tangent of these angles are. So I'm gonna lay out a table here and I'm going to start to fill out the table from the from the top, left to the top, right? And then down the table. So looking at going across, finding out what the sign of these various angles are, I'm don't even have to think about him. I'm going to go across this 0 30 45 60 90 degree row, and I'm going to put 01234 in the boxes. And right now, it doesn't mean anything, but it will buy time. We're finished. So you start off by listing the angles 0 30 45 60 90 Then you're gonna find Sign that you're gonna put in the box 01234 The next thing I'm going to do is divide those numbers by four. So that will generate the fractions. 0/4 1/4 to over 4 3/4 and 4/4, which is still meaningless. But we're not finished yet because the next thing I'm going to do is find a square. Root off those fractions, and if we find a square root of zero over for, it's still zero and the square root of one of 1/4 is 1/2 the square root of 2/4 is one over route to, and the square root of 3/4 is Route 3/2, and the square root of 4/4 is one. And guess what? These are the answers for finding the sign of 0 30 45 60 and 90 degrees. So that was fairly simple. You didn't have to memorize any of the special angles ratios you just started off with, putting 01234 in the boxes, dividing by four and then finding the square root. And that gives you your answer. So now I want to do something as easy as that for the second role. In other words, finding out with the co sign of 0 30 45 60 and 90 degrees are. But again, I'm not going to try and remember with the special triangles or unique triangles or angles are I'm going to take all of the numbers that I have found in the first row, and I'm going to repeat them in the second role, but in reverse order, I'm In other words, I'm going to take the one in the last roll, and I'm gonna put it in the one in the first place that I'm gonna take root 3/2 and I'm gonna place it in the second place that I'm gonna take one over route to put it in the third place and 1/2. I'm going to put it in 1/4 place and then the zero I will put in the fifth place when filling out the third row, the tangent feta. We are going to have to remember one simple trick identity. And that is the fact that Tan Theta is equal to sign data over Coast data. And that goes for every one of our unique angles. So But that's once we've remembered that all we need to do is pull from the 1st 2 rows, starting with the fact that we're gonna look at sign Fatal worth eight a zero degrees sign data is zero coast data is one. So tamp ada is 0/1 For data equal to 30 degrees. Science data is 1/2 coast data is Route 3/2. So we have a fraction over a fraction, which we can reduce to one over the route of three for 45 degrees. Its scientific is one over route to and co sign Athena, or 45 Degrees is one over route to which is one over route to all over one over to which again is a fraction over a fraction. And that's pretty simple, because the numerator is equal to the denominator. And if that's the case, then that fraction is one for the 60 degrees. We're going to have Route three over to all over one over to which is again, a fraction over a fraction. But this time we can reduce it to just Route three. And in the case of 90 degrees, scientist is one. Coast data is zero, so that is going to be 1/0, and we know that that has infinite possibilities. But that's just leave it as 1/0 for now. Now we're on to the fourth row, and in looking at the Cosi can see Canton Co tangent, we can remember that the these terms are just the inverse of the sign co sign and the tangent. So the cosi canned oven angle is one over the sine of the angle, and the sea can't is one over the co sine of the angle, and the co tangent is one over the tangent of the angle, and that is for every one of the angles. So now are we going to do is plug in the values for one all over sign Fada and sign Fado or sign of Zero is zero. So the 1st 1 becomes 1/0, which, of course, has infinite possibilities in its infinite. However, let's just leave it as 1/0. For now, looking at the Cosi can't of 30 degrees, it's just the inverse of 1/2 which is to over one. And the inverse of one over route to for 45 degrees is route to over one, and the inverse of Route 3/2 for 60 degrees is to over route three and 4 90 degrees. The inverse of one is just one looking at the seek hand of zero degrees. It's the inverse of the co sign of zero degrees, which is one, and that is one as well. Going across the row, we see that the next C can't of 30 degrees is gonna be the inverse off Route 3/2 or 2/3. The next one for 45 degrees, is the inverse of one over Route two, which is root to all over one, and the inverse of 1/2 is to over one. And for the last term, the inverse of zero is one over the co sign of zero sorry for one over the co sign of 90 and the co sign of 90 is zero. So it's 1/0 for the C can't of 90 degrees now. That, of course, is going to be infinite, but we will leave it as 1/0 for now, Going to the Last roll co tangent. The tangent of Athena is 0/1, So the inverse is just 1/0, and the inverse of one over Route three is going to be Route 3/1. The inverse of one of what is one? Of course. The inverse of Route three is one over Route three, and the inverse of 1/0 is 0/1. No, that is the answer to our table for sign co sign tangent. Cosi Can't See, Can't and Co Tangent of the Angles. 0 30 45 60 and 90 degrees. So let's clean up the table a little bit and put on Lee the significant values in there that we calculated. And there you have the answer to all of the special angles or unique angles, and we didn't have to memorize the ratios at all. All we had to do is remember how to fill out the table, and that's a little bit of fun. I don't think you'll have to take that to the banker, take it to your next exam. But it's helpful to go through it, and it's a good little exercise. The next thing we will be looking at is something called the Law of the Signs. And in order to explain that I'm going to start off with just I'm a cute triangle, and it doesn't have to be an acute triangle. However, it demonstrates very nicely for us. I'm going toe label the angles ABC in upper case and I go no letter the sides ABC in lower case and the sides opposite the angles are gonna have the same letter, except the sides will have lower case letters and the angles will have upper case letters. Now I'm gonna drop a perpendicular from angle B to side Lower case be and I'm going to label that perpendicular line I m And we know that the sign of a Nangle is given by the opposite All over the high pot in use And for the little triangle on the left hand side of em The sign of angle A is going to be em all over. See end. The sign of angle see is going to equal em all over a And that's for the right angle triangle. Let's to the right of my perpendicular line em. I can rewrite the first equation and have everything equal to em. And I'm going to rewrite the second equation having everything equal m. Now, both of those equations are equal. Tow em. In other words, both of the left hand sides of those equations are equal to em. So the left hand sides of the equation must be equal to each other. In other words, see, Sign A is equal to a sign C and I can cross multiply that equation and I get signed A all over a is equal to sign. See all over. See that IHS sign angle A his all over side A is equal to sign angle see all over the side . See, now I'm gonna take that same triangle with the same letters and I'm this time I'm going to drop a perpendicular from angle a to side a and I'm going to still call that perpendicular em. And in this case, sign of angle B is equal to em all over C. C being more case and that is the side of the triangle. But it's also the high pontin use of the right angle, triangle and side See, or side angle C is equal to em all over side. Lower case be and Justus In the first set of equations, I can solve four i m. And if both those left hand sides of the equation are equal to em, then they must be equal to each other. In other words, the side see sign angle B is equal to be sign angle. See, we now have two equations that the left hand sides are both equal to the same thing. Sign angle see all over side See now, having gone through these two sets of calculations for this triangle, both with different perpendicular xem What we have ended up with is two equations both having the left hand side equal to the same thing. Signed C all over side. See, So that means that they all must be equal to each other. In other words, sign angle A all over side A is equal to sign. Angle be all over. Side B is equal to sign angle see all over side See And this is the law of signs. As you might suspect because there was a law of signs, there exists a law of CO signs as well, and that becomes handy when you're given two sides and the contained angle. You can calculate the third side using the law of co signs, and we start out much the same way with ah triangle. This one is an acute triangle, and we're gonna label it much the same way. But I'm only gonna label angle a with a capital letter are uppercase a and the sides are in lower case. And a, for example, is now colored in Burgundy. And it is the side opposite angle A and I'm gonna drop a perpendicular from the top vertex angle down to meet side see perpendicular, and I'm gonna call that line em, and I'm going to give it a dimension of one dimension of length M and that perpendicular is going to divide side See into two components are actually two SIG line segments, and I'm going to call them D and E. Now I have a right angle triangle. 9. Ch 08 Trig Identities: Chapter eight Trig identities. So far, we've been introduced to a smattering of trig identities, and I'm going to list them here on this one slide, and then I'm going to go on to add a fume or to this page and then continue on with a few more. After that, the first rial trig identities that we were introduced to was called the Reciprocal Identities, and that is when we discovered the Cosi can't see can't and co tangent were equal to the inverse of the sign co sign and tan, and you can reverse those and say that a sine theta is equal to one over the Cosi camp. Ada co scientist is one over the sea Can't data and tan theta is one over the coup tension data we then discovered would call the quotient identities, and that was where we discovered that the tangent oven angle is equal to the sign of that angle over the co sign of that angle. And if that's the case, then a co tension is the reciprocal of that co tangent. Fada would be equal to Coast data Oliver Sein data. We then went on to prove the Pythagorean theorem and we discovered that the sine squared data plus co square data was equal to one. And we can mathematically rearrange that equation such that one minus coast quart ADA was equal to sine squared data and one minus sine squared data was equal to co squared data. I don't want to look at something called the negative angle identities, and I'm going to do that with the help of the rectangular coordinate plain. And if we describe the right angle triangle on that rectangular coordinate plain with the, uh, angle that we're looking at, will call it feta. The high pot news I've designated as H and the opposite side as old and the adjacent side as a. And in this case, we have discovered that the high pot news is always positive, so sine theta would be equal to oh, all over H. Now, if we move that angle such that it was not positive but negative, we would be in the fourth quadrant and H, as I said, is always positive. But oh, in this case will be negative. So I have indicated that as a negative role, if you wanted a riel number in there, you could put it in, but we'll just call it negative. Oh, because the why coordinate would be negative. So if we're to look at the sign of that negative angle, it would be equal to Negativo all over h, which is, if you wanted to equate that to look equation above sign. Negative data is equal to negative Positive all over positive H, which is equal to negative sign data. So the negative angle, the sign of the negative angle is equal to the negative of sign data. Now, the co sign of Fada, this is the positive data now is going to be given by the adjacent side all over the high pot news, both of which in this case are positive. And where if we were to look at the co sign of the negative angle, that also is equal to a all over H because A is still positive. So in this case, co sign of a negative angle is equal to the co sign of the positive angle and taking it to the next step. We're gonna look at the tangent. The tangent is equal to sign data all over Coast data tan Negative data is equal to sign negative feat of over coast negative feta. And we learned from the equations above that, that is just negative. Sign data all over coast data, which is equal to the hole. Fraction signed. State over coast data being negative, which is equal to negative tan theta. So these are the negative angle identities, and we'll list them on our page sign. Negative data is equal to negative. Scient, Aita Khost. Negative. Athena is equal to coast. Positive data. Tan. Negative data is equal to negative. Tan feta. And if you wanted to carry that on to the Cosi Can't see can't Inco tangent. You would have very similar results. Cosi can't. Negative. Thena, then is just the reciprocal of sign negative data, which is negative. Cosi camp data and see Can't Negative data is equal to see Can't data because the reciprocal of co sign data is the positive co scientist. So negative. C can't. Data is equal to see. Can't data and co tangent. Negative data is equal to negative co tangent data. We're not gonna look at it. A couple of new identities, actually, four of them and this one is called the some formula because what it's stating is that sign or the sign of the sum of two angles in this case angle, X and angle. Why? The sign of the some of those two angles is equal to the sign X, Times Coast. Why plus Coast X times sign why we're gonna go about proving this now and I'm going to do it starting with a right angle triangle and I'm gonna call at writer label at right angle Triangle A, B and C and we're gonna look at the contained angle. Why? Which is just a left of the right angle. And I'm going to draw a second right angle triangle placing, uh, it right on top of the first right angle triangle. And I'm going to have a contained angle there opposite the right angle. We're gonna call it X, and I'm gonna label that right angle triangle a c d. I'm now going to drop a perpendicular from D such that it meets the line A be at 90 degrees . Then I'm going to draw another perpendicular lying from C so that it meets the first line d f that I just drew their, uh, at 90 degrees so we have the significant points on this diagram that we're going to look at A, B, C, B, E and F with contained angles, ex and why. The last thing I'm going to do is I'm going to say, Let the side a D the iPod news of the second right angle triangle. I want that to equal one. And again, it doesn't matter whether it's one mile or one foot, as long as we relate all of the other links in the same denomination, so we'll just call it one for now. And I'm going to do some analysis of the second right angle triangle soy shaded in yellow so that we could follow it a little bit easier. And we're gonna look at what is the sine of the angle X, while a sine of the angle X is the opposite all over the high Ponte News and the opposite is side D. C. And that is all over one, which is I pot news and the one is in. That equation is superfluous, so we could just say that D. C is equal to sign X. That means that that side is equal to sign next. Now remember that Sine X is just a number. It's going to be a number between zero and one, so it is just a number. But in order to find it, we would have to say that it is equal to the side. D. C. Similarly, co sine of the angle X is given by a C all over one, where a C is the adjacent side and again, the we don't need the one in that equation. So we can say that a C is equal to Coast X, and we can move co sex down and label the diagram. The lines A F and E C are parallel, and the reason we can say this is because e. C meets D F at 90 degrees and a f meets DF at 90 degrees. So those two lines are parallel that may. That means that lying A C is a trans versatile and the opposite angles of a trans versatile line we discovered were equal. So that means that angle E. C. A. Is equal to why I would now like to concentrate on this right angle triangle, and if we look at the angle life shaded in blue angle D C E. It is going to be 90 minus. Why? Because angle a C D was a right angle. And if we take why often we're left with just D C E. So D C is equal to 90 minus Why and the other angle of that right angle triangle because the contained angles of ah triangle have to add 280 degrees that a Google is going to be equal to 180 degrees minus the other two angles which happened to be D C and B C E. And we know that d. C is 90 degrees. And we just found out what d c e is. It's 90 minus. Why? So if we put those into our formula, we have 180 degrees minus 90 minus the quantity 90 minus. Why taking the brackets away? We have this equation which boils down to this that C D E is equal to Why no. If we look at lying d f, we see that it is made up of line segments D e and E f such that DF is equal to d E plus e f. And if we look at the fact that E F is part of ah rectangle, and the other side of that rectangle is CB weaken substitute CB for e f. Such that DF is now going to be equal. Did d E plus C B. If we go back to the original right angle triangle that we started out with, and we wanted to find a sign of why it would be taking the opposite over the high pot in use. And the opposite side is C B and the high pot news we already calculated out to be Coast X . We can rewrite that equation and come up with the the fact that C B is equal to co sex times. The sign of why, If we look at that little triangle up at the top and I've highlighted in yellow, we can see that coast. Why of that? Middle triangle is D ead all over sine X, which is the opposite all over the high pot use, and we can now say that d E then is equal to sign X Times Coast. Why, if we look at the big angle of that right angle triangle that have highlighted in yellow the big angle is made up of the some of X and y and if we take the sign of X plus why it is the opposite over the iPod in use and the opposite is D F. And the high pot news is one so that can be rewritten in terms off DF, if you would Ah, you can say that DF is equal to d E plus C b and we can see that the EU we've already calculated out to be sine x coast. Why and c b we calculated it out to be co sex time sign. Why? So the equation for the sign of X plus y is going to be sign Axe Coast Y plus co sex time sign Why? Which is what we set out to prove now We just looked at the sign of the sum of two angles. Now I want to look at the co sign of the sum of two angles and we're gonna prove that Coast X plus why is going to be equal to Cose? X Times Coast. Why minus sine x times signed Why? And I'm going to go back two using the same diagram. I'm gonna go back to where we just came up with the fact that angle C D. E was equal toe. Why? Because all of the stuff that we did just prior to that remains the same and not gonna change it, however, were not going to be dealing with the co sign. So if we want to take the co sign of that big angle, which is X plus why and I've shaded in yellow, it's going to be a f all over one, which means that a f is going to be cose x plus. Why, if we look at lying a B, we can see that it is made up of two line segments A, F and F B or B F. Either way, you look at it. Um, Anyway, if we wanted to find out what a f waas, we would take the line a B and subtract FB from it. So putting that in mathematical terms a f is equal to a B minus f b and because where we have ah rectangle on that area, we can see that E C is the same length as F B. So we could say that a F is equal to a B minus E c going back to our original right angle triangle. We can see that coast Why is given by a bee all over co sex, which is the high pot news. And that means a B is equal to cose X times coast. Why, And looking at the little triangle, right, angled triangle and looking at the angle Why, if we take the sign of that angle, the opposite is E C. And the high pot news is s a sine X meaning that e c is equal to sign x times sign why we can see that a b is part of the equation that we developed when we were looking at a f We can see that E c is part of the same equation and we can see that a f is part of that equation too. So we can substitute those values that we just calculated such that Coast X plus Why is going to be equal to coast X times coast y equal to sign X times sign why you can see the substitution is there and that is what we were setting out to prove. I now want to look at the difference formula this time we're taking the difference of two angles. In other words, the sign of X minus. Why is what we're looking for? And we'll find that it's going to be equal to sign X Times Coast y minus co sex time. Sign why and to develop that proof. We're going to start out the same way, and we're going to start with the right angle triangle, and we're going to draw a second right angle triangle. However, this time we're gonna put the two high pot news together, and they're going to be the same length. It almost is like that Red Triangle is the upside down version of the one we did before. However, we got two right angle triangles that have the same high pontin news. I'm going to drop perpendicular czars from one point to the other and then another one across, as we did in the first place. And I'm going to designate the significant intersections and points on this diagram now as a B, C, D, E, f and G. Now I'm going to up make another assumption. Justus, we did before is I'm going to assume that the too high pot news of these two right angle triangles. They're equal, and I'm gonna have them equal one. I'm going to designate the contained angle opposite the right angle of our red right angle triangle as why and I'm gonna designate the big angle of both of the right angle triangles together as X. And if that is X, then the contained angle to the left of the right angle of our original triangle is going to be X minus. Why? And as we did before, we can see that because of the perpendicular is that we have drawn that E c is going to be parallel to a F. And if that's the case than a C is a trans versatile of the parallel lines. So those two angles x minus y are going to be equal. In other words, E. C A. Is going to be equal to X minus. Why the same as C A B s? Let's concentrate on the second right angle triangle. And I'm going to say that the sine of the angle why is going to be the opposite over the high pot news and the high pot news is one and the opposite side is D. C. And of course, one in the denominator is superfluous. So we that makes the fact that D. C. Is equal to sign. Why? And I could move that over and put it on a diagram. Similarly, Coast Why is going to be D a all over one making D a equal to coast? Why? And I could move that over and put it on our diagram as well. Let's look at this right angle triangle now that has contained angle of X minus y and a right angle, 90 degree angle as well and a g f The other angle of that triangle, because the contained angles of any triangle have to add up to 180 degrees, that angle would be 100 and 80 degrees, minus 90 the right angle and minus the angle X minus y. And if we subtract 90 from 180 we are left with 90 degrees minus that angle X minus. Why I can remove the brackets and that's going to give us 90 minus X plus Why, and that is the angle a G f. And that angle is a contained angle of two intersecting lines, which have highlighted there. And those two intersecting lines say that the opposite angles have to be equal. We already discovered that before, so that means that angle 90 minus X plus why has to be equal to C G. D. As well. And I put it on the diagram. Now I'm removing one of the highlights because I want to look at the straight line D F and the angles have to on that straight line. Have to add up to 180 degrees, cause it's a straight line in a straight line is 180 degrees. So if I wanted to find out what the angle A g d waas, I would start out with 180 degrees and subtract that angle 90 degrees minus X plus. Why? And that's what I've shown in the equation. And if I remove the brackets and subtract 90 or from ah, 180 degrees, that's going to leave us with 90 degrees plus X minus. Why? I'm gonna put the other intersecting line back in, because the other opposite angle CGF also has to be equal to 90 degrees plus X minus y, and I'm writing it in the diagram. Now, if we look at the angle a d f, we can see that the contained angles of that angle have to add to 180 degrees, and the contained angles are 90 degrees and X. So if I want to find out the third angle, I would have to take 180 degrees. Subtract 90 then subtract X, which I have done in the equation, and that leaves us with 90 degrees minus X, which I have placed on our diarrhea now looking at our little right angle triangle, and I want to find the angle indicated there with a shaded pie shape that's angle E D C E D . C is going to be 90 minus 90 minus X because the angle A D. C. Is a 90 degree angle. And if I want to find what's left for our angle of our a shaded portion there e. D. C. It's going to be 90 minus the angle 90 minus x, which just leaves us with X equal to E. D. C, which I placed on our diagram looking at the perpendicular line that we drew at the beginning of the operation here. The D F. It's made up of two line segments D, E and E F. So if I wanted to find the length of E f, I would take DF in subtract D E E or in mathematical terms, E f is equal to d F minus d e. And if we look at the rectangle that we have with created with our perpendicular, we can see that E f is equal to C B. So I can say that C B is equal to DF minus d e. Looking at our original right angle again, the angle in the contained angle that we have opposite the right angle is X minus y. And if we take the sign of that, we have take the opposite over the iPod news, and the opposite is C B. And the high pot news is one. So sign X minus Y is equal to C B. Looking at this right angle triangle and the contained angle this time is X opposite. The right angle and sine X is the opposite over the iPod news, and the opposite is DF, and the high Pontin uses Coke's Why so? DF is equal to sign X TIMES COAST Why Looking at our little right angle triangle at the top and finding the co sign of X. It's equal to the opposite over that pot news and the opposite is D E and the high pot news is signed. Why? So we can say that D e e is equal toe coast X times sign why? And if we substitute for CB in our equation we see that sine x minus y is equal to DF minus d e While we just discovered that DF was equal to sign X times Cose why? And we discovered that d e was coast X times sign Why leaving us with this equation which happens to be what we set out to prove in the first place now Ah, we talked about the sign of the in a difference formula. Now I want to look at the co sign as a difference formula and we're gonna prove that Coast X minus Why is equal to co sex Times Coast y plus sine x times sign why? And I don't want to redevelop all the first part of our diagram again. We're going to use the same diagram, and we're gonna maintain everything the same up until the point where we discover that E. D. C. Was equal to X, and I'm going to move on from there looking at our original right angle triangle. Weaken now calculate the co sign of that X minus y, and that is a be all over one because one is the iPod in use in the adjacent side is a B. That leaves us with the fact that course X minus y is equal to a B. And if we take a look at the line A B, it's made up of two line segments A, F and F B. So mathematically we can say a B is equal to a F plus f B. And because we have ah rectangle created by our perpendicular, we can see that E. C. Is equal to F. B. So we can say that a B is equal to a F plus e c. No looking back at our right angle triangle that I've highlighted in yellow, the co sign of X is going to be a f all over the high pot news, which is coast. Why making the fact that a F is equal to coast. X Times Coast. Why in the little triangle, the top sine X is equal to E. C all over. Sign whyy CB in the opposite and sign wide being night Ponte News. That leaves us with the fact that E. C. Is equal to sign. X Times sign why we can substitute for a B. The fact that it's equal to Khowst X minus y. We can substitute in Khost. X Times Coast wide for a F and weaken substitute for E C sine X sign. Why giving us this equation? Which is what we set out to prove. So, going back to our trigger identity slide, we can now add the sum and difference formulas and the some indifference formulas. I know we proved them using angles, X and Y, but we could have substituted numbers or anything else for it. But the fact of the matter is, we're going to be using Greek letters here, Alfa and Beta. And it's the same is using X and why they're just letters. So the sum and difference formulas tell us that we have these four equations for the sum and difference formulas. Now we went through Ah, somewhat of an exotic proof for the difference formulas. And I'm gonna walk through another proof it's This comes up with the same answers which you have to do in mathematics. Doesn't matter what path you take. The end results have to be the same, but I think it's an interesting exercise, and it kind of highlights some of the things that we haven't dealt with in the in the proof that we just looked at. So what I want to prove here in the difference formula using the sign sine x minus Why? I want to prove that sine x that's equal to sign X coast y minus co sex times sign why we already proved that. But I'm gonna prove it again. And I'm going to start with the some formula first, which states that sine X plus y is equal to sign X coast. Why plus co sex times sign why? And what I'm going to do is I'm going to say let why equal minus why so everywhere I see why I'm gonna put minus why in and let's see what happens if you look at the left hand side of that equation and I put in minus why you're gonna have X plus minus. Why, in a plus minus, Why leaves us with minus y. So that would give us a sign X minus y. And on the right hand side, the first term while sine x doesn't change because we're only dealing with why, But we now have post minus why? And remember, in our trig identities, we found that coast y was equal to coast minus. Why? So I can substitute now Course. Why? So I have sine x coast. Why? And in the last part of that equation, I've got sign minus Why now? Remember when our trig identities we found out that sign minus on angle is equal to minus the sign of plus that angle. So, Aiken, substitute sign minus y for minus sign. Why? And if I have minus sign, Why Times co sex. That means I'm gonna end up with a minus sign in front of the co sign. So that's gonna leave us with minus co sex times. Sign why, which is what we set out to prove. And it's just a different way of proving it a little bit simpler too. And let's carry that forward to the difference formula for the co sign. And we're going to go through the same exercise. We're going to start out with the some formula, which you see there, and I'm going to substitute all of the wise. I'm going to say let why equal minus Why So if I do that in the equation Coast X plus minus , Why is going to leave us with just Coast X minus y and coz x times Coast minus. Why? While we've already discovered that the co sign of why is equal to the co sign of Minus Why so I can just say co sex Times Coast why and sign minus y is equal to minus sign. Why? Which will change the the sign in front of that last term, giving us plus sine X sign why which we set out to prove. So how useful is all of these trig identities? Well, as you come across them in scientific notation or discovery, we'll use them off and on. But let's just go through an exercise. It's pretty simple exercise to show you that there is some use for these summoned difference formulas. I want to find the exact value of coast 105 degrees. Now let's assume we don't have a calculator that figures out co signs and signs, and we don't have a set of tables. But we would like to come up with unanswered to our question here. So what we conduce you is let's break 105 degrees into two um two angles, 60 degrees and 45 because 60 plus 45 adds up to 105 so we can take co sign of 105. That's going to be equal to co sign of the angle. 60 degrees plus 45 degrees and the some angles. The sum of angles comes out to coast 60 times coast 45 minus sign 60 times Sign 45 and that is equal to 1/2 times route to over two minus root, 3/2 times route to over to and they all have a common denominator. While it doesn't really matter in this case, but they both the known ah, meters multiply out to four, and the numerator multiply out to route to and Route six. That leaves us with 1/4 of the quantity route to minus Route six. So we did come up with someone oven answer. We can find the square root of two in the square root of six that we had to, and let's try one more the exact value of signed 15. While we can use the difference formulas this time you say signed 45 minus sign Sarif signed 45 minus 30 which is equal to sign 45 times. Khost 30 minus coast, 45 times Signed 30 which is equal to root to all over two times. Route 3/2 minus root to all over two times. 1/2 which, if you multiply those out is Route six all over four minus route to over four, which is equal to 1/4 of the quantity route of six minus the root of two, and that ends this chapter. 10. Ch 09 Trig Identities: Chapter nine product to some formula and the trick functions that we're going to be looking at have the product of Trig functions on the left and the sum of trig functions on the right. And we're about to prove these four Trigana metric product to some formulas. And in starting to do the proof, we're going to use these formulas that we've already developed, which are the some formulas for signs and co signs. And to start with, we are only going to look at the co sign functions, the additions and the subtractions of two angles. I want to subtract the bottom equation from the top equation, and we have to deal with the left hand side in the right hand side separately. But let's start with the right hand side first. If we subtract the bottom equation from the top of the pop equation, we can see that if we subtract coast X Times Coast. Why from coast X, Times Coast Why we're gonna be left with nothing. So those not function disappears from the right hand side of the equation. And we're left with just subtracting the bottom equation from the top of equation here and This is what we have left. Then we have coasts X minus. Why? Minus cose X plus y That's equal to if you subtract minus sign X sign Why? From plus sine X sign why we essentially change the sign of the bottom +12 Positive. So we have to sign X sign why? And if we reverse put the left hand side on the right hand side in the right hand side of the left hand side. This is what we end up with. And if we divide both sides by two were left with this equation sign. Next time sign y is equal to 1/2 the quantity coast X minus Y minus coast x plus. Why? And that is what we're trying to prove for the first product to some formula now, going back to our to some indifference co sign functions, we subtracted them and were able to prove the the first product of some formula. Now I would like to add those two equations together, and if we add those two equations together looking at the right hand side, we will see that if we add minus sine x time sign why two plus sine X sign why those figures will disappear because they at 20 leaving us with these two equations, which we are going to add right now. And we end up with this co sign X plus y plus co. Sign X minus Y is equal to two co Sex Coast. Why? And if we change the left and the right together we will have the same equation, just aligned a different way. And if we divide both sides of the equation by two, were left with this equation. Co sex times Coast y is equal to 1/2 the quantity co sign X plus Why plus Coast X minus Why ? And that proves the second product to some formula. Now let's go Ray back to the beginning, And this time I want to use the some indifference of the sine functions. And we're going to start off by adding these two functions together. And you probably concede the pattern developing here. If we add the the two equations together and we look at the right hand side, the plus and minus cose Aynak sign why they will add to zero, leaving us with just this, which now will be sine X plus. Why plus sine X minus y is equal to two sine x coast. Why reversing them? They are. It's the same equation just written a different way and weaken. Divide both sides by two. And we're left with that with that equation, which actually proves the third product to some formula. And you probably guess what we're going to do now is we're going to instead of add adding knows two equations together. We're gonna subtract the to. And if we subtract the two, then you can see the on the right hand side, the sine x coast. Why function disappears because they had add to zero. Leaving us was just this. And if we add or subtract the left hand side, we're left with sine x plus y minus sine X minus. Y is equal to two co sex sign why? And we can change that around. We're just writing in a different way, putting the left hand side on the right hand side on the right hand side on the left hand side, and then we're gonna divide by to again, and we're left with this equation, which proves the last product to some formula now these look rather tedious. And you might wonder, while why are we going through all this exercise? Well, as I said before, as you start to develop formulas and work in different areas of science and electricity, you will come across the, uh, the opportunity to change the equations around. And you will use these product to some formulas. And we just went through the exercise of proving them now. So if you run into them, you can always come back to this presentation and see where these formulas came from. The next set of trig identities that I want to look at our called us some to product formulas, and they involve taking the sum of trig functions and changing them to the product of trig functions. And we're gonna go through an exercise now to prove these four trig identities. So these are the identities that we want to prove. And in starting the proof, I want to use our product to some formulas that we just developed. And by the way I did say we would use them for proof later on and in other functions. While we're going to use the proof right now to prove the some two product rules. So I'm just going to say, rather than using X and Y in this equation, I'm replacing X and Y with A and B. It doesn't really matter which variable you use, but our product to some formula stated that if you had signed a Times Coast, B A and B being angles, of course, that would is, this is equal to 1/2 the quantity sign a plus B plus sign, a minus beat. We just went through that proof. Now I'm going to do another trick. I'm going to say Let a the angle a equal x plus y all over to on. I want to let be equal X minus y all over to X and y. Of course, being variables as well, I can say Let them be anything because A and B are variables. So what I'm going to do now is I'm going to take wherever I have C c. A. In the equation just to the left. I'm going to replace it with X plus y over two, and wherever I see the angle B, I'm going to replace it with the angle X minus y all over too, Then that is going to give us this. And all I've done is replace the A and the B with X plus y over to an X minus y all over, too. So that was a simple step. Now, if you look at the right hand side of the equation, you can see inside the brackets for the sign formulas. You have X plus, while over two plus X minus y all over two. And then the 2nd 1 is X plus y over two minus X minus y all over too well, they have a common denominator. It's too right now. So I'm gonna replace those two fractions with a single fraction. And that would be X plus y plus X minus y over two. And in the second sign function is going to be X plus y minus X plus y all over, too. Now, you can see enough in the first sign that the two Wise will cancel out because of plus, why and a minus y. At +20 and in the second sign function, you can see that the excess cancel out because you have a plus x, an a minus. Why now? The equation is gonna look like this and you can see that the to sign functions are two X over two and two. Why all over too? And, of course, the twos will cancel out. Just leaving us with X and Y And the equation then can be rewritten toe look like this. I can swap the left hand side of the right hand side and the equation then will become this , which proves the first formula that we were setting out to prove in the first place. Continuing on with the proof, I'm going to use another product to some formula, and I'm gonna use the variables A and B again. I'm going to start with this formula and I'm going to go through the same exercise. And this time it's going to be the same thing. I'm gonna let a equals X plus y over two and be equal X minus y all over, too. I'm going to replace the A and the bees in our formulas sold at the equation. The product of some equation is now gonna look like this. And if you look at the to sign functions, the same thing is gonna happen. I they have a common denominator and the wise will cancel out in the 1st 1 and the exes will cancel out in the 2nd 1 And we will be left with this equation, which reduces to this because two X over two is just acts in two I over to is just why swapping the left and right hand sides. We end up with this, which actually proves the second formula that we're after. And you probably noticed the pattern. Now, this time I'm going to use a different product, some formula. I'm still gonna substitute for the A and the b rewrite the equation now to give us this again. We've got to come to nominator This time. They're coast functions doesn't really matter because inside the brackets with a weaken replace the two fractions with just one with a common denominator. And we can very easily see that the wise cancel out in the first co sign function and the X's cancel out in the second Khost function, leaving us with this equation and the can be reduced down to this equation swapping the left and the right hand sides. We are left with this equation which is actually the proof for the third function. And the same thing is gonna happen with the, uh, the next, uh uh next some to product formula. We want to prove we're going to start off using the product of some formula that you see here, and we're going to go through the same exercise, replacing in the A and B with exes and wise. Then we're going to rewrite the equation and it's gonna look like this. We're then gonna replace the four fractions with a common denominator of to, and that's going to give us X plus y minus X plus y over two. And in the second co sign function, it's going to be X plus y plus X minus y. And that means in the first co signed functions. The exes will cancel out in a 2nd 1 The wise will cancel out, and that's going to leave us with this equation, and we're gonna have in the right hand side of the equation, co sign to I over to minus co sign two x all over to which, of course, the twos cancel out, and that's going to leave us with to sign X plus y over two times Sign X minus y over two is equal to coast. Why minus Coast X and that can be rewritten chain exchanging the left side for the right side. And we're going to do one more thing. And that is multiply both sides of the equation by minus one, and the equation now will look like Coast X minus Coast Y is equal to minus two sine x plus , while over two times sign X minus y all over, too, which proves the last equation in our set of four. I would now like to look at co function identities, and there are six of them three main ones and three that are basically the inverse quantities of the 1st 3 And these co function identities are the fact that sine X ah is equal to co sign of 90 minus X, and the co sign of X is equal to the sign of 90 minus X, and the co tangent of X is equal to tangent of 90 minus X and the other three trig identities. Cool function identities are just the inverse of the main ones. So in looking at the proof, it's pretty simple. We're going to start with a right angle triangle, and I'm gonna label the sides of the right angle triangle A B, N. C. C. Being the high pot news. And I'm going to look at two angles of this right angle triangle. They are the other two angles other than the 90 degree angle, and that is ex and why. And we know the interior angles of a triangle are related because they must add to 100 needy degrees. And we've seen this before that if that is the truth than the angle, why would have to equal 90 minus X? And we know that the co sign of why is the adjacent over the high pot news and the adjacent side is be and the high part muses C. But that is also equal to Sign X because Sine X is the opposite all over the high pot news and the opposite of some of X is B and the high pot muses still see. So we have a relationship between coast why and sine X as well as we have a relationship between the two angles So we can substitute for why in that equation, because why is equal to 90 minus X so co sign of 90 minus X is equal to Sign X, which is one of the identities that we had we had to prove. And if we look at the sign of why it's the opposite all over the high pot in use and the opposite for why angle? Why is a high? Pontin uses C, and that's equal to the co sign off X, and we can substitute for Why, again and why is 90 degrees minus X for a sign of 90 degrees, minus X is equal to Coast X, and that is the second call function identity that we had to prove carrying on the tangent . Why is given by the opposite all over the adjacent and the opposite of why the angle? Why is a and the adjacent side of why is be and that is equal to co tangent X. And we know that because if you take the tangent of angle X, it is be all over A. But a over B is the inverse, which is co tangent of X so tangent of why or tangent of 90 degrees minus X is equal to co tangent X which is the third call function identity that we set out to prove we can work through the other three ah functions as well. See, can't. Why is see all over B and that's equal to the Cosi can't acts. So we can say that the seek and of 90 degrees minus X is equal to the co seek end of X and the Cosi can of why SC over a and that's equal to the C can't of X and the Cosi can't of 90 degrees minus X is equal to the C can't of X. And lastly, the co tangent of why is be Over a, which is equal to the tangent of X. That means that the co tangent of 90 Degrees minus X is equal to the tangent of X, which completes the six co function identities, and this ends the chapter 11. Ch 10 Trig Identities Double & Half Angle Formulas: Chapter 10 double and half angle formulas when looking at double angle from Those were looking at two angles that are identical, and we're placing them side by side, so they have won a Jason site. We're gonna call the Angles X, and there's two of them. So there's two X. If you put them together, they form a large angle. That is, two X is faras. The number of degrees is concerned, and in dealing with double angles, we're going to use the some Trigg identity, which we discovered already. Only this time it's gonna be easy, because why is going to be equal to X so we will have the sign off. Two X is equal to the sign of X Plus X, and if you plug in the exes into the formula that we proved, we have sine x Times co sacks plus co. Sex times, sign X, and that is equal to two sine x times Coast X. And if we're going to find the co sign of a double angle, we're gonna we'll have to use the formula that we discovered for the sum of two angles using co sacks. And that was coast X plus y is equal to co sex Times Coast. Why, in a sign next time, sign why? And it's easy again, because why is equal to X So coast to x is equal to Coast X plus X, which is equal to co sex co sex minus sine X sine X, which is equal to Khost squared X minus sine squared X. And we can have that formula in a different way if we would like because of the formula, the Pythagorean identity that we proved coast squared X plus sine squared X is equal to one so sine squared X is equal to one minus. Coast squared acts so we can substitute for the sine squared X in our formula, and we'll find that coast to X is equal to coast squared X minus the quantity one minus coast squared X Simplifying it down. We have co squared minus co squared X minus one plus co squared X What's comes out to two co squared X minus one, which means we have nothing but co signs in our formula, and we could do the same thing. Getting rid of the co signs on the right hand side of the equation by solving our Pythagorean equation in terms of sign quantities. In other words, co squared X is equal to one minus sine squared X. So going back up to where coast to X was equal to coast squared X minus sine squared X, and we substituted for the coast squared X and simplified the equation. We would come out to the fact that coast to X is also equal to one minus two times sine squared x. We're now going to look at the tangent of ah, double angle, and that angle we're gonna call it two X and two X is just an angle itself. So any time we have a tangent oven angle, it's equal to sign of that angle all over coasts of that angle, which is sign two x all over Coast to X, which we have already looked at before. And we can say that Sign two x is equal to sign. X Times Co sacks plus Co. Sex times sign X, all over co squared X minus sine squared X. We're now going to multiply the right hand side of that equation by one or, in other words, coast squared X all over Coast squared X, and the result looks a little bit confusing. It's a couple of fractions over a fraction, however. What we have done is divided the numerator by Coast squared X, which gives us two fractions separated by a plus sign, and we divide the denominator by co squared X. It becomes a fraction itself co squared, X minus sine squared X all over coast Squared X. Looking at the numerator, you can see a common term that we can remove by essentially canceling it out. Nadia's Co sign of X. In the first term, CO sex will remove the coast squared accidentally, nominator. It'll just be one in the numerator. Similarly, in the second part of the numerator, the Coast X and Co squared X cancel each other out. So we're left with sine X over co sex plus sine X over co sex and sine X over co Sex, of course, is tangent Tangent of X. So we have to 10 X. And if we split the denominator out into two fractions, which weaken? Do you can see that the first part of those two fractions co squared? X over Coast Board X is just one and sine squared X over co squared X is just the tangent squared, so the denominator now becomes one minus tan squared X. So for a double angle, we have found a solution for the sign co sine and tangent of that double angle. In terms of a single angle, we're now gonna look at half angle formulas basically going the other direction. And the way to look at it is we're going to start with untangle called X. We're gonna divide that angle in half so that it produces two more angles, which are X over two X over two. And what we want to do is have a function, whether it's a sign co sign or a tangent of X over to in terms off the angle X, whatever that turns out to be an arithmetic or triggered a metric functions, we're gonna gonna find how to calculate the half angle from the hole angle. And since we already discovered how to analyze the double angle, we're going to use that formula to continue on and calculate a formula or a function for the half angle. So we're going to start with coast to a we said it was coast to X before, but doesn't matter what the variable is, and you'll see the reason why I'm going to use a in a few minutes. But what the the double angle formula told us was that coast to a was equal to one minus two sine squared a and very quickly I'm going to let a equal X over two. So I'm gonna put X over two in our double angle formula anywhere I see a So now we're back to using X as the variable angle. And our double angle formula says that Khost two times x over two is equal toe one minus two sine squared x all over, too. We can simplify that equation if we look at what's happening with the left hand side of that equation. On angle that is two times X over two is really just x so we can simplify the equation. It now becomes CO sex is equal to one minus two sine squared x all over, too. And I'm just gonna switch about the equation because I want to get my X over two quantities on the left hand side of my half angles on the left hand side and my full angles on the right hand side. No, I can just divide both sides of the equation by two, and I'm left with sine squared. X over two is equal to one minus coast ex all over to, And if I find a square root of that, it's a sign. X over two is equal to plus or minus the square root of one minus co. Sex all over, too, and the plus or minus sign mathematically really means, and you'll see this a little bit later. Depends on what quadrant of the rectangular coordinates that you're in. You can actually have two answers to this equation, depending on which quadrant that you are in now, going back to our original formula and instead of having coast to a is equal to one minus two sine squared A. I'm gonna use the other equation that we developed. And that was before the double angle formula that is, that coast to a is equal to two coast, where day minus one. And again I'm gonna let that the A equal X over two plug that back into my formula, and that gives us to co squared X over two minus one is equal to Coast X, and that simplifies down if we go through the same process to coast X. Over two is equal to plus or minus the square root of one plus CO. Sex all over, too. So we developed a formula or a function for the half angle formula for the sign and the co sign. The sign and the co sign of 1/2 angle are all in terms of the full angle. That leaves us to discover what's happening with the tangent. So we're going to start with the tangent. And this time we're going to use the the identity that the tangent of any angle is equal to the sign of that angle over the co sign of that angle. And the angle we're looking at in this in this term is X over two. So tan X over two is equal to sign X over two all over coast X over two, and we just discovered what the The answer to the sign. The sign of the half angle on the coast of the half angle is so we're going to use that and we discovered that it was the plus or minus square root of one minus co sex over to and the coast X over two is plus or minus the one plus square root of one plus coast ex all over, too. And I'm just gonna rewrite that equation in just a little bit different form. I'm gonna keep the square root in the nominee a separately from the numerator and the denominator. So the square root of one minus coast ex all over to is the same thing as the square root of one minus co sex all over the square root of two. And the same thing can happen in the denominator. We're gonna end up with the square root of one plus CO sex all over the square root of two . And we can simplify that further by if you want to call across multiplying but week, instead of having a fraction over a fraction, change it to just one fraction. And that is going to be the square root of one minus co sex times the square root a tube all over one plus the square root of one plus co sex times a square root of two. And of course, that's plus or minus both in the the numerator and the denominator. The fact now we can see is that we have a square root in the numerator and the denominator so we can essentially cancel that out. And that leaves us with the square root, a one minus coast X all over the square root of one plus Khost X, and that can be plus or minus. And we could essentially leave that equation there cause we already have, uh, the half angle in terms of a full angle. But it is still rather cumbersome with the square root in the numerator and the denominator . So let's try and simplify it a little bit further. And we can do that by multiplying that equation by one that one being the square root of one minus co. Sex all over the same thing or square root of one minus co. Sex. But when we use that form of one, our equation becomes simplified as you'll see and in simplifying these equation, and we have to go back and try and try to remember our binomial fear ums and the binomial theorem. If you look at the product of two Bino meals, one minus co sex and one minus co. Sex is equal to one minus co sacks, all squared and one plus co sex times one minus co. Sex is actually equal to one minus coast squared X. So if we remove the brackets and replace them with square root signs, it's the same thing. We're still going to multiply two by no meals in the numerator and to buy normals in the denominator. The numerator is the one minus co sex, which will become one minus cose X squared under the square root sign, and the denominator is gonna be one minus coast square decks under the square root sign. Now, if we find a square root of one minus coast X squared that's just one minus co sacks and the square root of one minus co squared X becomes just sine squared X because one minus Khost X is sine squared axe. Now we confined the square roots of both the denominator and the numerator very easily. It works out to one minus coast X in the numerator and just sign X in the denominator. And of course, that's plus or minus. Also, depending on where we started with our original angle in which quadrant we're gonna end up with. It could even be plus or minus. However, we have successfully reduced our tangent to a simple original angle of X for the tangent of the half angle. So let's do a couple of exercises just to see how we can use these double angles and half angles. And let's start with a right angle Triangle Lee 345 triangle and the contained angle. We're gonna call it X for now, and we don't know it. Let's assume we don't remember what the angle of the 345 triangle is, but we do know the sides, so we do know the ratios. We know the science that co signs off angle X and let's double that angle and we're gonna have a Nangle. Let's two X uh, in a Sfar as the size is concerned. So it's two times X of the new angle, and our exercise will be defined what that angle is. In other words, find the co sign of two X. Well, we know the co sign of X. It's the adjacent over the high pot news, which is 3/5, and we know what the sign of X is its 4/5, and we know what the co sign, a formula. The co signed two exes. It's Ah ko squared X minus sine squared X. So all we gotta do is plug in the values for acts, and we find that the Coast squared X is 3/5 squared, and the sine squared X is 4/5 squared. So if we take the first value minus the second value, first of all, finding the square of the first value in the second value. So the square of 3/5 says 9 25th and the square of 4/5 is 16 25th and we can subtract the second from the first value because there's a common denominator, and that's going to give us minus 7 25th which, if calculated, is actually minus 0.28 And you might ask yourself, Why do we have a minus sign here? Well, if you look up at the rectangular coordinate plain, it's easy to see that we are in the second quadrant. So when you're in the second quadrant, the co sine of the angle is definitely going to be a minus Let's do one more and let's find a co sign of 50. Now we could look it up or punching in our calculator, but it's more fun going through the exercise. So we know that 15 is the, uh, angle or the half angle of 30 degrees, which is the same thing as saying co sign of 30 all over, too. And we know that the angle, the whole angle of the half angle, is 30 degrees. And we already know that the co sign of the half angle is a plus or minus square root of one, plus the angle over two. So that's one plus Khost 30 all over, too, and going back to our triangle right, angled triangle or 30 60 90 triangle, we can see that the sides are in that ratio. So the co sign of 30 degrees we know what that is. It is Route three all over, too. So now our equation for co sign of the half angle or co sign of 15 degrees is plus or minus one plus the square root of three over to all over, too, and we can simplify that by multiplying by one or square root of two over the square root of two gives us plus or minus in the numerator, the square root of two, plus the square root of three all over, too. And if we look at 15 degrees, we know that the co sign is positive for 15 degrees, so we can forget about the minus sign because we're in the first quadrant and we consol of that equation just by by calculating the numbers. And it works out two plus 0.966 What happens if you have a trig racial butter asked to find what the angle is in days gone by and even today, if you like, you could use or had to use what we call trig tables such as this, which are designed to give us the signs, coke, sines and tangents. For a given angle, however, you can use the table in reverse. For example, let's say you're given the sign of an unknown angle and that that sign value is 0.7 if you start at the angle zero. We know that the sine of the angle increases as the angle increases, so all you had to do or have to do is follow the column until you reach the closest value for which that is in the table, and that value is in our case here, 45 degrees, which is point are 0.707 which is the sign ratio for 45 degrees. You can see that 450.7 is about halfway between 44 45 so the answer would be 44 a half degrees. Today, all you have to do is grab your calculator or your smartphone and use the inverse trig function keys. Notice that they are sometimes above the sign co sine and tangent buttons. As you see here, there is a sign to the minus one coast to the minus, one in tan to the minus one. These are your inverse triggered a metric functions, also known as Arc sine Arc Co sign and Arc Tangent. If you use these buttons in conjunction with your tree gray show, you will get the angle measurement that you are looking for the next few slides, arm or oven example rather than a learning situation, Joe. So just sit back and observe and enjoy what's gonna happen, because it's just kind of Ah ah, prelude to something that's going on in the other courses. And I said before that the reason I put this trig course together is quite often you're going to run into these triggered a metric functions when you're analyzing electric circuits. So I just want to show you what how we use these trick functions. And when you come across them, a few decided to take some of the other courses here. You will actually see the next couple of slides, which I'm going to show you, is kind of a preview of what's what's out there now. This is also kind of ah, prelude to the next chapter, which is going to be on the time to mean function, but we're gonna have a little preview of it rate. Now, at this point in the electrical power course we're observing, the current voltage and power delivered to an impedance zed in the circuit that you see in the upper right hand are upper left hand corner. The upper right hand diagram shows the vector or phaser representation for the current and voltage in the circuit. The grafts along the time domain indicate the current the voltage with respect to time as while is a power consumption by the impedance over that same period of time, a couple of things to notice. And don't bother memorizing this because it will be dealt with in a course where it's more relevant. The current and voltage are out of phase, and the power flow in the circuit is at times positive and at times negative. And it is constantly changing. What we're about to figure out, though, is that the average we want to find out what the average power flow is because that's in dealing with the revenue for utilities. They don't measure, uh, or don't bill you on instantaneous power. They bill you on average power, and that's kind of the average power that you consume over a month. So it doesn't matter that the power is going back and forth from sometimes negative, sometimes positive. It's the average that the power company wants to build on. And we proved that by taking the the formula for instantaneous power, which is just the voltage times current, and we're dealing with Sinus soil of voltages and current. So this is the actual formula for the instantaneous power. Now I'm not going to go through this in detail, explained every move. All I want to do is highlight where we're bringing in the Trigana Metric functions and the Trigana metric identities just to show you that we do use them. And the 1st 1 that we're going to use is the product to some formula that you see here. We're gonna let the X equal Omega T minus phi. And why be Omega T? And if we do that, our formula becomes this weaken, then satisfy or simplify the formula, tell it looks like this, then we are going to use the difference formula this time letting X equal to omega T. And why equal to Fi. And if we plug those values in or substitute those, that equation then becomes this and we then rely or go back and use what we are have learned as the negative angle identity. In this case, Coast minus X is equal to coast plus X. That will help us simplify the equation to this and ultimately to this which we can say that over a period of time, that being a month, which is a ah lot longer time period than 1/60 of a second. We can get rid of the ought, the Sinus soil part of this equation again, this is theory that you don't have to remember, but we we at this stage now we can start to analyze the the equation that we have and it will reduce to this giving us the average power is the peak voltage times that peak current all over, two times a phase angle between them, which is a fundamental or the fundamental power equation. If you know anything about power in a sea circuits the actual value peak values of the village in the current VM and I am putting them over to a za Samos putting VM I am all over route two times route to, which means that VM over route to is root, mean square of the voltage and I am or two is the root, mean square current. So it's actually the voltage times a current times a phase angle between them if you were using the root mean square voltage. So now that I got you all excited about the electrical courses that we can be taking, you can go back to the site where the streaming is occurring and you can take advantage of the course material related to the A C power in a sea circuits anyway, that ends this chapter. 12. Ch 11 Trig Functions in the Time Domain: Chapter 11 in the time domain. The time domain refers to the analysis of mathematical functions such as physical signals, voltage current etcetera with respect to time in the time domain, the signal or functions value is known for all real numbers for the case of continuous time or at various separate instances. In the case of discrete time Ana. So scope is a tool commonly used to visualize riel world signals. In the time domain, the function may be written as an equation. In this case, a repetitive signal or a function is written in terms of a sign you sidle function where Omega represents the angular velocity and degrees per second and T is in time and seconds. In terms of North American electrical system, Omega represents 60 cycles a second, which is 60 times 360 degrees, which equals 21,600 degrees per second. Sine omega T. When described in the form of a unit circle where omega T is the angle formed by a moving radius, the red line, which is the high pot news of a continuously changing right angle triangle sine omega T is the trig function formed by the fraction of the length of the green Line divided by the length of the red light, which will vary from zero to plus one back to zero, then two minus one, then back to zero again and repeats itself over and over and over again. This graph, when shown on a rectangular coordinate plain the horizontal is still time, but instead of marking it in seconds, we're marking it in terms of Omega T, which is degrees. The vertical axis represents the function in which this case, we are saying that it is voltage. However, it could be current flocks or even the pendulum distance of ah, hanging pendulum. This is voltage in this example, when t is equal to zero, the beginning of our observation sine omega T A zero subsequently So a low voltage B zero, when T is equal to 90 degrees, the voltage V will be at a plus B the maximum and when it when t is equal to 180 degrees sign ah, 100 needy degrees. Zero. Subsequently, the voltage will also be zero when t is equal to 270 degrees sign. 270 degrees is minus one. Subsequently, the voltage V will be minus V maximum negative when t is equal to teak or 3 60 signed 3 60 is again zero subsequently solar voltage B zero. This time we're looking at the same function in the time domain. But this time time the horizontal instead is being marked. Instead of being marked in seconds were marking it in terms of radiance per second. And the horizontal line will then be a radiance. When t is equal to zero at the beginning of our observation or zero radiance, then the voltage is also zero. When t is equal to a 1.57 radiance or pi over two radiance. Subsequently the voltage will be plus B at a maximum. When when the radiance equals 3.14 or pi radiance sign up, I radiance zero. Subsequently, the voltage will be zero. And when t is equal to 4.7 radiance or three pi over two radiance, then we're in a maximum negative level for the voltage. And when t is equal to 6.3, radiance were gone through one complete cycle and we're back to zero again. Now what happens if we multiply this function by a magnitude of a value greater than one. The amplitude of the function, which is sine omega T, increases in both a positive and the negative direction to a value of plus ve and minus the and let me put the original sine omega T back in just for comparison. If the magnitude V is a value or a constant of less than one, the amplitude of the function, which is sine omega T, decreases in both the positive and negative direction to a value of the constant times plus one and minus one. Next, let's zoom in and see what happens when we add an angle displacement of data degrees to the function. This means that AT T equals zero degrees. The wave has already advanced by fada degrees, which looks like we have given the function I had start such that it now leads the function sine omega t by fate a degrees. Next, let's see what happens when we subtract an angle of displacement off of a data degrees from Omega T. And in the function it's going to be included in the number of degrees that we're taking the sign off This means that the sine wave doesn't start to begin or be positive until T is equal to fated degrees, which looks like we have delayed the function such that it now lags the function by sine omega T or legs, the function of sine omega t by Theda degrees. In other words, the function is zero when the angle is zero or when omega team minus data is zero, which is when Omega T is equal to plus they don't degrees. We have to keep in mind that the function also exists for values of times less than zero. In other words, we're looking at this signal at a snapshot in time. So we're assuming the signal has been going for some time, and it's continuing to go for some time now. Let's apply on angle displacement of 90 degrees, two sine omega T. This will shift the way form 1/4 cycle to the left, 90 degrees. Now let's look at the function Costa Omega T With regards to our unit circle, remember that in the unit circle, the radius of the circle is one and looking at a right angle triangle with a high pontin, use equal to the radius than the co sign of the contained angle will be the length of the adjacent site of the right angle triangle over one. So as we start with omega T equal to zero cose, omega T will be at one. And as Omega T increases cose omega T decreases until omega T equals 90 degrees, at which time Costa Omega T will equal zero etcetera. What we see is that Costa Omega T the burgundy curve now follows exactly the sine omega T plus 90 degrees the green curve. We might conclude that cose Omega T is equal to sign Omega T plus 90. So before jumping to the conclusion that Costa Omega T is equal to sine omega T plus 90 let's look at a proof and on and make sure that that equation is solidly proved. Now we're gonna look at the rectangular coordinate system, and we're going to use the unit circle in, uh, the rectangular coordinates system. And I'm going to place a right angle triangle rate on the X axis with, ah, the the opposite angle to the right angle, and I'm going to call it X and in a place that Vertex of that angle right at the origin. And I'm gonna call the, uh, Triangle A, B and C, And we know previously that in the unit circle that Coast X is actually the length a sea of that right angle triangle. Now I'm going to rotate that angle about a about the origin so that the side A C is perpendicular. In other words, it runs right along the vertical axis. And because we're in a unit circle or are the circle goes through the Vertex be the new triangle. I'm gonna label a B primed see crime the a point of the Vertex be primed. It's going to still be on the unit circle. And I want to look at the big angle right now. And I've shaded that in yellow, as you can see there. And that angle is actually X plus 90 because we've rotated it the angle X by 90 degrees. So that big angle that you see there is X plus 90 degrees. And if we took the sign of that angle, it is going to be the vertical distance of be primed all over the high pot. Unease, which is one and the vertical distance is actually the length of the orange line that goes from be primed down to the horizontal axis you see there. And because that's a perpendicular, it will be exactly the same length as a C. So we can say that sine X plus 90 degrees is equal to a primed. Now we have to triangles, which are identical except that one is placed 90 degrees to the other. And because they're identical triangles, the side of a C is the same as a C primed, so a see the length. A sea of that right angle triangle is equal to the length of a C, primed of the other right angle triangle. So we can say that Cose X is equal to the quantity X plus 90 degrees. Now if we left X equal omega T, then we have coast Omega T is equal to sine omega T plus 90 degrees, which we were trying to prove in the first place. There's another way of proving this, and I'm just going to go through it very quickly. We went through this trick identity of the some of angles which sine X plus y is equal to sign X Times Coast y plus Co. Sex times Sign why and we're gonna use it to prove this idea proved this equation Cose Omega T is equal to sine omega t plus 90 And if we let X equal Omega T and why equal 90 degrees? Then if we plug goes into our formula, we'll have sine x plus Why reading sine omega t plus 90. And just before doing, making that substitution in the equation, we have to I want you to remember that sign of 90 degrees is one and a co sign of 90 degrees is zero. So now, if we rewrite that identity equation well now have sine omega t plus 90 degrees is equal to sine omega T only GTI times coast, 90 degrees plus coast omega T times sine 90 degrees and the first part of the right hand of that equation the first term the co sign of 98 0 So that term, that whole term is going to be zero. And in the last term of the right hand side sign of 90 is one. So that means that sign 90 will disappear and just leave Costa Omega t. So we just seen that the first term is going to be zero. So that's not gonna add anything. So we can get rid of that. And we can get rid of the 90 degrees, a sign of 90 degrees, because that's one leaving us with. Just coax Omega T. And we can rewrite that equation by just flipping the right hand side with the left hand side. And that, too, will prove our equation, and that ends this chapter.