Essential Trigonometry Concepts to Know

Let me show you the top 10 most important things for you to know about trigonometry. Must know number one, similar triangles. Similar triangles are triangles that have the exact same shape, but might have different sizes. Let me draw two triangles so you can see what I mean. Now I've started by drawing these two triangles the exact same size and shape, so they're congruent. If I were to take triangle DEF and make it smaller, without changing any of its angles, so keep the shape the exact same, it is still similar to triangle ABC. And similar triangles have two very important properties. Their pairs of corresponding angles are equal. So in these two triangles, you can see that angle A and D are equal to each other, B and E are equal, and C and F are equal. The other important property is that ratios of corresponding sides are also equal. So the ratio of side AB to side DE would be equal to the ratio of side AC to DF, which would be equal to the ratio of side BC to EF. And we can use these properties to solve for missing sides and angles in triangles that are similar to each other. Let me show you an example. In this question, we're asked to solve for the length of X. In order to do that, I'll have to use these two similar triangles that I see, the big triangle and the smaller triangle within it. We'll have to start by proving they're similar, which there are three ways to prove similarity. You can do angle-angle similarity, just find two pairs of equal angles. Side-side-side similarity, where you show that the ratios of all three pairs of sides are equal. Or side-angle-side similarity, where you find two equal ratios of pairs of sides, and one matching pair of angles. For this question we can do angle-angle similarity. If you know your angle theorems, we have two parallel lines here, connected by this side. I know that the angles inside of that F pattern called corresponding angles are equal to each other. So I could say that angle D is equal to angle B. Same thing on the other side of the triangle. I know that these angles in here are equal to each other. So angle E equals angle C. And then of course, the two triangles both share angle A. So based on all three pairs of angles being equal, I know the triangles are the same shape. So I could say that triangle DEA is similar to triangle BCA. And because I know that they're similar, I know that the ratios of corresponding sides have to be equal to each other. So I know that the length of BC divided by the length of DE is going to be equal to the length of BA divided by the length of DA. And then if I sub into this equation what I know, I can then solve this equation using algebra. I can cross multiply, 10 times 6 is 60, and 4 times x plus 6 is 4x plus 24. Solving this for x, I'd have 36 equals 4x, divide the 4, I have x equals 9. There you go, now you can see how to use properties of similar triangles to solve for unknowns. Must know number two, SOHCAHTOA. This is an acronym that helps you remember the three primary trig ratios. Let me draw a right triangle. In this right triangle, I've indicated the 90 degree angle, and then I've labeled one of the acute angles. angle theta. We'll call that the reference angle. All other right triangles with this same reference angle theta would be similar to each other, right? No matter how much I change the size of this triangle, as long as I keep that angle theta the same, it has the exact same shape, meaning the triangles are similar, which means based on what we know about similar triangles, they would have the same ratios of corresponding sides. In any right triangle, there are three sides. Let's name them. The longest side of a right triangle is always called the hypotenuse, I'll label that h. The side across from our reference angle theta, we'll call that the opposite side, I'll call it o. And the side touching our reference angle theta, we'll call that the adjacent side because it's right beside our reference angle, I'll label that a. Because there are three sides, there are three different pairs we could make of these three sides. That's why you see three ratios. in SOHCAHTOA. If we find the ratio of the side opposite from the reference angle divided by the hypotenuse, we call that the sine ratio. Cosine of the reference angle is the adjacent side divided by the hypotenuse, and tan is a function that means we're looking at the ratio of the opposite side divided by the adjacent side. Now these primary ratios can be used to find a missing side or a missing angle of a right triangle. Let me show you both examples. Let's look at this first example. where we're going to solve for a missing side, X. I notice it's a right triangle with a reference angle of 40. All right triangles with a reference angle of 40 are similar to each other. They all have the same ratios of sides. So whatever x divided by 8 is, will be the same ratio of all other right triangles with a reference angle of 40. So let's communicate that. From 40, the side we're looking for is adjacent to it, and 8 is the longest side of the right triangle, so that's the hypotenuse. The function that involves adjacent and hypotenuse is cosine. So I know cosine of 40 degrees equals the adjacent side, x, divided by the hypotenuse, 8. Your calculator has stored within it what the ratio of adjacent over hypotenuse is for all right triangles with a reference angle of 40. If you typed cosine of 40 degrees on your calculator, it'll give you that ratio. And then we can just multiply the 8 to the other side of that ratio to isolate x. SOHCAHTOA can also be used to find a missing angle inside of a right triangle. If we're looking for this angle here, notice that 8 is opposite from it. and 17 is a cross from the right angle, so that's the hypotenuse. The ratio that involves opposite and hypotenuse is the sine ratio. So I could say sine of the angle equals the opposite side, 8, divided by the hypotenuse, 17. Now all right triangles that have an opposite over hypotenuse ratio of 8 over 17 are similar to each other, which means they've all got the same angles, so we can solve for angle theta. Your calculator will be able to tell you if you use the inverse sine function, which is sometimes called arc sine, of 8 over 17. By using the inverse sine function, you're telling your calculator that you are inputting the ratio and looking for the angle, and your calculator would tell you that the angle is about 28.07 degrees. So that's how you use SOHCAHTOA to find a missing side or a missing angle. Must know number three, sine law and cosine law. Now, not all triangles you work with are going to have a 90 degree angle. If they don't, the only way you'll be able to solve for a missing side or angle is by using sine law or cosine law. Cosine law has two versions. One version if we're solving for a missing side, and one version that is just a rearranged version of it where we're solving for a missing angle. And then there's sine law, which just tells you the ratio of any side divided by sine of its opposite angle is equal. Cosine law has two scenarios where it's used, and sine law has two scenarios where it's used. So I'll show you each scenario of both of these. The scenario where we use cosine law to find a side length is when you know two sides and the angle contained by those two sides in a triangle. So notice in this triangle we know two sides and the angle contained by those two sides, so we could solve for the opposite angle using cosine law. If I follow the pattern of cosine law, it tells me that the unknown side squared equals the sum of the squares of the other two sides minus 2 times the product of the known sides times cosine of the angle contained by the two sides. That will tell you what x squared equals. To get x, make sure you square root whatever this value is. And in this question it's about 19.96. Let me shrink this and let's do another question. This rearranged version of cosine law is used to solve for an angle if you know all three sides of an oblique triangle. If I'm looking for this angle theta, when I use this cosine law, the important thing to notice is that the side that starts in the numerator is the one opposite from the angle you're looking for. So if I'm finding cosine of theta, the side opposite from theta is 13. So I must start with 13 squared, and then subtract the squares of the other two sides, and then divide by negative 2 times the second two sides, 14 and 16. This will give us what cosine theta is equal to. and when we know the ratio and want the angle, we would do inverse cosine of that ratio and it'll give us the angle, and we would get an angle of about 50.82 degrees. Sine law can be used when you know two sides and an angle opposite from one of the two sides. Based on sine law, I know that the ratio of 64 divided by sine of its opposite angle would be equal to the ratio of 42 divided by sine of its opposite angle. And then algebraically we could rearrange this to isolate sine theta. and then do inverse sine of that ratio and you would get the angle, which is about 60.93 degrees. The other scenario where you would use sine law is if you know two of the angles and one of the sides. In this triangle, I know that x divided by sine of its opposite angle, 47, would be equal to 5 divided by sine of its opposite angle, 51. If you rearrange to isolate x, you would find out that the value of x is about 4.71 units. And there you go, now you know how to use sine law and cosine law to solve for missing sides and angles of oblique triangles. Must know number four, the special triangles. There are two special triangles that are going to allow you to find exact values of sine, cos, and tan for a few special angles. Let me construct those triangles for you. The first one is an isosceles right triangle. If this triangle is isosceles, I know that the two legs of the right triangle are equal in length. I could assign them both a length of 1. And I know these two angles would have to be equal to each other, and since they both add to 90, I know they're both 45 degrees. If I use Pythagorean theorem to solve for the length of the hypotenuse, I would get the square root of 2. And then I could use these dimensions to figure out the sine, cos, and tan ratios for 45. Now not all right triangles with a reference angle of 45 are going to have lengths of 1, 1, and root 2, but their ratios of lengths would all be the same, because all the triangles, no matter how big or small, would have the same shape, meaning the same ratios. Let's assume that this 45 is our reference angle. That would make... this opposite, this adjacent, this hypotenuse. Let me write the three primary ratios for 45 degrees. Sine of 45 degrees is opposite over hypotenuse, so 1 over root 2. Cosine of 45 is adjacent over hypotenuse, which is also 1 over root 2. And tan of 45 is opposite over adjacent, which is 1 over 1, which is just 1. Now the sine and cosine ratios have a square root in their denominators. you should rationalize those denominators, and you can do that by multiplying top and bottom of both of those fractions by root 2. So those are the exact values for sine, cos, and tan of 45 degrees. The other special triangle is a half equilateral triangle. If this equilateral triangle started off with all side lengths being 2, and of course all the angles being equal, so they would all be 60. If I cut this equilateral triangle in half, that 60 degree angle that was at the top of this triangle gets bisected into two 30 degree angles, and the side at the bottom also gets bisected, so this length is 1. Now I don't actually need the left half of this triangle, so I'm going to erase it, and then I could use Pythagorean theorem to solve for that side, and I would find out its length is equal to the square root of 3. And now I notice there are two angles in here that I could find the exact values of their sine, cos, and tan ratios. If I start by finding the ratios for the 30 degree reference angle, So sine of 30 is opposite over hypotenuse, which is a half. Cos of 30 is adjacent over hypotenuse, which is root 3 over 2. And tan of 30 is opposite over adjacent, which is 1 over root 3. But if we rationalize that by multiplying top and bottom by root 3, you get root 3 over 3. Now if we make 60 the reference angle, sine of 60 would be opposite over hypotenuse, so root 3 over 2. Cos of 60 is adjacent over hypotenuse, so 1 over 2. and tan of 60 is opposite over adjacent which is root 3 over 1 which is just root 3. So there you go if you ever want a primary ratio for either 30, 45, or 60 you can use your special triangles to find the exact values of those ratios. Must know number five, cast rule and the unit circle. These tools are going to broaden your understanding of trigonometry and allow you to start using it for angles bigger than angles you would find inside of a triangle. Let's start with the unit circle. I've started by drawing a cartesian grid, and on this cartesian grid I'm going to label the positive x-axis 0 degrees, and then moving in a counterclockwise direction I'm going to label 90, 180, 270, and then coterminal with 0 degrees would be 360 degrees. The reason why I label them like that is because when we think of angles now, we're not going to think of them as angles in a triangle, but we're going to think of an initial arm, which starts on the positive x-axis, and then rotate some angle counterclockwise in a circle away from that positive x-axis. So if you rotate a 90 degrees, you would be up here, 180 here, 270 here, 360 back to where we started, which is why we say it's coterminal with zero degrees. Now, wherever that initial arm finishes, we call that its terminal arm. And if we know where that terminal arm intersects a circle that has a radius of 1, that actually tells us a lot about its sine, cos, and tan ratios. Let me show you by drawing a circle that has a radius of 1. And I should also mention that this unit circle whose radius is 1, it's centered at the origin. Now this yellow terminal arm has rotated some angle theta. So I'm going to label that angle theta, and then I'm going to construct a right angle triangle inside of this circle. This point that it intersects the unit circle at is some x, y point. If this is point x, y, then I know to get to that point, I would go x units that way, and then y units that way. So I could label this triangle side lengths of x and y. And because this is a unit circle, I know the radius of that circle is 1. Now if you remember Soh Cah Toa, you would be able to write the sine, cosine, and tan ratios for angle theta based on the side lengths of this triangle that we see inside of the unit circle. Sine of theta equals the opposite side, y, divided by the hypotenuse, 1, and y divided by 1 is just y. Cosine of theta equals the adjacent side divided by the hypotenuse, so x divided by 1, which is just x, and tan of theta is opposite over adjacent, so y divided by x. So the important thing you need to notice is the sine ratio of angle theta is equal to the y coordinate of where you intersect the unit circle. and cosine of theta is equal to the x coordinate of where the terminal arm intersects the unit circle. So if we know the point where the terminal arm intersects the unit circle, then we in fact know the cosine and sine ratios for angle theta. So I could say that point is equal to cos theta comma sine theta. And what points on this unit circle will we know the coordinates of? Well because the radius is 1, I know moving one unit in all those four directions, I could label four points on this unit circle. This point would be the point. Up here is the point. Here's and here's. So if I asked you to find sine of 90 degrees, you wouldn't have to ask a calculator. If you rotated 90 degrees, your terminal arm would intersect the unit circle at that point right there. and the sine ratio is equal to the y coordinate of where you intersect the unit circle, so sine of 90 is equal to 1. Let me test you with one more. What if I asked you to find cosine of 270 degrees? If you rotated 270 degrees, your terminal arm would intersect the unit circle at this point right here, and the cosine ratio is equal to the x coordinate, so cosine of 270 is equal to 0. Let's look at another tool which is very directly related to this unit circle. This tool is called the CAST rule, and it's just an acronym that helps you remember what ratios are positive in each quadrant. And for the CAST rule, you write C, A, S, T in each of those quadrants. I've also heard people say, all students take calculus to help you remember that acronym. But what these letters tell you is, in each of these quadrants, which ratios are positive. A stands for all. And then of course s is for sine, t is for tan, and c is for cosine. And then why does this make sense? Well, if we think back to the unit circle, I know that the y-coordinate of where you intersect the unit circle is the sine ratio. Where are y-coordinates positive? Above the x-axis in quadrant 1 and 2. That's why the cast rule tells us in quadrant 1, sine is positive, and in quadrant 2, sine is also positive. and where y coordinate's negative, that's below the x-axis in quadrant three and four. That's why in those quadrants, sine would be negative. Only tan is positive in quadrant three, and only cosine is positive in quadrant four. And cosine is equal to the x-coordinate of where you intersect the unit circle, and where x-coordinate is positive in quadrant one and four. And Castrol tells us in quadrant one cos is positive, in quadrant four cos is positive, but in quadrant two and three, Only sine and tan are positive, meaning cosine would be negative in both those quadrants. Hopefully you understand how the unit circle and cast rule can be useful, and in the next couple sections, I'll show you how we can actually use them. Must know number six. You need to be able to find the exact value of ratios for angles that are bigger than 90 degrees. To do that, you're going to need to remember your special triangles and the unit circle. Let me draw those quickly as a reminder. Let's say you were asked to find the exact value of the sine ratio of 150 degrees. The first thing you want to do is figure out where that principal angle of 150 degrees falls on a Cartesian grid. A principal angle of 150 degrees means from the positive x-axis rotate counterclockwise 150 degrees. And that would land you about right here. The angle from positive x counterclockwise to that terminal arm is 150. Now each principal angle... has a reference angle. The reference angle is the angle between the terminal arm and the closest x-axis. The reference angle is the angle between the terminal arm and the closest x-axis. The angle between 150 and 180 would be 30 degrees. So I could say that the reference angle is 30 degrees. This reference angle and the cast rule is going to help me find the sine ratio for 150. And as a reminder the cast rule tells me that in quadrant number two, where the terminal arm is, the sine ratio is positive. For that reason I know that sine of 150 is going to be equal to the positive sine ratio of the reference angle. So it'll be equal to sine of 30 degrees. And from my special triangle I know that sine of 30 Well, sine means opposite over hypotenuse. So sine of 30 is 1 over 2. So my final answer for sine of 150 is that it's equal to a half. Let me take a second and explain to you why sine of 150 is equal to sine of 30. If I were to draw a principal angle of 30 degrees, that would mean rotate 30 degrees counterclockwise from the positive x, and that will land you in quadrant number 1. The reference angle is 30 degrees. and if we look at where both of those terminal arms intersect the unit circle, if we look at the points where those terminal arms intersect the unit circle, notice that they have the exact same y coordinate. And if you remember, if we look over here, the y coordinate of where we intersect the unit circle is equal to the sine ratio of the angle. So of course sine of 150 and sine of 30 have the same value, because they intersect the unit circle at two points that have the same y value. and the y value of those points would of course be a half. Let me do one more example quickly for you. I'll shrink this to make room. Let's find cosine of 225 degrees. I'll start by drawing my Cartesian grid and figuring out where 225 degrees falls. If I were to rotate 225, I would go past 180 into quadrant number 3, so the principal angle is 225, and the reference angle, which is the angle between the terminal arm and the closest x-axis, So the angle between 225 and 180 is 45 degrees. And don't forget, based on cast rule, in this quadrant, only tan ratios are positive, which means any cosine ratio of an angle in that quadrant will have to be negative. So I can change the principal angle of 225 to the reference angle of 45 as long as I make the cosine ratio negative. So I can say this equals the negative cosine of the reference angle, 45 degrees. And cosine of 45 from your special triangle, you would notice cosine is adjacent over hypotenuse. So I would have negative 1 over root 2, which if I rationalize that would be negative root 2 over 2. That would be the exact value of cosine of 225. There you go. Now you know how to use your special triangles, the unit circle, and the cast rule to find exact values for trig ratios. Must know number 7. sine and cosine as functions. Let's start with sine. If we were to think of sine as a function, so y equals sine x, shows the relationship between x, which is our angle, and y, which is equal to the whole sine ratio of the angle. We could create a bunch of x-y pairs and graph and see what the function looks like. So that's what I'll do. I'll make a table of values where my independent variable x are going to be angles, and the y values are the sine ratios of those angles. The angles that I'm going to pick are at 90 degree intervals starting at 0. So 0, 90, 180, 270, 360. In order to figure out the sine ratios for each of these angles, it'll be helpful if you remember the unit circle. Based on the unit circle, hopefully you remember that the point where we intersect the unit circle The x-coordinate tells us the cosine ratio for the angle, and the y-coordinate tells us the sine ratio for the angle. So if I want sine of 0, I look if the terminal arm intersected the unit circle at 0 degrees, the y-coordinate would be 0. So sine of 0 is 0. And then just continue looking at the y-coordinates as you rotate around the unit circle, and you would figure out the sine ratios go 0, 1, 0, negative 1, 0. And if this terminal arm kept rotating around, and... we're going to have that same pattern of y values that just keeps repeating over and over again. That's why this is called a periodic function. It has a pattern of y values, which we see here, that just repeat over and over again. Let me graph that for you. If I plot the five points that I have in my table, I get one cycle of this sine function, and it looks like a wave that oscillates up and down between y values of negative one and one. If I were to continue this pattern, I would get another cycle of this function. The domain of this function is infinite, this pattern just keeps continuing forever in both directions. Let me now show you what a cosine function looks like. So I'm just going to shrink this so I have room. Let's graph the function y equals cosine x. where the x value is the angle, and the y value is the cosine ratio of that angle. I'll choose the same x values, and to get the y values, I can just look at the x coordinates as I rotate around the unit circle, since that's what the cosine ratio of the angle is equal to. So as I rotate around, my x values are 1, 0, negative 1, 0, and 1. And that pattern of y values would continue. This will just give me one cycle of the graph, and let me show you that. If I wanted another cycle... I would just continue the pattern. Let's talk about a couple key properties of these graphs. The two things we need to look at are things called the amplitude and period of these functions. The definition of amplitude is that it's half the distance between the max and min y values of a periodic function. Well for both sine and cos, the maximum y value is 1, and the minimum y value is negative 1. So the full distance between max and min is 2, half of that is 1, therefore the amplitude is 1. The period of a periodic function is the horizontal length of one cycle. If I map off one cycle of sine and one cycle of cosine, notice the horizontal length of both of those cycles is 360 degrees. There you go! Hopefully you now understand how we could think of sine and cosine as functions, and you have a general idea of what the graphs look like. Must know number eight, radians. Up until now, all the angles we've been working with were measured in degrees. But there's another unit of measurement we could think about for angles, and that is radians. Let me show you what a radian is. If I draw a circle, it'll help you visualize it. Let me draw an angle that subtends an arc length of this circle. In this circle that I've drawn, I have the radius here, and this right here is what we call the arc of the circle. If we were to measure this angle in degrees, we would just figure out how many... 360th of this circle this section occupies. But if we're going to measure it in radians, we're going to figure out how many radiuses have we moved around the arc of this circle. And how we figure out how many radiuses we've moved around the arc is we use the formula, the arc length divided by the radius. So whatever the ratio of this arc length divided by this radius is, that's the measure of that angle in radians. So what does one radian look like? Well, if this arc length was exactly equal to the length of this radius, that would be one radian, right? In the angle calculation, when we do arc over radius, if the arc and radius are equal, really what I'm doing is the radius divided by the radius, which is of course one. So how many radians are in a full circle? Well, do you remember the relationship between the circumference of a circle and its radius? The relationship is that the circumference equals 2 pi times the radius. So that tells us that the radius fits around the circumference of the circle 2 pi times. Therefore, there are 2 pi radians in a circle. So in a circle, we could say there's 360 degrees or 2 pi radians. Those are equivalent to each other. So let's pick a common degree to convert to radians. How about 30 degrees? If I want to convert it to radians, well, how many radians are in each degree? In this equation I made here, let's change this 360 degrees to 1 degree by dividing both sides by 360. If I do that, you'll see that 1 degree equals pi over 180 radians. To change 30 degrees to radians, I would have to multiply by pi over 180. 30 divided by 180 reduces to 1 over 6, so this would equal pi over 6 radians. So all of the trigonometry that we did up till now in degrees, we could easily convert all of it to radians by just multiplying any degree measure by pi over 180. Must know number nine, trig identities. First of all, what is a trig identity? A trig identity is an equation that is true for all values of the variable. And there are three fundamental identities that you should be familiar with. The reciprocal identities, the quotient identities, and of course the Pythagorean identity. The reciprocal identities are that cosecant of any angle is equal to the reciprocal of, so 1 over, sine of that same angle. Secant is the reciprocal of cosine, and cotangent is the reciprocal of tangent. The quotient identities tell us that tan x is equal to the quotient of sine x and cos x. And since cotangent is the reciprocal of tangent, that must mean that cotangent is the reciprocal of sine over cos, which is cos over sine. And the Pythagorean identity is that sine squared of x plus cos squared of x equals 1. And there are common ways to rearrange that. Now any of these equations are true no matter what value of x you sub in, so sub in anything you want for these x's into sine squared x plus cos squared x, and it's always going to equal 1. Now what we can do with these identities is prove other identities. So let me make some room and then we'll prove an identity. Let's see if we can prove that tan squared x minus sine squared x is equal to sine squared x times tan squared x. If this is an identity, it'll be true for all values of the variable. So how we prove that is separating into left side and right side of the equation, and work with both of these sides completely separately from each other. So I'll draw a line down the middle and start rewriting these expressions in equivalent ways to reveal that they are in fact the exact same expression. So I'm not allowed to change the value of either of them. What I could do on the left, I know that based on the quotient identity, tan of any angle equals the quotient of sine over cos, so I can change tan squared x to sine squared x over cos squared x. And then I'm subtracting sine squared x, but remember that sine squared x is technically over 1. I can multiply top and bottom of that by cos squared x to get a common denominator. I haven't changed its value, I've just multiplied it by 1. So now I have a common denominator of cos squared x with both of those terms, so I can combine those into a single fraction. In the numerator, I have two terms. and both of those terms have a sine squared x, so I could common factor that out. And now I have a quotient of sine squared x and cos squared x based on the quotient identity. I know that that's equal to tan squared x, and that's being multiplied by a 1 minus cos squared x, which based on the Pythagorean identity I know is equal to sine squared x. And now notice, that's the exact same expression as I have on the right side of the equation. So the two sides of this equation are actually the exact same expression. they're exactly equivalent to each other for all values of x. So we've proven the identity, we can say left side equals right side or QED. Must know number 10, solving trig equations. I'm going to give you two trig equations to solve. The first one is sine x equals negative one over root two. And for this one, I only want the solutions between zero and two pi. which means only the solutions in the first cycle of this function. Since sine is a periodic function, there's going to be an infinite number of times it's equal to negative one over root two, so by restricting the domain, I'm only asking you for two of the answers in this case. So let's go ahead and find those two answers. You're going to have to remember your special triangles, so let me draw those off to the side. When solving a trig equation, I like to draw a Cartesian grid, and based on cast rule, I like to figure out where my answers are going to be. Since the sine ratio is negative, I know sine ratios are negative in two quadrants, in quadrant 3 and in quadrant 4. So I'll place a terminal arm in both of those quadrants. In order for the angles that get to these terminal arms to have the same sine ratio, I know that those terminal arms must have the same reference angle. And what is the reference angle that creates a ratio of 1 over root 2? Well, sine of pi over 4 equals opposite over hypotenuse 1 over root 2. So I know the reference angle for both of these terminal arms is pi over 4. and now I just have to calculate the two angles to get to those terminal arms. So my first angle, I'll call this x1, I'm looking for the angle that goes from positive x counterclockwise all the way to that terminal arm. Well notice, it went past pi by pi over 4. So I could say that x1 is equal to pi plus pi over 4, which is 5 pi over 4. And to get to x2, I would have to go not a full 2 pi, I would have to go pi over 4 less than that. So I can do 2 pi minus pi over 4 to get my second principal angle. And 2 pi minus pi over 4 is 7 pi over 4. So there are two answers to this equation between 0 and 2 pi. The answers are 5 pi over 4 and 7 pi over 4. Let's do a second equation where I don't restrict the domain. Let's say we have 2 sine squared x minus 3 sine x plus 1 equals 0. I would start by factoring this quadratic expression by finding numbers that have a product of 2 times 1, so a product of 2, and a sum of negative 3. The numbers that satisfy that product and sum are negative 2 and negative 1. There's lots of factoring methods you could use. I'll just split the middle term of this quadratic into negative 2 sine x minus sine x. And now I would factor by grouping. I would take a 2 sine x from the first two terms and take a negative 1 from the last two terms. I can take out this common binomial of sine x minus 1. And after I take that out, I'm left with 2 sine x minus 1 as my second factor. Now this product would be 0 if either of these factors were 0. The first factor would be 0 if sine x was equal to 1, and the second factor would be 0 if sine x was equal to a half. Now sine x equaling 1, if you know your unit circle, that answer will be easy. If you rotate an angle of pi over 2, it intersects the unit circle at a y coordinate of 1, and I know the sine ratio is equal to the y coordinate, so the angle that would make the sine ratio be 1 is pi over 2. So the first answer to this equation is pi over 2. Now because the domain of this equation has not been restricted, I know that I could continue rotating around this circle, and every two pi radians I rotate around, I'm going to be back intersecting at that same point. So how I could communicate that is I could say that I can add 2 pi k times where k has to be an integer. So there's a set of answers that come from the first factor being 0. Now how can this second factor be 0? Let me draw a quick Cartesian grid. I have a positive sign ratio. Well, sign ratios are positive in quadrant 1 and 2, so there'll be a terminal arm in both of those quadrants. Both of those terminal arms have the same reference angle, and their reference angle is going to be pi over 6. and I know that because sine of pi over 6 is 1 over 2, opposite over hypotenuse. So I'll place the reference angle of pi over 6 in both of those quadrants. And now I have to calculate the principal angle to get to each of those terminal arms. In quadrant one, the principal angle is just equal to the reference angle, so it's pi over 6. And my second terminal arm, which is my third answer for x overall, to get to there, I would go pi over 6 less than pi. Pi minus pi over 6 is 5pi over 6. and from either of those terminal arms you could rotate 2 pi radians and be back to the same spot. So to communicate that, I would say to either of these answers I could add 2 pi k where k has to be an integer. So these three sets of x values give me all the solutions to this original equation. Hopefully this video helped you gain more confidence in your understanding of trigonometry. Let me know about what you want a top 10 of next.

Let me draw two triangles so you can see what I mean. Now I've started by drawing these two triangles the exact same size and shape, so they're congruent. If I were to take triangle DEF and make it smaller, without changing any of its angles, so keep the shape the exact same, it is still similar to triangle ABC.

And similar triangles have two very important properties. Their pairs of corresponding angles are equal. So in these two triangles, you can see that angle A and D are equal to each other, B and E are equal, and C and F are equal.

The other important property is that ratios of corresponding sides are also equal. So the ratio of side AB to side DE would be equal to the ratio of side AC to DF, which would be equal to the ratio of side BC to EF. And we can use these properties to solve for missing sides and angles in triangles that are similar to each other. Let me show you an example.

In this question, we're asked to solve for the length of X. In order to do that, I'll have to use these two similar triangles that I see, the big triangle and the smaller triangle within it. We'll have to start by proving they're similar, which there are three ways to prove similarity.

You can do angle-angle similarity, just find two pairs of equal angles. Side-side-side similarity, where you show that the ratios of all three pairs of sides are equal. Or side-angle-side similarity, where you find two equal ratios of pairs of sides, and one matching pair of angles.

For this question we can do angle-angle similarity. If you know your angle theorems, we have two parallel lines here, connected by this side. I know that the angles inside of that F pattern called corresponding angles are equal to each other. So I could say that angle D is equal to angle B. Same thing on the other side of the triangle.

I know that these angles in here are equal to each other. So angle E equals angle C. And then of course, the two triangles both share angle A. So based on all three pairs of angles being equal, I know the triangles are the same shape. So I could say that triangle DEA is similar to triangle BCA.

And because I know that they're similar, I know that the ratios of corresponding sides have to be equal to each other. So I know that the length of BC divided by the length of DE is going to be equal to the length of BA divided by the length of DA. And then if I sub into this equation what I know, I can then solve this equation using algebra.

I can cross multiply, 10 times 6 is 60, and 4 times x plus 6 is 4x plus 24. Solving this for x, I'd have 36 equals 4x, divide the 4, I have x equals 9. There you go, now you can see how to use properties of similar triangles to solve for unknowns. Must know number two, SOHCAHTOA. This is an acronym that helps you remember the three primary trig ratios. Let me draw a right triangle.

In this right triangle, I've indicated the 90 degree angle, and then I've labeled one of the acute angles. angle theta. We'll call that the reference angle.

All other right triangles with this same reference angle theta would be similar to each other, right? No matter how much I change the size of this triangle, as long as I keep that angle theta the same, it has the exact same shape, meaning the triangles are similar, which means based on what we know about similar triangles, they would have the same ratios of corresponding sides. In any right triangle, there are three sides.

Let's name them. The longest side of a right triangle is always called the hypotenuse, I'll label that h. The side across from our reference angle theta, we'll call that the opposite side, I'll call it o. And the side touching our reference angle theta, we'll call that the adjacent side because it's right beside our reference angle, I'll label that a. Because there are three sides, there are three different pairs we could make of these three sides.

That's why you see three ratios. in SOHCAHTOA. If we find the ratio of the side opposite from the reference angle divided by the hypotenuse, we call that the sine ratio.

Cosine of the reference angle is the adjacent side divided by the hypotenuse, and tan is a function that means we're looking at the ratio of the opposite side divided by the adjacent side. Now these primary ratios can be used to find a missing side or a missing angle of a right triangle. Let me show you both examples.

Let's look at this first example. where we're going to solve for a missing side, X. I notice it's a right triangle with a reference angle of 40. All right triangles with a reference angle of 40 are similar to each other.

They all have the same ratios of sides. So whatever x divided by 8 is, will be the same ratio of all other right triangles with a reference angle of 40. So let's communicate that. From 40, the side we're looking for is adjacent to it, and 8 is the longest side of the right triangle, so that's the hypotenuse. The function that involves adjacent and hypotenuse is cosine.

So I know cosine of 40 degrees equals the adjacent side, x, divided by the hypotenuse, 8. Your calculator has stored within it what the ratio of adjacent over hypotenuse is for all right triangles with a reference angle of 40. If you typed cosine of 40 degrees on your calculator, it'll give you that ratio. And then we can just multiply the 8 to the other side of that ratio to isolate x. SOHCAHTOA can also be used to find a missing angle inside of a right triangle.

If we're looking for this angle here, notice that 8 is opposite from it. and 17 is a cross from the right angle, so that's the hypotenuse. The ratio that involves opposite and hypotenuse is the sine ratio.

So I could say sine of the angle equals the opposite side, 8, divided by the hypotenuse, 17. Now all right triangles that have an opposite over hypotenuse ratio of 8 over 17 are similar to each other, which means they've all got the same angles, so we can solve for angle theta. Your calculator will be able to tell you if you use the inverse sine function, which is sometimes called arc sine, of 8 over 17. By using the inverse sine function, you're telling your calculator that you are inputting the ratio and looking for the angle, and your calculator would tell you that the angle is about 28.07 degrees. So that's how you use SOHCAHTOA to find a missing side or a missing angle. Must know number three, sine law and cosine law. Now, not all triangles you work with are going to have a 90 degree angle.

If they don't, the only way you'll be able to solve for a missing side or angle is by using sine law or cosine law. Cosine law has two versions. One version if we're solving for a missing side, and one version that is just a rearranged version of it where we're solving for a missing angle. And then there's sine law, which just tells you the ratio of any side divided by sine of its opposite angle is equal.

Cosine law has two scenarios where it's used, and sine law has two scenarios where it's used. So I'll show you each scenario of both of these. The scenario where we use cosine law to find a side length is when you know two sides and the angle contained by those two sides in a triangle.

So notice in this triangle we know two sides and the angle contained by those two sides, so we could solve for the opposite angle using cosine law. If I follow the pattern of cosine law, it tells me that the unknown side squared equals the sum of the squares of the other two sides minus 2 times the product of the known sides times cosine of the angle contained by the two sides. That will tell you what x squared equals.

To get x, make sure you square root whatever this value is. And in this question it's about 19.96. Let me shrink this and let's do another question.

This rearranged version of cosine law is used to solve for an angle if you know all three sides of an oblique triangle. If I'm looking for this angle theta, when I use this cosine law, the important thing to notice is that the side that starts in the numerator is the one opposite from the angle you're looking for. So if I'm finding cosine of theta, the side opposite from theta is 13. So I must start with 13 squared, and then subtract the squares of the other two sides, and then divide by negative 2 times the second two sides, 14 and 16. This will give us what cosine theta is equal to.

and when we know the ratio and want the angle, we would do inverse cosine of that ratio and it'll give us the angle, and we would get an angle of about 50.82 degrees. Sine law can be used when you know two sides and an angle opposite from one of the two sides. Based on sine law, I know that the ratio of 64 divided by sine of its opposite angle would be equal to the ratio of 42 divided by sine of its opposite angle.

And then algebraically we could rearrange this to isolate sine theta. and then do inverse sine of that ratio and you would get the angle, which is about 60.93 degrees. The other scenario where you would use sine law is if you know two of the angles and one of the sides. In this triangle, I know that x divided by sine of its opposite angle, 47, would be equal to 5 divided by sine of its opposite angle, 51. If you rearrange to isolate x, you would find out that the value of x is about 4.71 units. And there you go, now you know how to use sine law and cosine law to solve for missing sides and angles of oblique triangles.

Must know number four, the special triangles. There are two special triangles that are going to allow you to find exact values of sine, cos, and tan for a few special angles. Let me construct those triangles for you. The first one is an isosceles right triangle. If this triangle is isosceles, I know that the two legs of the right triangle are equal in length.

I could assign them both a length of 1. And I know these two angles would have to be equal to each other, and since they both add to 90, I know they're both 45 degrees. If I use Pythagorean theorem to solve for the length of the hypotenuse, I would get the square root of 2. And then I could use these dimensions to figure out the sine, cos, and tan ratios for 45. Now not all right triangles with a reference angle of 45 are going to have lengths of 1, 1, and root 2, but their ratios of lengths would all be the same, because all the triangles, no matter how big or small, would have the same shape, meaning the same ratios. Let's assume that this 45 is our reference angle.

That would make... this opposite, this adjacent, this hypotenuse. Let me write the three primary ratios for 45 degrees. Sine of 45 degrees is opposite over hypotenuse, so 1 over root 2. Cosine of 45 is adjacent over hypotenuse, which is also 1 over root 2. And tan of 45 is opposite over adjacent, which is 1 over 1, which is just 1. Now the sine and cosine ratios have a square root in their denominators.

you should rationalize those denominators, and you can do that by multiplying top and bottom of both of those fractions by root 2. So those are the exact values for sine, cos, and tan of 45 degrees. The other special triangle is a half equilateral triangle. If this equilateral triangle started off with all side lengths being 2, and of course all the angles being equal, so they would all be 60. If I cut this equilateral triangle in half, that 60 degree angle that was at the top of this triangle gets bisected into two 30 degree angles, and the side at the bottom also gets bisected, so this length is 1. Now I don't actually need the left half of this triangle, so I'm going to erase it, and then I could use Pythagorean theorem to solve for that side, and I would find out its length is equal to the square root of 3. And now I notice there are two angles in here that I could find the exact values of their sine, cos, and tan ratios.

If I start by finding the ratios for the 30 degree reference angle, So sine of 30 is opposite over hypotenuse, which is a half. Cos of 30 is adjacent over hypotenuse, which is root 3 over 2. And tan of 30 is opposite over adjacent, which is 1 over root 3. But if we rationalize that by multiplying top and bottom by root 3, you get root 3 over 3. Now if we make 60 the reference angle, sine of 60 would be opposite over hypotenuse, so root 3 over 2. Cos of 60 is adjacent over hypotenuse, so 1 over 2. and tan of 60 is opposite over adjacent which is root 3 over 1 which is just root 3. So there you go if you ever want a primary ratio for either 30, 45, or 60 you can use your special triangles to find the exact values of those ratios. Must know number five, cast rule and the unit circle. These tools are going to broaden your understanding of trigonometry and allow you to start using it for angles bigger than angles you would find inside of a triangle. Let's start with the unit circle.

I've started by drawing a cartesian grid, and on this cartesian grid I'm going to label the positive x-axis 0 degrees, and then moving in a counterclockwise direction I'm going to label 90, 180, 270, and then coterminal with 0 degrees would be 360 degrees. The reason why I label them like that is because when we think of angles now, we're not going to think of them as angles in a triangle, but we're going to think of an initial arm, which starts on the positive x-axis, and then rotate some angle counterclockwise in a circle away from that positive x-axis. So if you rotate a 90 degrees, you would be up here, 180 here, 270 here, 360 back to where we started, which is why we say it's coterminal with zero degrees.

Now, wherever that initial arm finishes, we call that its terminal arm. And if we know where that terminal arm intersects a circle that has a radius of 1, that actually tells us a lot about its sine, cos, and tan ratios. Let me show you by drawing a circle that has a radius of 1. And I should also mention that this unit circle whose radius is 1, it's centered at the origin. Now this yellow terminal arm has rotated some angle theta. So I'm going to label that angle theta, and then I'm going to construct a right angle triangle inside of this circle.

This point that it intersects the unit circle at is some x, y point. If this is point x, y, then I know to get to that point, I would go x units that way, and then y units that way. So I could label this triangle side lengths of x and y.

And because this is a unit circle, I know the radius of that circle is 1. Now if you remember Soh Cah Toa, you would be able to write the sine, cosine, and tan ratios for angle theta based on the side lengths of this triangle that we see inside of the unit circle. Sine of theta equals the opposite side, y, divided by the hypotenuse, 1, and y divided by 1 is just y. Cosine of theta equals the adjacent side divided by the hypotenuse, so x divided by 1, which is just x, and tan of theta is opposite over adjacent, so y divided by x. So the important thing you need to notice is the sine ratio of angle theta is equal to the y coordinate of where you intersect the unit circle.

and cosine of theta is equal to the x coordinate of where the terminal arm intersects the unit circle. So if we know the point where the terminal arm intersects the unit circle, then we in fact know the cosine and sine ratios for angle theta. So I could say that point is equal to cos theta comma sine theta. And what points on this unit circle will we know the coordinates of? Well because the radius is 1, I know moving one unit in all those four directions, I could label four points on this unit circle.

This point would be the point. Up here is the point. Here's and here's.

So if I asked you to find sine of 90 degrees, you wouldn't have to ask a calculator. If you rotated 90 degrees, your terminal arm would intersect the unit circle at that point right there. and the sine ratio is equal to the y coordinate of where you intersect the unit circle, so sine of 90 is equal to 1. Let me test you with one more.

What if I asked you to find cosine of 270 degrees? If you rotated 270 degrees, your terminal arm would intersect the unit circle at this point right here, and the cosine ratio is equal to the x coordinate, so cosine of 270 is equal to 0. Let's look at another tool which is very directly related to this unit circle. This tool is called the CAST rule, and it's just an acronym that helps you remember what ratios are positive in each quadrant.

And for the CAST rule, you write C, A, S, T in each of those quadrants. I've also heard people say, all students take calculus to help you remember that acronym. But what these letters tell you is, in each of these quadrants, which ratios are positive. A stands for all.

And then of course s is for sine, t is for tan, and c is for cosine. And then why does this make sense? Well, if we think back to the unit circle, I know that the y-coordinate of where you intersect the unit circle is the sine ratio.

Where are y-coordinates positive? Above the x-axis in quadrant 1 and 2. That's why the cast rule tells us in quadrant 1, sine is positive, and in quadrant 2, sine is also positive. and where y coordinate's negative, that's below the x-axis in quadrant three and four. That's why in those quadrants, sine would be negative. Only tan is positive in quadrant three, and only cosine is positive in quadrant four.

And cosine is equal to the x-coordinate of where you intersect the unit circle, and where x-coordinate is positive in quadrant one and four. And Castrol tells us in quadrant one cos is positive, in quadrant four cos is positive, but in quadrant two and three, Only sine and tan are positive, meaning cosine would be negative in both those quadrants. Hopefully you understand how the unit circle and cast rule can be useful, and in the next couple sections, I'll show you how we can actually use them.

Must know number six. You need to be able to find the exact value of ratios for angles that are bigger than 90 degrees. To do that, you're going to need to remember your special triangles and the unit circle. Let me draw those quickly as a reminder. Let's say you were asked to find the exact value of the sine ratio of 150 degrees.

The first thing you want to do is figure out where that principal angle of 150 degrees falls on a Cartesian grid. A principal angle of 150 degrees means from the positive x-axis rotate counterclockwise 150 degrees. And that would land you about right here.

The angle from positive x counterclockwise to that terminal arm is 150. Now each principal angle... has a reference angle. The reference angle is the angle between the terminal arm and the closest x-axis. The reference angle is the angle between the terminal arm and the closest x-axis.

The angle between 150 and 180 would be 30 degrees. So I could say that the reference angle is 30 degrees. This reference angle and the cast rule is going to help me find the sine ratio for 150. And as a reminder the cast rule tells me that in quadrant number two, where the terminal arm is, the sine ratio is positive. For that reason I know that sine of 150 is going to be equal to the positive sine ratio of the reference angle. So it'll be equal to sine of 30 degrees.

And from my special triangle I know that sine of 30 Well, sine means opposite over hypotenuse. So sine of 30 is 1 over 2. So my final answer for sine of 150 is that it's equal to a half. Let me take a second and explain to you why sine of 150 is equal to sine of 30. If I were to draw a principal angle of 30 degrees, that would mean rotate 30 degrees counterclockwise from the positive x, and that will land you in quadrant number 1. The reference angle is 30 degrees.

and if we look at where both of those terminal arms intersect the unit circle, if we look at the points where those terminal arms intersect the unit circle, notice that they have the exact same y coordinate. And if you remember, if we look over here, the y coordinate of where we intersect the unit circle is equal to the sine ratio of the angle. So of course sine of 150 and sine of 30 have the same value, because they intersect the unit circle at two points that have the same y value. and the y value of those points would of course be a half.

Let me do one more example quickly for you. I'll shrink this to make room. Let's find cosine of 225 degrees. I'll start by drawing my Cartesian grid and figuring out where 225 degrees falls. If I were to rotate 225, I would go past 180 into quadrant number 3, so the principal angle is 225, and the reference angle, which is the angle between the terminal arm and the closest x-axis, So the angle between 225 and 180 is 45 degrees.

And don't forget, based on cast rule, in this quadrant, only tan ratios are positive, which means any cosine ratio of an angle in that quadrant will have to be negative. So I can change the principal angle of 225 to the reference angle of 45 as long as I make the cosine ratio negative. So I can say this equals the negative cosine of the reference angle, 45 degrees.

And cosine of 45 from your special triangle, you would notice cosine is adjacent over hypotenuse. So I would have negative 1 over root 2, which if I rationalize that would be negative root 2 over 2. That would be the exact value of cosine of 225. There you go. Now you know how to use your special triangles, the unit circle, and the cast rule to find exact values for trig ratios.

Must know number 7. sine and cosine as functions. Let's start with sine. If we were to think of sine as a function, so y equals sine x, shows the relationship between x, which is our angle, and y, which is equal to the whole sine ratio of the angle.

We could create a bunch of x-y pairs and graph and see what the function looks like. So that's what I'll do. I'll make a table of values where my independent variable x are going to be angles, and the y values are the sine ratios of those angles.

The angles that I'm going to pick are at 90 degree intervals starting at 0. So 0, 90, 180, 270, 360. In order to figure out the sine ratios for each of these angles, it'll be helpful if you remember the unit circle. Based on the unit circle, hopefully you remember that the point where we intersect the unit circle The x-coordinate tells us the cosine ratio for the angle, and the y-coordinate tells us the sine ratio for the angle. So if I want sine of 0, I look if the terminal arm intersected the unit circle at 0 degrees, the y-coordinate would be 0. So sine of 0 is 0. And then just continue looking at the y-coordinates as you rotate around the unit circle, and you would figure out the sine ratios go 0, 1, 0, negative 1, 0. And if this terminal arm kept rotating around, and... we're going to have that same pattern of y values that just keeps repeating over and over again. That's why this is called a periodic function.

It has a pattern of y values, which we see here, that just repeat over and over again. Let me graph that for you. If I plot the five points that I have in my table, I get one cycle of this sine function, and it looks like a wave that oscillates up and down between y values of negative one and one. If I were to continue this pattern, I would get another cycle of this function. The domain of this function is infinite, this pattern just keeps continuing forever in both directions.

Let me now show you what a cosine function looks like. So I'm just going to shrink this so I have room. Let's graph the function y equals cosine x. where the x value is the angle, and the y value is the cosine ratio of that angle. I'll choose the same x values, and to get the y values, I can just look at the x coordinates as I rotate around the unit circle, since that's what the cosine ratio of the angle is equal to.

So as I rotate around, my x values are 1, 0, negative 1, 0, and 1. And that pattern of y values would continue. This will just give me one cycle of the graph, and let me show you that. If I wanted another cycle... I would just continue the pattern. Let's talk about a couple key properties of these graphs.

The two things we need to look at are things called the amplitude and period of these functions. The definition of amplitude is that it's half the distance between the max and min y values of a periodic function. Well for both sine and cos, the maximum y value is 1, and the minimum y value is negative 1. So the full distance between max and min is 2, half of that is 1, therefore the amplitude is 1. The period of a periodic function is the horizontal length of one cycle. If I map off one cycle of sine and one cycle of cosine, notice the horizontal length of both of those cycles is 360 degrees.

There you go! Hopefully you now understand how we could think of sine and cosine as functions, and you have a general idea of what the graphs look like. Must know number eight, radians.

Up until now, all the angles we've been working with were measured in degrees. But there's another unit of measurement we could think about for angles, and that is radians. Let me show you what a radian is. If I draw a circle, it'll help you visualize it.

Let me draw an angle that subtends an arc length of this circle. In this circle that I've drawn, I have the radius here, and this right here is what we call the arc of the circle. If we were to measure this angle in degrees, we would just figure out how many...

360th of this circle this section occupies. But if we're going to measure it in radians, we're going to figure out how many radiuses have we moved around the arc of this circle. And how we figure out how many radiuses we've moved around the arc is we use the formula, the arc length divided by the radius.

So whatever the ratio of this arc length divided by this radius is, that's the measure of that angle in radians. So what does one radian look like? Well, if this arc length was exactly equal to the length of this radius, that would be one radian, right?

In the angle calculation, when we do arc over radius, if the arc and radius are equal, really what I'm doing is the radius divided by the radius, which is of course one. So how many radians are in a full circle? Well, do you remember the relationship between the circumference of a circle and its radius?

The relationship is that the circumference equals 2 pi times the radius. So that tells us that the radius fits around the circumference of the circle 2 pi times. Therefore, there are 2 pi radians in a circle. So in a circle, we could say there's 360 degrees or 2 pi radians.

Those are equivalent to each other. So let's pick a common degree to convert to radians. How about 30 degrees? If I want to convert it to radians, well, how many radians are in each degree? In this equation I made here, let's change this 360 degrees to 1 degree by dividing both sides by 360. If I do that, you'll see that 1 degree equals pi over 180 radians.

To change 30 degrees to radians, I would have to multiply by pi over 180. 30 divided by 180 reduces to 1 over 6, so this would equal pi over 6 radians. So all of the trigonometry that we did up till now in degrees, we could easily convert all of it to radians by just multiplying any degree measure by pi over 180. Must know number nine, trig identities. First of all, what is a trig identity? A trig identity is an equation that is true for all values of the variable.

And there are three fundamental identities that you should be familiar with. The reciprocal identities, the quotient identities, and of course the Pythagorean identity. The reciprocal identities are that cosecant of any angle is equal to the reciprocal of, so 1 over, sine of that same angle.

Secant is the reciprocal of cosine, and cotangent is the reciprocal of tangent. The quotient identities tell us that tan x is equal to the quotient of sine x and cos x. And since cotangent is the reciprocal of tangent, that must mean that cotangent is the reciprocal of sine over cos, which is cos over sine. And the Pythagorean identity is that sine squared of x plus cos squared of x equals 1. And there are common ways to rearrange that.

Now any of these equations are true no matter what value of x you sub in, so sub in anything you want for these x's into sine squared x plus cos squared x, and it's always going to equal 1. Now what we can do with these identities is prove other identities. So let me make some room and then we'll prove an identity. Let's see if we can prove that tan squared x minus sine squared x is equal to sine squared x times tan squared x. If this is an identity, it'll be true for all values of the variable.

So how we prove that is separating into left side and right side of the equation, and work with both of these sides completely separately from each other. So I'll draw a line down the middle and start rewriting these expressions in equivalent ways to reveal that they are in fact the exact same expression. So I'm not allowed to change the value of either of them. What I could do on the left, I know that based on the quotient identity, tan of any angle equals the quotient of sine over cos, so I can change tan squared x to sine squared x over cos squared x. And then I'm subtracting sine squared x, but remember that sine squared x is technically over 1. I can multiply top and bottom of that by cos squared x to get a common denominator.

I haven't changed its value, I've just multiplied it by 1. So now I have a common denominator of cos squared x with both of those terms, so I can combine those into a single fraction. In the numerator, I have two terms. and both of those terms have a sine squared x, so I could common factor that out. And now I have a quotient of sine squared x and cos squared x based on the quotient identity.

I know that that's equal to tan squared x, and that's being multiplied by a 1 minus cos squared x, which based on the Pythagorean identity I know is equal to sine squared x. And now notice, that's the exact same expression as I have on the right side of the equation. So the two sides of this equation are actually the exact same expression.

they're exactly equivalent to each other for all values of x. So we've proven the identity, we can say left side equals right side or QED. Must know number 10, solving trig equations.

I'm going to give you two trig equations to solve. The first one is sine x equals negative one over root two. And for this one, I only want the solutions between zero and two pi. which means only the solutions in the first cycle of this function.

Since sine is a periodic function, there's going to be an infinite number of times it's equal to negative one over root two, so by restricting the domain, I'm only asking you for two of the answers in this case. So let's go ahead and find those two answers. You're going to have to remember your special triangles, so let me draw those off to the side.

When solving a trig equation, I like to draw a Cartesian grid, and based on cast rule, I like to figure out where my answers are going to be. Since the sine ratio is negative, I know sine ratios are negative in two quadrants, in quadrant 3 and in quadrant 4. So I'll place a terminal arm in both of those quadrants. In order for the angles that get to these terminal arms to have the same sine ratio, I know that those terminal arms must have the same reference angle. And what is the reference angle that creates a ratio of 1 over root 2? Well, sine of pi over 4 equals opposite over hypotenuse 1 over root 2. So I know the reference angle for both of these terminal arms is pi over 4. and now I just have to calculate the two angles to get to those terminal arms.

So my first angle, I'll call this x1, I'm looking for the angle that goes from positive x counterclockwise all the way to that terminal arm. Well notice, it went past pi by pi over 4. So I could say that x1 is equal to pi plus pi over 4, which is 5 pi over 4. And to get to x2, I would have to go not a full 2 pi, I would have to go pi over 4 less than that. So I can do 2 pi minus pi over 4 to get my second principal angle. And 2 pi minus pi over 4 is 7 pi over 4. So there are two answers to this equation between 0 and 2 pi.

The answers are 5 pi over 4 and 7 pi over 4. Let's do a second equation where I don't restrict the domain. Let's say we have 2 sine squared x minus 3 sine x plus 1 equals 0. I would start by factoring this quadratic expression by finding numbers that have a product of 2 times 1, so a product of 2, and a sum of negative 3. The numbers that satisfy that product and sum are negative 2 and negative 1. There's lots of factoring methods you could use. I'll just split the middle term of this quadratic into negative 2 sine x minus sine x.

And now I would factor by grouping. I would take a 2 sine x from the first two terms and take a negative 1 from the last two terms. I can take out this common binomial of sine x minus 1. And after I take that out, I'm left with 2 sine x minus 1 as my second factor.

Now this product would be 0 if either of these factors were 0. The first factor would be 0 if sine x was equal to 1, and the second factor would be 0 if sine x was equal to a half. Now sine x equaling 1, if you know your unit circle, that answer will be easy. If you rotate an angle of pi over 2, it intersects the unit circle at a y coordinate of 1, and I know the sine ratio is equal to the y coordinate, so the angle that would make the sine ratio be 1 is pi over 2. So the first answer to this equation is pi over 2. Now because the domain of this equation has not been restricted, I know that I could continue rotating around this circle, and every two pi radians I rotate around, I'm going to be back intersecting at that same point. So how I could communicate that is I could say that I can add 2 pi k times where k has to be an integer.

So there's a set of answers that come from the first factor being 0. Now how can this second factor be 0? Let me draw a quick Cartesian grid. I have a positive sign ratio.

Well, sign ratios are positive in quadrant 1 and 2, so there'll be a terminal arm in both of those quadrants. Both of those terminal arms have the same reference angle, and their reference angle is going to be pi over 6. and I know that because sine of pi over 6 is 1 over 2, opposite over hypotenuse. So I'll place the reference angle of pi over 6 in both of those quadrants.

And now I have to calculate the principal angle to get to each of those terminal arms. In quadrant one, the principal angle is just equal to the reference angle, so it's pi over 6. And my second terminal arm, which is my third answer for x overall, to get to there, I would go pi over 6 less than pi. Pi minus pi over 6 is 5pi over 6. and from either of those terminal arms you could rotate 2 pi radians and be back to the same spot.

So to communicate that, I would say to either of these answers I could add 2 pi k where k has to be an integer. So these three sets of x values give me all the solutions to this original equation. Hopefully this video helped you gain more confidence in your understanding of trigonometry. Let me know about what you want a top 10 of next.

Transcript for:Essential Trigonometry Concepts to Know

Transcript for:
Essential Trigonometry Concepts to Know