 the video introducing you to the various trigonometric functions and how they should be applied at the various angles. There are six trigonometric functions total. There is sine, cosine, tangent, cosecant, secant, and cotangent. The abbreviations for each of these can be seen in the parentheses. This is often how we will write these trig functions with their abbreviations. Now, consider a right triangle. A right triangle has a right angle in it. The side across from the right angle is called the hypotenuse. So I'll label side r as the hypotenuse. And then the two sides that meet the form of the right angle, side x and side y, those are called legs. So y is a leg, x is a leg. What we do with the trigonometric functions, sine, cosine, tangent, cosecant, secant, and cotangent, is we apply them to an angle. In this case, we call that angle theta. Typically, we will use Greek letters to represent angles. So theta is just a generic angle. Sine, sine of theta. Sine of theta means you take the opposite leg and put it over the hypotenuse. So that would be y over r. Cosine of theta. Cosine. Think about co-workers, the word co. You want the side that's adjacent that's closest to the angle, the leg, x. And you want to put it over the hypotenuse r. Adjacent over hypotenuse x over r. And tangent is opposite over adjacent. So you go to your angle theta, opposite is y, adjacent is x. The hypotenuse is not used here. Now there is a chant that can help you remember sine, cosine, and tangent. And that chant is actually sa, ka, toa. What does this mean? Well, s is for sine. Sine is opposite. Opposite over hypotenuse. Ka. C is for cosine. Cosine is adjacent over hypotenuse. And then toa. T is for tangent. O is opposite. A is adjacent. Because tangent is opposite over adjacent. So you can use this chant to help you remember these first three trig functions. Now if you know the first three trig functions, it might be a little bit easier to do these last three trig functions. Cosecant. Cosecant and sine are reciprocals of each other, meaning they are flipped versions. If sine is opposite over hypotenuse, then cosecant is hypotenuse over opposite. So instead of y over r, it'll be r over y. Secant is the reciprocal of cosine. So you have this sc pairing that's happening here. Secant, cosine are reciprocals of each other. Cosine is adjacent over hypotenuse. Secant is hypotenuse over adjacent. We have x over r. Well, now we're going to have r over x. Cotangent is the reciprocal of tangent. Tangents opposite over adjacent. Cotangents adjacent over opposite. So we originally had y over x. We will now have x over y. So these are the six trigonometric functions and how they are found whenever you apply them to an angle. Let's work out our first example. So it says the point negative six negative five is on the terminal side of an angle in standard position. Find the exact value of each of the six trigonometric functions. Well, the first thing I see here is I see a point negative six negative five. So I'm actually going to go in and I'm going to graph that point. So negative six, negative five. I'll plot a point at negative six, negative five. And to make your right triangle here, from the point to the origin, you will draw your diagonal line. Then from the point toward the x-axis, you will draw a vertical line. Notice that we now have a triangle. The angle that is closest to the origin is going to be our angle theta. That is our angle theta. So notice that we have a side here of the triangle of length negative six, so to speak, and a height here of negative five. Now, I will be finding the trig functions values at this angle. However, the issue is I don't know this hypotenuse. This is the hypotenuse because, yes, there is a right angle here. So the hypotenuse is across from the right angle. How do I find the hypotenuse of a right triangle if I know the two legs? Well, simple, right? Pythagorean theorem. So we have to dig back into our algebraic toolbox here and use Pythagorean theorem, which states that negative five squared plus negative six squared is equal to, and you can just put a placeholder here for this side. We'll just call it r, I guess, and we'll find out what r is. So negative five squared plus negative six squared is equal to r squared. That means 25 plus 36 is r squared. 61 is equal to r squared. This is a quadratic equation that can be solved by taking the square root of those sides. So I have square root of 61 equals r. So my hypotenuse measure is actually going to be 61 in this case. Now we are ready to, we will find sine of theta. Sine, sine was opposite, so negative five over hypotenuse, square root of 61. This is negative five over square root of 61. Now we have to rationalize this because we have a radical in the denominator. So I will multiply by the square root of 61 over itself. And this will give me negative five square root of 61 over 61. Now you can leave the negative sign in the numerator or you can bring it out front of the fraction. Either way is perfectly fine. Next, how about cosine of theta? Cosine, cosine is adjacent over hypotenuse. So I have your angle theta, the leg that's adjacent or side that's adjacent is negative six, hypotenuse is square root of 61. So I have negative six over square root of 61. Guess what we are going to do again? We are going to rationalize. Multiply by square root of 61 over itself. So you have negative six square root of 61 over 61. Now you can also once again choose to bring the negative sign out front, but once again that's just personal preference. Next, the third one, tangent of theta. Tangent is opposite over adjacent. So negative five over negative six, the hypotenuse is not used here. So negative five over negative six. That is actually going to be just positive five sixth. Next, cosecant. Remember cosecant is the reciprocal of sine. So sine is negative five over square root of 61. Cosecant will be square root of 61 over negative five. Once again you can bring that negative sign out front if you would like. Next, secant. Time for secant. Secant is the reciprocal cosine. I had negative six over square root of 61. So now I'll have square root of 61 over negative six. I'm going to go ahead and bring that negative six out front. And then there's cotangent. Cotangent is the reciprocal of tangent. So let me just go to my answer five over six and I'm going to flip it and make it six over five. So these are the six trigonometric functions values at the angle theta in this particular example. So there's six different answers. Please take note that there are a total of four of them that are negative and two of them that are positive. We'll talk about that a little while in the future. Now we're actually going to learn how to find the trig functions values at specific angle measures. So first we're going to start off nice and simple. We're going to talk about how to find the trig functions values at angles that are zero, 90, 180, 270, or 360 degrees. In radians, remember this is zero, pi over two, pi, three pi over two, or two pi. This diagram I have over here on the right hand side is your unit circle. It's called the unit circle because the origin is at zero zero and it has a radius of one. So notice that these ordered pairs. So I have zero degrees, 90 degrees, 180, or 270. And then zero, I go back to 360 degrees here if I was just to keep going around and around forever. In radians this would be zero, pi over two, pi, three pi over two. And if I was to keep going I would come back to two pi. Now notice that these points I have one zero, zero one, negative one zero, zero, negative one. They behave just like ordered pairs. Well the cool thing about these points is that the first number in each of these ordered pairs is actually cosine's value at that specific angle measure. And then the second coordinate in these ordered pairs is actually sine's value at each of these ordered pairs when the trig function is applied to the angle. So let's actually do some trig function evaluations at some specific angle measures. Cosine of 90 degrees. So go to your unit circle diagram here, find 90 degrees. I'm at zero one. Where is cosine here? Cosine is the first coordinate in the ordered pair. Cosine is the first coordinate in the ordered pair. Sine is the second. An easy way to remember this is that C comes before S, cosine before sine. So cosine of 90 degrees is zero. Tangent of three pi over two. So that brings me down here to 270 degrees. That is three pi over two. Now the issue is we didn't even talk about tangent being one of the numbers in the ordered pair. It's just cosine and sine. Well there's a trick with tangent that you're going to learn a little bit later. And that trick is that tangent is actually going to be sine over cosine. So we wrote tangent in terms of sine and cosine. Sine of three pi over two over cosine of three pi over two. Looking at our ordered pair zero negative one. Cosine is first. Sine is second. So you actually get negative one over zero, which is undefined. Now we're at cosecant of 540 degrees. That is a huge angle measure. And we'll talk more about this in the next video. But every trig function repeats itself every 360 degrees. So I can actually not change the value of this trig expression by simply subtracting 360 degrees from the angle. When you subtract 360 degrees from the angle, you actually end up getting at 180 degrees. So I have cosecant of 180 degrees. Once again we have an issue. I only know cosine and sine at each of these key points as they're called. So cosecant. What is cosecant in terms of sine? Well it's the reciprocal. So cosecant of 180 degrees is one over sine of 180 degrees. This is one over, so I get 180 degrees. Cosine's negative one, sine is zero. So my sine of 180 degrees is zero. One divided by zero, that's undefined. So that is evaluating trig functions values at key points. Now what if you don't have a key point? What if your angle measure is not 0, 90, 180, 270, 360, and so forth? Well now we're going to talk about how to find the trig functions values. Now angles with measures 30 degrees, 45 degrees, and 60 degrees. So this is a little bit of throwback to geometry for those of you that might even vaguely remember this, but I have here a 45, 45, 90 triangle, and a 30, 60, 90 triangle. Alright so looking at the 45, 45, 90 triangle here, we have a total of two different side measures. The side across from the 45 degree angle is one in both cases. The side across from the 90 degree angle is square root of two. So this is always the relationship of a 45, 45, 90 triangle. Next, 30, 60, 90 triangle. The side across from the 30 degree angle is one, 60 degree angle is square root of three, and 90 degree angle is two. An easy way to remember this is there is a theorem that goes with triangles that says that the smallest side is always across from the smallest angle, and the largest side is always across from the largest angle. So my largest angle here is definitely this 90 degree angle. My largest side is two. My smallest angle is definitely 30 degrees. My smallest side measure is one, which leaves 60 degrees with square root of three. It's important to learn these relationships. Example time, find the value of each of the following. I will begin with cosine of pi over four. If you don't like it when they give something to you in radians, it's okay if you secretly think about it in terms of degrees. Pi over four is 45 degrees. So cosine of 45 degrees. Well, I think I need to draw me a 45, 45, 90 triangle. 45, 45, 90 triangle. So go ahead and label your side measures. You have one, one, and then the hypotenuse is square root of two. Let's think about cosine. Cosine, cosine of 45 degrees. Cosine, I believe, when we talked about it was adjacent over hypotenuse. That's what cosine is. So if I go to a 45 degree angle, it doesn't matter which one. Adjacent is one, hypotenuse is square root of two. Remember, when we say adjacent or opposite, referring to the legs of the triangle, not the hypotenuse, hypotenuse has its own specific label. So cosine of 45 degrees is one over square root of two. One over square root of two. Please rationalize by multiplying by the square root of two over itself. So you'll have square root of two over two. Next, three cotangent of 30 degrees. All right, so the three stays out front. It'll be multiplied by whatever cotangent of 30 degrees is. First off, what was cotangent? Cotangent is the reciprocal of tangent. The tangent's opposite over adjacent. I believe cotangent would have to be adjacent over opposite. 30 degrees is my angle I'm concerned with. So let me draw a 30, 60, 90 triangle. It doesn't matter where you put the 30 degree angle and where you put the 60 degree angle. As long as you put one across from 30, square root of three across from 60, and two across from the 90. Cotangent of 30 degrees. Adjacent over opposite. Square root of three over one. So we have to multiply these two numbers together. You have three over one times square root of three over one. That gives you three square roots of three. Now time for another example. We're seeking a pi over three plus cosine of pi over four. Now if you don't like radians, convert to degrees. Pi is the same thing as 180 degrees. 180 divided by three is 60. And 180 divided by four is 45. Now remember cosine is adjacent over hypotenuse, which would mean the reciprocal, which is secant, would be hypotenuse over adjacent. Let's evaluate. Secant of 60 degrees. So go to your 60 degree angle. Hypotenuse over adjacent. Two over one. Then cosine of 45 degrees. Adjacent's one. Hypotenuse is square root of two. So you have one over square root of two. We're going to go ahead and combine these two fractions. So the common denominator is square root of two. So my first fraction will become two square roots of two over square root of two plus one over square root of two. So this is actually going to give you two square roots of two plus one over square root of two. If you haven't guessed it already, yes, we do need to rationalize this. So I need to multiply top and bottom both by square root of two. So I'll have square root of two times two square roots of two plus one over square root of two times square root of two is square root of four which is two. Now we have to go through and distribute over the numerator. We have two times the square root of four. So you have two square roots of four two times two which is four. Then we have square root of two times one which is just square root of two. So you have four plus square root of two over two. That is your final answer. It is fully rationalized. Now what happens when they give you angles that are greater than 90 degrees because we just talked about 30, 45, and 60 degree angles? Well anytime they give you an angle greater than 90 degrees you actually just have to go and take the given angle that's in quadrants two, three, or four and move it to quadrant one using the following formulas. This is called finding the reference angle of the angle they give you. So if your angle is in quadrant two then the reference angle would be 180 degrees minus the given angle theta or if you're talking radians pi minus the given angle theta. If they give you an angle in quadrant three the reference angle is the given angle minus 180 degrees or the given angle minus pi. If they give you an angle in quadrant four then it's 360 degrees minus the given angle or two pi minus the given angle. That is how you find the reference angle. So while for moving an angle back to quadrant one what was the point in them giving it to us in quadrant two, three, and four? And what the quadrant of an angle indicates is what the sign is of that trig function at that specific angle. So what is the sign going to be of that specific trig expression? And that's determined by what angle the original what quadrant the original angle was in. So consider the following I have my coordinate playing in quadrant one I have all then I move to quadrant two students take calculus. Whether you think it's true or not this is a good way to remember what the signs of each of the trig functions are in each of the quadrants. So all in quadrant one all trig functions are positive all of them every single one. In quadrant two you have students students in quadrant two sign and it's inverse cosecant are positive. So for the trig functions are negative and only sign and cosecant are positive in quadrant three tangent and it's inverse cotangent are positive. So think about take T is for tangent and it's inverse cotangent. Calculus C in quadrant four cosine and it's inverse secant are positive. So cosine and secant are the only positive trig functions in quadrant four. So anytime you have to find a trig functions value at a specific angle remember the following steps. So determine what quadrant the angle is in find the reference angle using the formulas on the previous page draw a right triangle with the indicated reference angle and then label the triangle using the 45 45 90 or 30 60 90 relationships. Evaluate the trig functions value at the angle measure and determine whether the answer is positive or negative depending on where the original angle was what quadrant was the original angle in. Guess we should work some examples now. Sign of 5 pi over 4 now once again if you don't like gradients go ahead and convert it to degrees that's perfectly fine but 5 pi over 4 5 divided by 4 is 1.25 so this is about 1.25 pi it's a little bit more than just 1 pi so this is actually going to be a quadrant 3 angle this angle is in quadrant 3 so in quadrant 3 is sine positive or negative well in quadrant 3 remember all students take calculus quadrant 3 tangent and cotangent are positive that's it therefore sine is negative sine is negative let's remember that don't forget that you're in your final answer you'll have to put a negative sign alright so now we need to find the reference angle so now I need to find the reference angle reference angle for quadrant 3 is going to be whatever your angle is minus pi minus pi so that's 5 pi over 4 minus 4 pi over 4 which is just pi over 4 or 45 degrees alright so now we're going to actually go in and we're going to evaluate sine of pi over 4 we're going to evaluate sine of pi over 4 and to do that I guess we'll draw a right triangle a 45, 45, 90 triangle so pi over 4 is 45 degrees label your sides 1, 1, and square root of 2 go to your 45 degree angle sine is opposite over hypotenuse 1 over square root of 2 go ahead and rationalize while you're at it here multiply by square root of 2 over itself and you get square root of 2 over 2 now this is sine of pi over 4 the reference angle of my original quadrant 3 angle well we already mentioned that in quadrant 3 sine is negative so literally take your answer sine of pi over 4, square root of 2 over 2 and just draw a negative sign in front of it you're done it's very important don't forget that last step of throwing in that negative sign next cotangent of 240 degrees what quadrant are we in? we are actually going to be in quadrant number 3 remember our quadrant tier we have all students take calculus t is for tangent and cotangent cotangent is positive and quadrant 3 so our final answer cotangent of 240 will be positive alright then I need to go ahead and do my reference angle how do you find a reference angle for an angle in quadrant 3? well you just take 240 and subtract 180 you'll pull this formula from the previous slides which means my reference angle is what 60 degrees right? so now I just have to go in and find cotangent of 60 degrees I suppose that means I need to draw me a triangle with a 60 degree angle in it a 30, 60, 90 triangle label the sides appropriately learn this relationship it'll benefit you in the long run cotangent well if tangent is opposite over adjacent then cotangent is adjacent over opposite so go to your 60 degree angle adjacent is 1 opposite is square root of 3 I have 1 over square root of 3 guess we should rationalize right? so multiply by square root of 3 over itself it turns out you get square root of 3 over 3 so that means that cotangent of 240 degrees is square root of 3 over 3 is it positive or negative? well cotangent is positive and quadrant 3 so keep your answer positive well how about a negative angle? well there's a few ways you could address this but one way you could do this is to just go ahead and add 360 degrees to the angle because that doesn't change anything it'll just give us a positive angle instead a positive angle of 330 degrees what quadrant are we in at 330 degrees? we are in quadrant 4 alright so quadrant 4 let's find out if secant is positive or negative all students take calculus alright so quadrant 4 cosine and secant are positive so secant is positive alright so secant is positive well now I just need to find the reference angle here so I have to look at my quadrant 4 reference angle formula and I believe it says that you take 360 degrees and subtract the given angle subtract the quadrant 4 angle subtract 330 degrees and you'll get 30 so my reference angle is 30 degrees so now the goal is going to be let's find secant of 30 degrees which means that we need to draw a triangle a 30, 60, 90 triangle alright so label your sides secant is the reciprocal of cosine if cosine is adjacent over a hypotenuse secant would have to be hypotenuse over adjacent hypotenuse is 2 adjacent is square root of 3 2 over square root of 3 I guess we got to rationalize right? so we're going to multiply by a square root of 3 over itself and then we'll have 2 square roots of 3 over 3 so my final answer is secant of 330 degrees I'm in quadrant 4, secant is positive so just keep the 2 square roots of 3 over 3 positive I'm boxing your answer so what you just learned is how to apply trig functions to various angle measures in the next video we'll actually look at some of the properties of trig functions and use those to our benefit so as we find the trig functions value at a specific angle measure thanks for watching