Common Core Algebra II.Unit 11.Lesson 11.The Reciprocal Functions

Hello, I'm Kurt weiler, and this is common core algebra two. By E math instruction. Today, we'll be doing unit 11 lesson number 11 on the reciprocal functions. All right? So so far, we've looked at the sine function, the cosine function, and the tangent function. But there are actually three more trig functions. All right? And all of these are defined in terms of the original three, which can then actually be defined in terms of just sine and cosine. All right? Let's jump into them and get a little work with them in this lesson. All right, so the three reciprocal trig functions are known as the secant. The cosecant and the cotangent. The secant function is simply defined as following. The secant of an angle is one divided by the cosine of the angle. And the cosecant is defined as the cosecant of an angle is one divided by the sine of the angle. Now let me just point out for the record how irritating it is that secant is one divided by cosine. But cosecant is one divided by sine. Those are really, really irritating. It'd be great if secant was one divided by side, and cosecant was one divided by cosine, but it's not the way that they turned out geometrically or algebraically. Cotangent is easy enough, it's one divided by tangent. But it is normally thought more commonly as cosine divided by sine. Remember tangent is signed divided by cosine. So cotangent is cosine divided by sine. All right, in this course, we want you to just get a little bit exposure to these three. You'll surely work with them more in whatever fourth year math course you take or in college level mathematics. So let's jump right into it. One of the things we want to be able to do is evaluate the sign in the cosine. Sorry, we want to be able to evaluate these three functions at any of the special angles given the unit circle. So this isn't too bad. The secant of 60° by definition is one divided by the cosine of 60°. That's just what it is. Now, the cosine of 60° is the X coordinate at 60°, which is one half. Now, as always, this is a complex fraction. So we could multiply the top and the bottom by two. Here those twos would cancel. And we just find that the secant of 60° is two. All right? So that's not too hard. Why don't you go ahead and try to do the cotangent of 150°. Well, we could think about it as one divided by the tangent of a 150. But it's probably most helpful to think about it as the cosine of a 150. Divided by the sine of a 150. All right? The cosine of 150 is negative root three divided by two. And the sine of 150 is positive one half. So again, simplifying that complex fraction as we often do by multiplying the top and the bottom by two. And we find the cotangent of 150° is negative. Root three. Now in part C, we have the cosine of three pi over four radians. If you know what that corresponds to in terms of your degree measurements great. If not, perhaps you come over here and you say, oh, I have three pi over four radians. Maybe I'll convert that into degrees, a 180° divided by pi radians. The radians would cancel the pi as we cancel. Four goes into a 180, 45 times. And that is a 135°. So cosecant of three pi over four is the same as the cosecant. Of a 135°. That's going to be one divided by the sine of a 135°. Which is going to be one divided by root two over two. Now we can simplify this by getting rid of the complex fraction, and that will give us two divided by root two. Some teachers, and some tests consider that simplified. We've talked about this before. Some ask you to do what's called rationalizing the denominator. Now, in this case, that turns out kind of nice. If I had multiply both top and bottom by root two over two, we get two root two root two times root two is two. And then those cancel. So really, this is a very nice way of stating the cosecant of three pi over four which is the square root of two. But in my mind, either one of those are acceptable. All right. Pause the video for a minute and write down anything you need to. Okay, let's clear out the text. Now exercise two, which of the following is closest to the value of secant of 52°. This may seem like a strange question to you, kind of like jeez, man. I mean, this seems really easy or really hard depending on how you look at it. 52° isn't one of those special angles on the unit circle. But why not just plug this into the calculator? Take a look at your calculator for a minute. What you might notice is that there's no secant button on it. It always boggles my mind on these TI calculators or other ones. How many different things they have on them. And yet they don't have secant, cosecant, or cotangent. Somebody should get onto that. And yet, no problem. Right? We know that the secant of 52° is the same as one divided by the cosine of 52°. All right. So try that on your calculator. Okay. Did you get one point 6 two rounded to the nearest hundredth? If you didn't, it means you're probably in. Radian mode. And I wouldn't be surprised if you're in radian mode. I mean, what have we been doing for the last like many, many lessons we've been graphing sign and cosine, and we did that all in radian mode. If you didn't get 1.62, pause the video, pop back into your calculator, put your calculator in degree mode and try it again. Okay, well, hopefully you got it this time. Hopefully you got it the first time. But it's very easy with this mode business. To not get it right. I'm going to clear this out. All right. Now, all of the reciprocal functions cosecant secant and cotangent. They all have variable denominators, which means there's a chance that we could end up dividing by zero in them. So exercise three says which of the following values of X is not in the domain of cosecant X in other words, which of these values of X, if I try to plug it in there, we'll make cosecant not exist. Does not exist. Well, let's take a look at something. Cosecant of X is one divided by sine of X and sine of X is the Y coordinate on the unit circle. So the question is, which one of those angles has a Y coordinate of zero on the unit circle? Think about it for a moment. All right, well, hopefully you chose choice one. If I do a real quick and not great job of sketching the unit circle, a 180° is right here at the point negative one comma zero. Sine of a 180° is zero, and therefore the cosecant would be undefined, and therefore X equals one 80, would not be in the domain. That's easy enough. Pause the video now, write down anything you need to. Okay, let's clear out the text. And let's go. Then backs out of the sheet. All right, now kind of like what we did earlier on in this unit. I'd like to determine whether or not code tangent C can't cosecant, all that stuff is positive or negative based on the quadrants. Okay? So let's think about cotangent of beta in quadrant two, right? So remember, unit circle, quadrant one, quadrant two, quadrant three quadrant four. So if I have an angle that terminates in the second quadrant, will cotangent be positive or negative. Let's play around with that. The cotangent of beta will be the cosine of beta divided by the sine of beta. Well, in the second quadrant, the X coordinate is negative. And that's cosine. But the Y coordinate is positive. That sign. And a negative divided by a positive is a negative. So cotangent of beta is negative for any angles in the second quadrant. Those would be angles and degrees between 90 and a 180. And of course, many other angles, depending on code terminology. Oh, that is a cool word code terminal. I don't even know if it's a real word. Anyway, letter B and letter C asks about secant and cosecant. Pause the video now and think about these two. All right. Let's take a look. Secant beta will be one divided by the cosine of beta. If we're in quadrant four, then the cosine is a positive number because it's the X coordinate. So we have one divided by a positive number. But once a positive number, so a positive divided by a positive is a positive. In fact, what that means is that C can't always. Has the same sign. As cosine. Because really, the difference between them is that they're reciprocals. They're just flipped fractions. So if a fraction is positive, when you flip it, when you find it's reciprocal, it's still positive. And if it starts off as negative and you flip it, you find its reciprocal. It's still negative, right? If I start off with cosine being negative one half, secant is negative two. Et cetera. All right, I think I gave the farm away for letter C let's take a look at it. Cosecant of beta will be one over the sine of beta. And in the third quadrant, I keep going back to this picture. Since the Y coordinate is negative, and we have a positive divided by a negative cosecant of beta will be negative. And just like over here, cosecant actually, let me write it all the way out. Cosecant will have the same sign. This sign the same sign. As side. That one. All right. And this is important to understand for a variety of different applications involving trig. Some of which we'll get to this year, some of which we'll have to wait until further study. Pause the video now, write down anything you need to. Okay, clearing it out. Exercise 5. If cotangent of theta is less than zero. And secant theta is greater than zero, then theta could be which of the following angles. I'll pause the video and think about this. Somehow this relates to the last problem that we did. All right, well, let's think about this for a second. Cotangent of theta will be cosine theta divided by sine theta. And we're being told it's less than zero. So that means it's either this scenario. Or it's this scenario. Either way, we're going to get a negative out of this. On the other hand, secant theta, which is one over sine theta, well, that's greater than zero. So that's got to be this deal. All right? That means sine has to be positive. So we've got to be in this scenario. We've got to be in a scenario where the sign is positive. And remember, that's the Y coordinate. And the cosine is negative. Remember, that's the X coordinate. So then the question becomes where or in what quadrant are we in? Where the Y coordinate is positive, but the X coordinate is negative. That's got to be quadrant two. And what angle, which one of these is a quadrant two angle? Well, that's choice three. All right. It's kind of cool. Pause the video now, write down anything you need to. It's really kind of amazing. How much of this work relies on your basic understanding of fractions and signed numbered division, right? A positive divided by a positive positive, et cetera. All right, let's take a look at exercise 6. It says a right triangle is shown below with sides of lengths a and B find the length of the hypotenuse in terms of a and B label it on the diagram. In other words, yeah, okay, I'll call it C but what is C equal to? Well, we know that a squared plus B squared is equal to C squared. So if we solve for C, we take the square root of both sides. And that gives me C is the square root of a squared plus B squared. And that's where you want to stop. Do not distribute that square root. Don't tell me that C is a plus B that violates the triangle inequality from last year. So don't do it. Now what we're going to do is we're going to write out every one of those ratios in terms of the constants a and B so what I'd like you to do is take a shot at this right now. All right. Well, we're going to do some right triangle trig first, sine of a is opposite over hypotenuse. The side opposite of a is little a and the hypotenuse is the square root of a squared plus B squared. Cosine is adjacent over hypotenuse. I should give it myself a little bit more room here. Adjacent over hypotenuse. Which is going to be B divided by. A squared plus B squared. The tangent of a, well, that's going to be opposite divided by adjacent. And that turns out to be the simplest of them all. A divided by B now, what's cool is that all of these are called the reciprocal functions for a reason. This cosecant of a is one divided by the sine of a and we could write them that down here, but the whole point of reciprocals of one divided by is that it flips the fraction. So cosecant of a square root of a squared plus B squared divided by a secant of a, which will be one divided by the cosine of a will be the square root of a squared plus B squared. Divided by B and finally the cotangent of a, this is one place where it's nice to know it's one divided by tangent. It's going to be B divided by a. All right, so pause the video now and write down anything here that you need to. Okay, let's clear out the text. Do one last problem. And finish up our unit on trigonometry. All right, last problem. If alpha is an angle whose terminal array lies in the fourth quadrant, you know, that's important. And cosine of alpha is one third. Determine the exact value of cosecant alpha. Show how you arrived at your answer. All right, the first thing that you want to do in this problem actually, let me pause. This is the last problem we're going to do in trigonometry. You pause the video and take a shot at it. Okay, let's go through it. The first thing that I want you to do in a problem like this or that I think you should do in a problem like this is to write out something very strange. Cosecant alpha. That's odd. Is one over the sine of alpha. All right? Now, what that immediately tells me is I need the value of sine alpha. I know the value of cosine alpha. As usual, in this case, what we do is we use the Pythagorean identity, cosine squared, plus sine squared is equal to one. So I'll get one third squared plus sine alpha squared is equal to one. That'll be one 9th. Plus sine alpha squared equals one, subtract one 9th from both sides. And I'll get sine alpha squared is equal to 8 9. Take the square root of both sides. And I'll get sine alpha is equal to plus or minus the square root of 8 knights. Now remember, square roots can distribute over division. So I'll leave the square root of 8 as the square root of 8, but I'll make the square root of 9 into three. Now it's got to be either plus or minus, which one is it? Well, we're in the fourth quadrant, and in the fourth quadrant, the sine of any angle is negative. Fourth quadrant, down here, signs the Y coordinate, Y is negative. So sine of alpha is actually negative root 8 divided by three. Which means the cosecant of alpha, which is its reciprocal, is negative three divided by the square root of 8. Now, that's the exact value I'm very, very comfortable with that. There's lots of different ways, though, that you could write this and again, I want to emphasize this in case your teacher would want you to do this. You could rationalize, all right, so this would be one way to leave your answer as well. And we could even simplify the square root of 8. Oh my goodness, as two times root two, and then we could divide that out. And we'd get negative three root two over four. So there are certainly plenty of math teachers that we consider this to be the simplest form. Now, notice this question didn't actually say give its exact answer and simplest form, but some of your teachers might say always put it in simplest form. So that's one option. All right. Pause the video now, write down anything you need to. Okay. Clear. Now the text. Let's wrap it up. So there are actually 6 trig functions. In order of importance, number one is sine number two is cosine. And there are close one two. Tangents definitely very important, especially for right triangle trig, less important for trigonometric modeling. But it's still definitely number three. The last three secant cosecant and cotangent are much less important, but they do get used in some higher level math, so it's good to get an initial exposure at the algebra two level. All right. Well, I'd like to thank you for joining me for another common core algebra two lesson by E math instruction. My name is Kirk weiler, and until next time, keep thinking. And keep solving problems.