All right, so I thought we could do some examples to really enforce our knowledge of these other trig functions. So a common question you're going to see is that if I know one angle, can I get the other six? And that's going to turn out to be true. So let's do an example here. Let's say I give you that sine of some value theta is two-thirds, so you're able to measure that the sine is two-thirds. And I also want to tell you that theta is inside of quadrant two, so I'll write that theta in Q2, so on the unit circle. Find the values of the other six functions. So here we go, find for me all the rest. So I know sine, so I want cosine of theta, I want tangent of theta, and I want secant theta, cosecant theta, and cotangent theta. Again, at its surface, this looks like a very intimidating problem. How can I get so much information from so little? The beauty of mathematics here, sine of theta is two-thirds, find the rest. Okay, oftentimes in this, remember, you gotta remember everything depends on sine and cosine. Really this question is, and we've done one of these before, is just find cosine. Once you have sine and cosine, you get all the rest for free. How do we get sine and cosine? Let's draw a little helper triangle here. So we'll draw a nice, beautiful right triangle, perhaps one I drew is not so beautiful, but we'll put theta on the corner. I don't know what theta is, I have no idea. All I know is that sine of theta is two-thirds. So I will remind myself how to apply these periodic functions. SOH CAH TOA, sohcahtoa, here we go, sine of theta is opposite over hypotenuse, two-thirds. So that means I can always find the missing third side using the Pythagorean theorem. So what is the adjacent side over here? So I have a squared plus 2 squared equals 3 squared, good old Pythagorean theorem. So that means that a squared is 9 minus 4, which is, of course, 5. And so a is the square root of 5, so that is my missing value on the floor, on the adjacent side. And then I use sohcahtoa to tell you what exactly cosine is. So cosine is adjacent over hypotenuse, root 5 over 3. Now, here's the catch, though, always a catch, right? In quadrant two, in quadrant two, cosine is the x value, and the x value is negative. So cosine is going to be negative, so just watch out for that. If you're going to make a mistake, it's probably going to happen right there. Once you have sine and cosine, again, check this like crazy, make sure you agree, the rest of them then come from just putting them together. Remember what tangent is, it's the quotient of sine over cosine. So I'll write it out for good measure, sine over cosine. And that's two-thirds divided by negative root 5 over 3. As always, when you have fraction divided by a fraction, we're going to keep the numerator, we're going to change division to multiplication, and then we can flip the denominator. And you'll see that the 3s cancel, and our final answer is negative 2 over root 5. If you want, you can certainly rationalize the denominator and do a root 5 over root 5 to get negative 2 root 5 divided by 5 as your final answer. And it's good that tangent's negative because we're in quadrant two, so it certainly should be negative. So there's my tangent value. I have my cosine value, I have my sine value, we're cooking along. One thing that students sometimes get a little lost in is they're like, well, what's theta? And the point is, you don't need to know. Remember, the output of these periodic functions, sine and cosine, they're ratios of any angle, well, of any triangle with this angle. So you don't need it, which is really a statement about sides. Secant, of course, is the reciprocal of cosine. So I'll take my answer for cosine and I will flip it, don't lose the negation, 1 over 3 root 5. Cosecant, that's, of course, 1 over sine. So I take my answer for sine, actually no answer there, they just gave it to me. That'll be two-thirds flipped, so that's 3 over 2, positive 3 over 2. And cotangent, of course, is the flipped 1 over tangent of theta, and this is negative 5 divided by 2 root 5. You can rationalize the values if you want. But this is where, it's an intimidating question, but with very little work, you can get all six of the values. Really the effort comes in finding sine and cosine. So make sure you know how to use sohcahtoa and Pythagorean theorem to get that. Second example, sometimes in addition to finding values, they're going to ask you do it in an algebraic expression. So let's solve, let's give you an expression here. How about 2 cosine x minus 3 tangent of x equal 0. Okay, solve for x. Remember I said everything about these periodic functions comes from having a good understanding of sine and cosine. You want things with sine and cosine, we want to hang out in quadrant one. Tangent's there to kind of scare you, it's there to kind of throw you off. And I would always recommend, always, always, always, converting anything that's not sine and cosine, and just rewriting it and saying, I'm not going to do that problem. I'm going to say instead of 3 tangent x, I'm going to write it as 3 sine of x over cosine of x is 0. And now I'm a little bit happier than I was before because now it becomes a little more manageable. What do I mean by that? So let's move the expression of the other side to get 2 cosine x equals 3 sine of x over cosine of x. Now I have a fraction equal to a fraction. What do you do then? Algebraically, you cross-multiply. So we have 2 cosine squared x, so I multiply top left times bottom right, is 3 sine of x. So notice, this is a little better, a little nicer. I do not have any fractions anymore, so that's a step in a nice direction. So now the challenge is, I have a cosine squared and a sine squared. Think about what you do here, how would you put this, is there any way, right? So this is usually step one, put it in terms of sine and cosine. Then think yourself, is there any way so that I can get this to be only one of these functions. Sine and cosine, putting them together a tricky, is there anything you can do? And the answer is yes, there is always a relation, the fundamental identity. Remember that? That says that cosine x plus sine squared of x is 1. So you can substitute out cosine squared of x with 1 minus sine squared. So let's do that next. So I put it all in terms of sine and cosine, and then I just use this one identity to put them all in terms of each other. So this becomes 1 minus sine squared of x equals 3 sine of x. This is where, if you just know identities involving sine and cosine, you can get pretty far and solve lots and lots of problems here. Let's distribute the two. And when I do that, I will get 2 minus 2 sine squared x. I'm going to move the 3 sine of x back over to left side to get 0. And I'm going to write this in a way that you normally see, where the square comes first. And so let's go to negative 2 sine squared of x minus 3 sine of x plus 2 is 0, so I put the constant last. One more cleanup we're going to do, I don't like the negative in front, so let's multiply everything by a negative. I can certainly do that. And I get 2, and I'm going to write sine squared of x, as a reminder, it's sine squared, plus, so we're going to change all negatives to a positive here, 3 sine of x minus 2 is 0. And I'm writing it in this way so that hopefully you can see where I'm headed. Sometimes the sine and cosine functions throw you off a little bit. You gotta remember, if this were like a u, if we did a substitution, this would be 2u squared plus 3u minus 2 is 0. Stare at that for a minute, see if you can do something with that. And I hope if you stare at this long enough, you realize that that factors. So you can totally factor this as 2 sine of x minus 1, 2, positive, sorry, sine of x plus 2, and that's all equal to 0. So 2 sine x minus 1 sine x plus 2 equals 0. Foil that back in your head, put it all together, just convince yourself that that is in fact true. And now I have two things that multiply together to give us 0. And I wish I could do that, I'll go do that at the top over here. So we get 2 sine of x minus 1 is 0. Move some things around, you get sine x is a half, or we have sine of x is negative 2. So we set the second thing to 0, move the 2 over, you get sine of x is negative 2. Hopefully you see this before I call it out, sine of x is negative 2, that can negative, that can negative happen? Try again, that can never happen. The maximum and minimum value, it's 1 and negative 1. You'll never get an x value that gets you less than negative 1. So there are no solutions to the second equation. So all your answer is they're going to happen when sine of x is one-half. What I'd like to do, I didn't say in the beginning, but I'll just do it now to make it easier, I want it on the interval 0 to 2 pi. So let's stay within one lap of the unit circle, and it's just so I don't have to list that infinitely many answers. If you want, think about when sine of x is one-half, this is definitely a value we know. In this particular case, this is pi over 6, or 30 degrees. That's one value, right, it says the y value equal to a half. There's going to be a value in quadrant one, then you're going to have a value in quadrant two. So it's reference angle in quadrant two is going to be 5 pi over 6, it's like pi minus pi over 6, so back it up a little bit. So you get two answers. So final answer here, two correct answers, pi over 6 and 5 pi over 6. Let's do another example here, and let's just evaluate, let's do some things without a calculator. So let's find, let's do like three of them, secant of pi, let's do cosecant of negative pi. And one more, let's see if we can come up with a scary one. How about cotangent of pi over 2? Ooh, scary, scary. All right, pause the video, see if we can work these out. My hint to you, of course, is always rewrite these in terms of sine and cosine to get back to a more familiar place. All right, ready? Here we go. Secant of pi, this is of course 1 over cosine of pi. Cosine of pi, of course I know that, we go over pi on the unit circle and land at the spot (-1,0). I of course remember that cosine is the x value, so this becomes 1 over negative 1, perhaps better written as negative 1. Secant of pi negative 1, don't need a calculator for that. Wonderful. All right, let's do another one, cosecant of negative pi. Here we go. So this is 1 over sine of negative pi. Now, there's two ways to think of this one. You could either start at your favorite point at (1,0) and go negative pi. So now that is clockwise, clockwise, back to 180. And once again, you'll land, purely coincidental, at the point (-1,0) on the unit circle. And say, well, sine is my y value, so this has to be 0, that's perfectly fine as well. Another way to do it is remember your symmetry from sine and say, negatives pop out because you even odd symmetry. And sine of pi, you could have also gone the other way and grabbed it, that's 1 over negative 0, 1 over 0. And of course, you can't write it 1 over 0, so we would say this is undefined. Did you see that coming? So it's perfectly fine. Remember, these are fractions, and it's the denominator 0, this is undefined. So turns out negative pi is not in the domain, so it's undefined, okay? Do not return 1 over 0, that's not good. All right, last but not least, cotangent of pi over 2. How do you want to write this? So this is 1 over tangent of pi over 2. Tangent, though, I don't want to know things about tangent, I only want to know things about sine and cosine. So 1 over tangent is sine pi over 2 over cosine pi over 2, and it's kind of weird having a fraction downstairs. So why don't we take its reciprocal, and we could write this as cosine pi over 2 over sine of pi over 2. So it's the reciprocal of tangent. Cosine pi over 2, this one I know. Go think about unit circle in your head, go pi over 2, or 90 degrees, and you land at the point (0,1). The y value is sine, so that's 1 downstairs, the cosine value's x, and that's 0 upstairs. 0 over 1, that's perfectly great. If I have no slices of pizza out of one, I have no slices of pizza, so that is a zero. Okay, so we have negative 1, undefined, and 0. If you got all three of those, that this fantastic. Let's do some more. Now Let's do more evaluations, but let's just practice with a calculator. Let's make up some numbers here, how about tangent of, I don't know, -44.6 degrees, and let's do maybe radians, secant of negative pi over 8. All right, I want decimals to four places, watch your rounding and be careful between radians and decimals. Pause the video, work this out. And I will go do the same. Are you ready? Here we go. When I do this, I get for 44.6 degrees, -44.6 degrees, I get -0.9861. Do you agree, looks good? Good, moving on, secant negative pi over 8. When you plug this in, again, make sure you convert to radians. If you're getting different numbers, you may not have converted to radians, 1.0824. Sometimes you need a decimal, you go grab your favorite calculator to do just that. Let's do another, let's do another example here. We'll make this one our last example, and then we'll let you guys go off and do some problems. Let's just try to mix it up so slightly, secant squared of 2 theta if theta is equal to pi over 6. Seems like a perfectly good one. So secant squared 2 theta, scary stuff, right? We know secant, but secant squared 2 theta, who knows what this is. This is just going to test some notation, do you know what we mean when we say the squared? Remember, when you put the squared in the function, this is the same as secant of 2 theta whole thing squared. So why don't we just plug in and see what that gets us? So secant of 2 times theta, well, 2 times pi over 6, of course, is secant of, 2 cancels, 6 becomes a 3, and you get secant of pi over 3. Secant of pi over 3, who knows what that is. But I do know, this is all squared, of course, still, secant of pi over 3 is, of course, 1 over cosine of pi over 3. So we're going to get out of secant into more familiar territory, and we're just going to keep coming back to sine and cosine. So cosine of pi over 3, remind yourself what that is, cosine of pi over 3 is a half. So this becomes, with all the cosines and sines gone, 1 over a half squared. Now we just do some arithmetic to clean this up. 1 divided by 1 minus the half, whenever you have a fraction downstairs in the denominator, that flips up and that becomes 2. We have 2 squared, so our final answer at the end of the day is a very beautiful 4. All right, so whenever you see things, remember, don't get scared, they kind of look scary at first. Convert them back into sine and cosine, keep the fundamental identity in the back of your mind as an option. And then you can always check things with an online calculator or graph and work with these things, okay? Practice makes perfect. Great job. I'll see you next time.
