Evaluate Composite Trig Functions - Video Tutorials & Practice Problems

Evaluate Composite Functions - Values on Unit Circle

Hey, everyone, if you were asked to find the inverse cosine of one half or maybe even the sign of some angle like say pi over three, you would be able to come over to your unit circle and get an answer rather easily. But what if instead you were asked to find the sign of the inverse cosine of one half that doesn't seem quite so simple. But here I'm going to show you that it actually is that simple because we're going to be able to deal with each of these functions separately using what we already know in order to get a solution for this entire your expression. Now here I'm going to walk you through exactly how to do that. So let's go ahead and get started. Now when dealing with functions such as this one, these are referred to as composite functions because we have one function inside of another function. And when dealing with composite functions, we only want to deal with one function at a time. So in order to do that, we're always going to evaluate the function that is inside of the parentheses first. Now in doing this, we're essentially going to be solving this composite function from the inside out, we're going to start with that inside function. Now with this one, the sign of the inverse cosine of one half our inside function is the inverse cosine of one half. So that's what we're going to deal with. First, just completely ignoring that outside function the sine. So here to find the inverse cosine of one half, remember when dealing with inverse trig functions, you can also think of this as OK, the cosine of what angle gives me a value of one half. Now coming over here to our unit circle, I see that for my angle of pi over three, I end with a cosine value of one half. So that tells me that my inside function, I get an answer of pi over three. Now that I've dealt with that inside function, I can now deal with my outside function of sine. Now I'm just left to find the sign of pi over three, which we can again do relatively easily just coming over to our unit circle. Now again, for my angle of pi over three, I have a S value of root 3/2. So the sign of pi over three is root 3/2. And that gives me my answer to this entire expression. The sign of the inverse cosine of one half is route 3/2. So now that we know the basics of solving these composite trick functions let's go ahead and come down here to solve some more examples together. Now looking at this first example, I have the cosine of the inverse tangent of zero. Now remember that when working with these composite functions, we want to look at our inside function first. Now our inside function here is the inverse tangent of zero. Now in working with inverse trick functions, remember you can always think of this as the tangent of what angle will give me a value of zero. Now coming over here to our unit circle, we can see that for our angle of zero, we also get a tangent value of zero. Then also for our angle of pi, we will end up with a tangent value of zero as well. But remember that when working with inverse trick functions, we have to consider the interval for which they're defined. So we can only use values within that correct interval. Now, for the inverse tangent I know that my interval for my angles goes from negative pi over two to positive pi over two. So coming back down to my unit circle, I only want to deal with angles in that correct interval. So this angle of pi is not within my interval from negative pi over two to pi over two. So that is not my solution here. And here my solution is zero for that inside trig function. So here the inverse tangent of that value of zero is my angle of zero. Now I'm just left to deal with that outside function the cosine. So here I want to find the cosine of my angle zero. Now, for my cosine of zero, I can come back down to my unit circle here, the cosine of zero is simply one. So that gives me my final solution of one for the cosine of the inverse tangent of zero. And we're done here. Now let's take a look at one final example here, we're asked to find the inverse cosine of the sign of pi over three. Now, something that you might notice here is that my inverse trig function is now on the outside rather than on the inside like it was an example. A now you'll see these either way, it doesn't matter which way because we're still going to solve these the same exact way by starting with our inside trig function. Now our inside function here is the sign of pi over three. So over to my unit circle, I want to find the sign of pi over three which looking at my angle pi over three, I know that the sign is the square root of 3/2. So that gives me my solution to that inside function root 3/2. Now I'm just left to find the inverse cosine of that value. Now remember when working with inverse trig functions, you can also think of this as the cosine of what angle gives me a value of route 3/2. But remember we are also considering our interval here. So since we're working with the inverse cosine, in this case, our interval for the inverse cosine goes from zero to pi when considering what angles we want to deal with. So on our unit circle, I only want to find an angle that is between zero and pi. So an angle between zero and pi that gives me a cosine value of the square root of 3/2. I see that for pi over six, the cosine is root 3/2 and that is within my specified interval. So that's my final solution here pi over six. So the inverse cosine of the sine of pi over three is pi over six. And we're done here now that we know how to evaluate composite trig functions. Let's get a bit more practice together. Thanks for watching and let me know if you have any questions.

Evaluate the expression.

$\cos\left(\sin^{-1}1\right)$

• A. $0$
• B. $1$
• C. $-1$
• D. $\frac12$

Evaluate the expression.

$\sin^{-1}\left(\cos\frac{2\pi}{3}\right)$

• A. $\frac{\pi}{6}$
• B. $\frac{5\pi}{6}$
• C. $\frac{\pi}{3}$
• D. $-\frac{\pi}{6}$

Example 1

Hey, everyone in this problem, we're asked to evaluate the expression the second of the inverse cosine of negative root 3/2. Now this function might look a little bit complicated to you, but we can still break this down and solve this the way that we already know how. So let's start with that inside function here, the inverse cosine of negative root 3/2. Now because this is an inverse trick function. Remember that we can also think of this as OK, the cosine of what angle will give me negative root 3/2. Now also remember because this is an inverse trick function we're looking for angle within a specified interval. Now, for the inverse cosine, I only want to work with angles that are between zero and pi. So I am looking for an angle between zero and pi for which my cosine is equal to negative root 3/2. Now, looking at my unit circle, I know that for five pi over six that will give me a cosine of negative root 3/2. So that tells me that this inside function is five pi over six and I am now left to find the sequent of that five pi over six. Now from working with these trig functions previously, you may remember that the C A is really just one over the cosine. So this C can of five pi over six can really be rewritten as one over the cosine of five pi over six. Now, we actually just saw what the cosine of five pi over six was we know that it's negative root 3/2. So this is really one over negative root 3/2. Now, whenever I have one over a fraction, I could really just flip that fraction. So this is going to give me negative two over the square root of three. Now this is technically a correct answer. But whenever we have a radical in our denominator, we really don't want that there. So we actually want to rationalize this denominator which we can do by multiplying by route three over route three. Now that will give me a final answer of negative two root 3/3. And that's my final answer. The second of the inverse cosine of negative root 3/2 is negative two root 3/3. Thanks for watching and let me know if you have questions.

Evaluate Composite Functions - Special Cases

Hey, everyone, as you work through problems dealing with composite trig functions, you may come across an expression that looks something like this one, the inverse cosine of the cosine of 11 pi over six. Now looking at this, knowing that trig functions in their inverse undo each other, you may be tempted here to cancel this inverse cosine and cosine leaving you with just that 11 pi over six as your final answer. But this would actually be entirely wrong because remember when dealing with inverse trick functions, you have to consider interval for which they're defined. Now you may be worried that this is going to make these sort of problems really tricky, but you don't have to worry about that because we're actually going to solve these the same exact way we would any composite trick function evaluating them from the inside out and considering the interval along the way. So here, I'm going to walk you through exactly how to deal with these types of problems and you'll be a pro in no time. So let's go ahead and get started. Now, like I said, you cannot assume that your argument is your final answer here. So in this first example, the inverse code sign of the cosine of 11 pi over six, we cannot assume that our final answer is 11 pi over six. And because of that, we're just going to solve this as we would any composite trick function starting with that inside function, which in this case is the cosine of 11 pi over six. Now in finding the cosine of 11 pi over six, I can just come right on down to my unit circle. And I see that for 11 pi over six, the cosine is the square root of 3/2. So for that inside trig function here I get it answer of root 3/2. Now that I've dealt with that inside trick function, I can deal with my outside trig function which is the inverse cosine of that square root of 3/2. Now remember when dealing with inverse trig functions, we can also think of this as OK, the cosine of what angle will give me a value of root 3/2. But remember also when dealing with these inverse trig functions, we have to consider their interval. So specifically when dealing with the inverse cosine our interval for our angles is between zero and pi. So looking at our unit circle, I only want to look for angles between zero and pi for which the cosine is root 3/2. Now looking at this unit circle within this specified interval I see that for my angle pi over six, my cosine is root 3/2. So that gives me my final answer here. The inverse cosine of root 3/2 is pi over six. And that gives me my final answer. And you see that indeed, this is not equal to 11 pi over six. So we have to be super careful here. Now let's go ahead and move on to our next example where we're asked to find the sign of the inverse sign of two. Now remember we're going to solve these the way we would any other composite trig function. So when you're tempted here to cancel this sign, an inverse sign, don't remember that we have to solve these the way we would any other composite trick function starting from the inside. So here our inside function is the inverse sign of two. Now remember when dealing with inverse trick functions, you can, I also think of this as OK. The sign of what angle gives me a value of two. Now coming over to our unit circle and going around looking at all of these different angles. I don't see one for which the sign is going to be too. So this seems a bit strange. But if we consider the interval for inverse sign function, I know that we can only take the inverse sign of values between negative one and one and two is not within that interval. So if I try to take the inverse sign of two. I can't do that because this is simply an undefined value. Now, even if you were to try to type this into your calculator, the sign of the inverse sign of two, you would end up with an error which is something that you might not have considered if you were to have just canceled this sign and inverse sign. So remember whenever you see functions that look like these and you're tempted to just cancel those functions out. Don't remember that you have to consider the interval of your inverse trick function. Thanks for watching and I'll see you in the next one.

Example 2

Hey, everyone in this problem, we're asked to evaluate the expression, the inverse sign of the sign of pi over six. Now here you might be tempted to just cancel the inverse sign with the sign and end up with your answer of Pi over six. But remember we have to be extra careful when working with these composite trig functions because our inverse trig functions are only defined for a particular interval. So let's go ahead and break this down the way we would any composite trick function starting with that inside function, the sign of Pi over six. Now coming over here to my unit circle, the sign of pi over six is equal to one half. So that inside function gives me a value of one half. And now I'm just left to find the inverse sign of that one half. And remember when working with inverse trick functions, we can also think of this as OK, the sign of what angle is equal to one half. But that angle can only be within our specified interval for the inverse sign which happens to be from negative pi over two to positive Pi over two. So you only want an angle within this interval for which the sign is equal to one half. Now, inside of this interval, where is my sign equal to one half. Well, the sign of pi over six is equal to one half and that is within my interval. So that actually gives me a final answer of pi over six. Now, here you might be confused because I just told you that you can't cancel the inverse sign with the sign. But here we actually now the reason that we could do that is because our angle from the beginning was within our specified interval. So when that does happen, you actually are able to effectively cancel the inverse sign with the sign. But you always want to be extra careful when working with these because remember that isn't always true. So always, always double check your interval and make sure you're getting a value only within your specified interval. Thanks for watching and let me know if you have questions.

Evaluate the expression.

$\cos\left(\cos^{-1}\left(-\sqrt3\right)\right)$

• A. $\frac{\pi}{3}$
• B. $\frac{2\pi}{3}$
• C. $\pi$
• D. Undefined

Evaluate the expression.

$\cos^{-1}\left(\cos\left(\frac{\pi}{2}\right)\right)$

• A. $0$
• B. $\frac{\pi}{2}$
• C. $\pi$
• D. Undefined

Evaluate the expression.

$\tan^{-1}\left(\tan\frac{2\pi}{3}\right)$

• A. $-\frac{\pi}{3}$
• B. $\frac{\pi}{3}$
• C. $\frac{2\pi}{3}$
• D. $\frac{5\pi}{3}$

Evaluate Composite Functions - Values Not on Unit Circle

Hey, everyone, we've been evaluating composite trig functions using the unit circle. But what if you were given a composite trick function like this one? The sign of the inverse tangent of 3/4 3 4th is not a value that is directly on our unit circle. So how can we go about evaluating this? Well, you may be worried that this sort of problem is going to be really tricky, but we can actually just solve this by drawing a right triangle. And then we can use the absolute basics of right triangles that we already know in order to come to a solution. So here I'm gonna you through exactly how to solve these problems step by step. And soon you'll be able to evaluate any composite trig function whether you can use the unit circle or not. So let's go ahead and get started here. Now here in this example, we're tasked with evaluating the composite trick function the sign of the inverse tangent of 3/4. Now remember for any composite trick function, we're always going to start with that inside function, which here is the inverse tangent of 3/4. So we can also think of this, remembering how we think about inverse trig functions as the of what angle theta gives me a value of 3/4. So for what angle is this true? Well, we can actually just make this angle by drawing a right triangle. Now here, my right triangle is in quadrant one which makes sense because we're taking the inverse tangent of a positive value. So if I draw my angle theta right here and I want the tangent of this angle theta to be equal to 3/4 I can just label my opposite side as three. My adjacent side is four and the tangent of it is 3/4. Now you'll notice here that I don't actually care about the value of the angle, but just that it's tangent is equal to 3/4. Now, from here, I am just left to find the sign of this angle and evaluate that outside function. Now, in order to get the sign based on, so Koa, I know that I need my opposite side and my hypotenuse. So I first need to find my hypotenuse which I can do using the Pythagorean theorem. Or here you might also notice that since my leg links are three and four, that means my hypotenuse has to be five because this is a 345 angle. Now, from here, I can take that opposite side three divided by my hypo no five. In order to get my final solution, the sign of the inverse tangent of 3/4 is equal to 3/5. Now, let's recap how we got there here, we took our inside function, used it to draw a right triangle and then found our missing side length allowing us to evaluate that outside function and come to a final solution. Now, not all of these problems will be quite as straightforward. So here I'm going to walk you through some steps that will work for, for any composite trig function for which you can't use the unit circle. So let's come down to this example here. Here, we're asked to evaluate the expression, the sign of the inverse cosine of negative 5/13. Now you'll notice here that our argument is negative and that's kind of what complicates this a little bit, but no worries, we're going to walk through some steps here. Now, if you're ever unsure whether or not you should use the unit circle or draw a right triangle, if you look at your composite trig function and you see that your inside function is an inverse function and your argument is a kind of a strange number that you don't recognize as being on the unit circle. That's usually a good hint that you should draw a right triangle because the unit circle is not going to help you here. So here seeing that both of those things are true, let's go ahead and get started with our steps and get to drawing this right triangle. Now, of course, we're going to start by using our inside function, this is true of any composite trig function. So in step one, we're going to take that inside trig function and we are going to use our interval in order to determine what quadrant we will eventually draw our right triangle in. So here my insight function is the inverse cosine. And for the inverse cosine, I know that my values have to be between negative one and one and my angles between zero and pi. Now looking at my coordinate system based on that angle interval, I know that my angle has to be in quadrant one or quadrant two because this is my interval from zero to pi. Now, here's where we want to consider the sign of our argument. And here we're taking the inverse cosine of a negative value. Our argument is negative here. Now, based on our intervals, we're either in quadrant one or quadrant two, but cosine values can only be negative in quadrant two. So that's where my triangle is going to go. And we've completed step number one, quadrant two is where we'll draw our triangle moving on to step two. So we're actually gonna draw this triangle and then we're going to use our argument in order to label our angle and our two sides. So here in quadrant two, I'm gonna go ahead and draw my right now, the orientation of your right triangle might change just depending on what quadrant you're in, but you're always going to draw it with one side length against the X axis so that your angle can be measured from that x axis. So here this is where my angle theta is and looking at my argument negative 5/13, this tells me that the cosine of my angle that I labeled has to be equal to negative 5/13 in order to satisfy this inside function. So using Soko Towa, I know that that means that my adjacent side has to be negative five and my hypotenuse has to be 13. Now, it might be kind of weird here that I labeled this side link as being a negative value. But if we just think about where we're located on our coordinate system, my X values are negative here in quadrant two. So it makes sense that this side link is negative. So always consider your location on the graph when you're looking at the signs of your values. So now we've completed step number two, we can move on to step number three and use the Pythagorean theorem in order to find that missing third side. Now here, our missing side is a light length. So we're either solving for A or B, it doesn't matter which one. So I'm gonna go ahead and set up my Pythagorean theorem A squared plus B squared equals C squared. And I'm going to be solving for a here So I have a squared, I'm gonna plug in that other side length of negative five square that and that is equal to my hypotenuse, which is 13 squared. Now, from here, we can just do some algebra A squared plus negative five squared gives me 25 and that's equal to 169. Now, subtracting 25 from both sides cancels on that left side leaves me with a squared equals 144 having subtracted that 25 from 100 and 69. Now, from here, my last step is to take the square root, giving me my side length of 12, which I can go ahead and label up here on my triangle that missing side length of 12. Now we've completed step number three and we can move on to our final step where we're going to use our triangle in order to evaluate that out side function. So moving on to that outside function, which here is the sign. So here we want to find the sign of our angle theta that we have labeled on our triangle here. And I know that using Soko Towa, that means I want to take my opposite side, divide it by my hypotenuse and I have all the information I need. So my opposite side here is 12 and then my hypotenuse is 13. So that gives me my final answer of 12/13. So that tells me the sign of the inverse cosine of negative 5/13 is 12/13. Now, you may notice here that our argument started out as negative. We are taking the sign of the inverse cosine of a negative value. But then we ended up with a final answer that's positive. But again, think about where we're located on the graph in quadrant two, my sign values are going to be positive. So this makes perfect sense. So now you should be able to evaluate any composite trig function, whether you can use the unit circle or not. Let's continue practicing together. Thanks for watching and I'll see you in the next one.

Example 3

Hey, everyone in this problem, we're asked to evaluate the expression, the sign of the inverse tangent of two thirds. Now looking at this composite trig function, my inside function is an inverse and that value is not something I recognize as being on the unit circle. So I know that I need to solve this by using a right triangle. So let's go ahead and jump into our steps here. Now step number one is to use our interval to identify our quadrant. Now we're of course working with our inside function first here. So our insight function is the inverse tangent and the inverse tangent has an angle interval from a negative pi over two to positive pi over two. So I know that my right triangle either needs to be in quadrant one or quadrant two to match that interval. Now, here we want to consider the sign of our argument. Now, our argument here is a positive value of two thirds. And I know that within this interval, my tangent values can only be positive in quadrant one. So that's where my triangle is going to go. Now we've complete step number one, we can move on to step number two. And now we want to go ahead and actually draw that right triangle and use our argument in order to label everything that we need to. So I'm gonna go ahead and draw that right triangle in quadrant one with that side link down against the X axis. And then I'm going to label my angle theta as that inside angle right there. Now, looking at my argument, I have two thirds and my inside function tells me that the tangent of that angle that I just la has to be equal to that two thirds argument. So based on, so Katoa, I know that my opposite side has to be two and my adjacent side has to be three. Now we've completed step number two, moving on to step number three, we want to use the Pythagorean theorem in order to find that missing third side. Now here are missing side length is our hypotenuse. So setting up our Pythagorean theorem here, a squared plus B squared equals C squared. I'm solving for C because my missing side is a hypo. No. So here I take two squared plus three squared, those two side links that I already know that's equal to C squared. Now two squared is four and three squared is nine, that's equal to C squared. Adding those two values together. I get that 13 is equal to C squared. Now my final step here in getting this missing side length is taking the square root of both sides. Now, route 13 cannot be simplified anymore. So that's just what my side L is. Route 13 is equal to my hypotenuse that missing side link. So I can go ahead and label that up here on my triangle and we have completed step number three. Now moving on to our final step, we are going to use our right triangle in order to evaluate that outside function. Now here, my outside function is the sign. So I am tasked with finding the sign of my angle theta which I know I need my opposite side and my hypotenuse based on Sokoto. So I have all the information I need here looking at my triangle. I'm going to take that opposite side of two and divide it by my hypotenuse, which is the square root of 13. Now, this is technically a correct answer. But whenever we have a radical in the denominator, we wanna go ahead and rationalize that denominator, which we can do here by multiplying this by route 13 over route 13. So that gets our radical out of that denominator leaving me with two route 13 on the top and just 13 on the bottom. So my final answer here is that the sign of the inverse tangent of two thirds is equal to two route 13/13. Thanks for watching and let me know if you have questions

Evaluate the expression.

$\tan\left(\cos^{-1}\frac{12}{13}\right)$

• A. $\frac{12}{13}$
• B. $\frac{12}{5}$
• C. $\frac{5}{13}$
• D. $\frac{5}{12}$

Evaluate the expression.

$\sin\left(\tan^{-1}\frac{15}{8}\right)$

• A. $\frac{8}{15}$
• B. $\frac{15}{17}$
• C. $\frac{8}{17}$
• D. $\frac{15}{289}$

Evaluate the expression.

$\cos\left(\sin^{-1}\left(-\frac{7}{25}\right)\right)$

• A. $\frac{7}{24}$
• B. $\frac{8}{25}$
• C. $\frac{24}{25}$
• D. $-\frac{24}{25}$

Example 4

Hey, everyone in this problem, we're asked to evaluate the expression the tangent of the inverse sign of X over the square root of X squared plus four. Now this problem might look really overwhelming because there are multiple variables, there's a square root, there's a bunch of stuff going on. But we can still attack this the same way we would any of these composite trig functions that are definitely not on our unit circle and X over the square root of X squared plus four, definitely not on there. So let's go ahead and get started with step number one. Now for step one, we're gonna take that inside function and we're going to use our interval to determine our quadrant. Now here looking at this inside function, I am faced with finding the inverse sign. And now when working with the inverse sign, I know that I am working with angles from negative pi over two to positive pi over two. So my triangle either has to be in quadrant four or quadrant one. Now looking at the sign of our argument here, argument is kind of strange, but we can still take a look at it SX over the square root of X squared plus four. This is a positive value. So that means that my triangle has to go in quadrant one or my S values are positive. So step one is done, we can move on to step number two, which is to draw our triangle and use our argument in order to label our angle theta and our two sides. So up here in quadrant one, let's go ahead and draw our right triangle. And with my right triangle, I'm gonna go ahead and label my angle theta right here. And with this argument, this tells me that the sign of my angle theta is equal to X over the square root of X squared plus four. Now, this seems strange but hang with me here using so Katoa, this tells me that my opposite side has to be X and my hypotenuse is the square root of X squared plus four. So we're still working through this the same way we've completed step number two, we can move on to step number three where we're going to use the Pythagorean theorem to find that missing third side. So let's go ahead and set our Python theorem up here. We have A squared plus B squared equals C squared. And we're going to be solving for one of our leg length. So either A or B here, I'm going to solve for A. So I'm going to keep A as a variable A squared. And then plus my B squared B here is X, I'm squaring that value and that's equal to C squared. Now, C here is this whole expression, the square root of X squared plus four and all of that gets squared. Now from here, let's simplify this algebraically. Now, here on my left side, I can't really do anything. I'm just left with a squared plus X squared. But on that right side, if I square, I'm gonna get rid of that square root. So I'm just left with X squared plus four. Now, from here, if I'm solving for A, I wanna go ahead and isolate A by subtracting X squared from both sides. Now, when I do that, it cancels on that left side, but it also cancels on that right side. So I'm simply left with a squared equals four. This looks like a much more friendly expression. So here we're just left to take this square of both sides. We're left with a equals the square root of four, which is two. So our missing side length here is two. Now we've completed step number three, we can move on to our final step, which is to use our triangle in order to evaluate that outside function. Now, here outside function is the tangent. And in order to find the tangent of our angle, theta using Soko Toha, we're going to take our opposite side, we're going to divide it by our adjacent side. We have all of that information here. Our opposite side is X, our adjacent side is two. So my final answer here is X over two. The tangent of the inverse sign of X over the square root of X squared plus four is equal to X over two. And we're completely done here. Thanks for watching and I'll see you in the next one.