Inverse Sine, Cosine, & Tangent - Video Tutorials & Practice Problems

Inverse Cosine

Hey, everyone, you may remember that just like a square root undoes something being squared, an inverse trig function like say the inverse cosine undoes its corresponding trig function. But that's pretty much where we last left off when dealing with inverse trig functions in the context of right triangles. But we now know much more about trig functions in general. And we also know about the unit circle. So we can dive much deeper into our inverse trig functions using the unit circle. And you'll now be asked to evaluate expressions like the reverse cosine of one half. Now I'm going to show you exactly how to get there. And here, I'm going to specifically break down everything that you need to know about the inverse cosine function. So let's go ahead and get started. Now when working with a basic cosine function, we start with an angle and we take the cosine to end up with a value now with the inverse cosine because we're undoing the cosine function, we instead start with that value and then take the inverse cosine to end back up with an angle. So it's important to note that when working with our cosine function, we start with an angle and we end up with a value. Whereas with the inverse cosine, we instead start with that value and then end up back at an angle. So when evaluating an expression like say the inverse cosine of one half, we want to find the angle that has the corresponding cosine value which we can do on our unit circle. So let's come down here to this example here and go ahead and evaluate this expression here. We're asked to find the inverse cosine of one half. Now another way you can think of this and this will work for any inverse strict function is for the inverse cosine of one half. I can also think of this as OK, the cosine of what angle theta will give me a value of one half. And I want to find the angle that gives me that corresponding cosine value. So coming over here to my unit circle, I know that in quadrant one for my angle pi over three, this has a corresponding cosine value of one half. So this pi over three represents an angle that gives me my corresponding value. Now on my unit circle as I continue around, I also know that in quadrant four, all of my cosine values are positive here as well. So for my angle of five pi over three, that also has a cosine value of positive one half. So five pi over three represents another angle that gives me my corresponding value. But both of these angles are not the solution to my original expression. Only one of them is. So how do we determine which one of these is actually my solution? Well, in order to do that, we actually need to dive a bit deeper into our inverse cosine function. So we're actually going to take a look at the graph of cosine. So here we have the graph of our cosine function. And remember to get our inverse function, we can simply take our original function and reflect it over the line Y equals X. But because my cosine is not a 1 to 1 function, if I were to reflect this entire cosine graph over that line of Y equals I would end up with something that is not a function at all and it would not pass the vertical line test. So because of that, we actually only want to reflect part of our cosine graph in order to get our inverse cosine. And that specific part is between zero and pi. So if I'd reflect just this part over the line Y equals X, I end up with the graph of inverse cosine. But the way that these graphs look is not the most important part, but rather the specific values for which my inverse cosine function is defined. Now, remember I said when working with the inverse cosine function, we put values into it and we get angles back out. Now the values that we put into our inverse cosine function have to be between negative one and one. And the angles that we get back out must be between zero and pi. So when faced with evaluating an inverse cosine expression, our solution can only be angles within the interval zero to pi. So with that in mind, let's come back up to our example here and determine what our final solution actually is. So for the inverse cosine of one half, looking at my unit circle here, I only want to consider angles between zero and pi my specified interval here. So those bottom two quadrants are not going to have any of my solutions. That's just not going to work. I only want angles in that top half. So five pi over three is right down here in quadrant four. But this cannot be a solution because it is not within my specified interval. So that meet with my final answer of pi over three, which is up here in quadrant one between zero and pi. So my final answer here is that the inverse cosine of one half is pi over three. So when evaluating inverse cosine expressions, we are still going to look for angles with our corresponding values on the unit circle, but only within our specified interval from zero to pi. So now that we know how to evaluate inverse cosine expressions, let's get some more practice. Thanks for watching. And I'll see you in the next one.

Example 1

Hey, everyone in this problem, we're asked to evaluate the expression the inverse cosine of negative root 3/2. Now whenever working with a inverse trig function, remember that we can also think of this as OK, the cosine of what angle is equal to negative root 3/2. And we want to find the angle for which that is true. Now, when working with the inverse cosine, we know that our value, our angles can only be in between zero and pi but we can actually get even more specific here because we know that all of our cosine values are going to be positive in quadrant one. And all of our cosine values in quadrant two are going to be negative. So whenever we're taking the inverse cosine of a positive number, we know that our solution has to be in quadrant one. And whenever we're taking the inverse cosine of a negative number, just like we are here, we know that our solution has to be in quadrant two. So here we already know that our angle has to be in this second quadrant. So for which one of these angles is the cosine equal to negative root 3/2. Well, I know that the cosine of five pi over six is equal to just that. So that represents my solution. My angle here is five pi over six and we are done here. Thanks for watching and I'll see you in the next one.

Evaluate the expression.

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

• A. $0$
• B. $\pi$
• C. $\frac{\pi}{2}$
• D. $\frac{3\pi}{2}$

Evaluate the expression.

$\cos^{-1}\left(0\right)$

• A. $0$
• B. $\frac{\pi}{2}$
• C. $\pi$
• D. $\frac{3\pi}{2}$

Evaluate the expression.

$\cos^{-1}\left(-\frac{\sqrt2}{2}\right)$

• A. $\frac{\pi}{4}$
• B. $\frac{3\pi}{4}$
• C. $\frac{5\pi}{4}$
• D. $\frac{7\pi}{4}$

Inverse Sine

Hey, everyone, we just saw that for our inverse cosine function, we were able to flip our cosine graph but only over a specific interval. Well, you'll also be faced with evaluating inverse sine expressions. And that may seem like it's going to be annoying or a bit difficult, but it's actually the exact same principle. So we're going to do the exact same thing here that we did with our inverse cosine function. And we'll see that our graph and our intervals will just work out a little bit differently. But I'm going to walk you through all of that here. So let's go ahead and jump right in now here we have our sine graph and remember that to get the graph of our inverse sine function, we need to flip this over the line Y equals X. But because this function is not 1 to 1, I only want to flip part of it. So that specific part that I want to flip is between negative pi over two and positive pi over two. So flipping that reflecting it over the line Y equals X I end up with a graph that looks something like this for my inverse function. But remember the graph is not the most important part but rather the values for which this function is defined. Now remember when working with any inverse trig functions, we put values into them and we get angles back out and the values that we put into our inverse sine function have to be between negative one and positive one. And the angles that we get back out have to be between a negative pi over two and positive pi over two. So now with this interval in mind, we can go ahead and evaluate inverse sign expressions with no problem. So let's come down here to our first example which asks us to evaluate the expression the inverse sign of one half. Now remember when working with inverse trade functions, we want to find angles that have corresponding values on the unit circle. So we can also think of this expression as asking us, OK, the sign of what angle theta gives me a value of one half. But remember we have to consider our interval here. So remember that for our inverse sine functions, the value that we put in has to be between a negative one and positive one which one half definitely is. And the angle that we get back out for our answer has to be between negative pi over two and positive pi over two. So let's come down to our unit circle here. Now, on our unit circle, we want to find an angle theta for which our sine value is one half. And looking at my unit circle here, I see that for my angle of pi over six, I end up with a sine value of one half. So this presents the solution to my expression pi over six. Now you may be tempted here to come over to five pi over six and think OK, isn't the sign of five pi over six also equal to one half. And it is, but the problem is that this is not within our specified interval from negative pi over two to positive pi over two. So I actually don't care about that half of my unit circle at all. When dealing with the inverse sign, I only want to consider angles on this half of my unit circle between negative pi over two and positive pi over two. So this solution pi over six is my final solution here. And it's my only solution to the expression inverse sine of one half. Let's come to our next example here which is the inverse sign of negative root 2/2. Now remember we can also think of this as OK, the sign of what angle theta gives me a value of negative root 2/2. And coming over here to my unit circle. I see that down here in quadrant four for negative of pi over four, I know that my sign value is going to be negative root 2/2. But something else that you might have noticed here is that all of these values in quadrant four are negative. And on a typical unit circle, this negative pi over four would actually be seven pi over four. So what's going on here? Well, the problem is is that seven pi over four is not within my specified interval of negative pi over two to pi over two. So whenever dealing with the inverse function, we consider this quadrant four as being measured clockwise from zero rather than counterclockwise. So that all of our values are within that interval, negative pi over two to pi over two. So here my solution is negative pi over four within that specified interval. So the inverse sign of negative root 2/2 is negative pi over four. So now that we know how to evaluate inverse sign expressions, let's get a bit more practice together. 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 negative one. Now remember when working with inverse trig functions, we can also think of this as OK, the sign of what angle is equal to negative one. And we're looking for the angle for which that is true. Now, in working with the inverse sign specifically, we know that our solution can only be angles between negative pi over two and positive pi over two. But we can get even more specific here because all of my sine values in this first quadrant quadrant one from zero to pi over two are going to be positive. Whereas in the fourth quadrant from negative pi over 2 to 0, all of these S values are going to be negative. So if I'm taking the inverse sign of a positive value, I know that my solution has to be in quadrant one. Whereas if I'm taking the inverse sign of a negative value like we are here, the inverse sign of negative one, we know that our solution has to be in quadrant four between negative pi over two and zero. So here, since we're taking the inverse of negative one. I already know that my solution has to be down here in this fourth quadrant. So where is the sign equal to negative one? Well, for negative pi over two, I know that my sign or my Y value is equal to negative one. So here my angle and my solution to this expression is a negative pi over two and we are done here. Thanks for watching and I'll see you in the next one.

Evaluate the expression.

$\sin^{-1}1$

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

Evaluate the expression.

$\sin^{-1}\left(\frac{\sqrt3}{2}\right)$

• A. $\frac{\pi}{6}$
• B. $\frac{3\pi}{2}$
• C. $\frac{\pi}{3}$
• D. $-\frac{\pi}{3}$

Evaluate the expression.

$\sin^{-1}\left(\frac{\sqrt2}{2}\right)$

• A. $\frac{\pi}{4}$
• B. $-\frac{\pi}{4}$
• C. $\frac{3\pi}{4}$
• D. $\frac{7\pi}{4}$

Inverse Tangent

Hey, everyone, we just saw that for our inverse sine and cosine functions, we only worked with values and angles within a specified interval. So of course, we have to do the same thing when working with our inverse tangent function. So we can work through this the exact same way. And here I'm going to show you what the interval is when dealing with the inverse tangent function and we'll evaluate some expressions together. So let's go ahead and just jump right into this. Now, here I have the graph of my tangent function and in order to get the graph of my inverse tangent function and I of course need to reflect this over the line Y equals X. But this graph is not 1 to 1 as you might expect. So we only want to reflect part of it. And that specific part is between negative pi over two and positive pi over two. So reflecting this over the line Y equals X I end up with a graph that looks something like this for my inverse tangent function. But remember the graph is not the most important part but the intervals are so the values that we put into our inverse tangent function are actually not restricted and they can be anything between a negative infinity and infinity. But the angles that we get back out are restricted, they must be in between negative pi over two and positive pi over two. So with this interval in mind, let's go ahead and just evaluate some expressions. Now, the first expression that we're tasked with evaluating here is the inverse tangent of the square root of three. Now remember for any inverse trig functions, we can also think of this as OK, the tangent of what angle theta gives me a value of root three. Now looking at our unit circle here, the specific interval that I'm working with is between negative pi over two and positive pi over two. So I really just don't care what's going on on this half of my graph at all. I am only looking for an angle for which the tangent is the square root of three on this half of my graph. And it's actually the same exact interval we already saw it when working with the inverse sign. So let's go ahead and find our angle for which the tangent is the square root of three. Now looking in this first quadrant here, I see that for pi over three, my tangent is the root is route three. So that tells me that my answer here is pi over three that is within my specified interval. And I don't have to worry about anything else. Here is my answer. Now we've completed example A we can move on to example B which asks us to find the inverse tangent of negative one. Now again, we can think of this as OK, the tangent of what angle theta gives me a value of negative one. And we want to find that angle within our specified interval. So in my inner, from negative pi over two to positive pi over two, where is my tangent negative one? Well, I know that for my angle negative pi over four, my tangent is going to end up being negative one. So this gives me my final answer. The inverse tangent of negative one is negative pi over four. And we're done here. Now, as you continue to work through problems, you may have to evaluate some inverse trick functions using a calculator either because you're explicitly asked you or because there's no other way to evaluate the function. But in order to do this, all you have to do is press the second button and then whatever your corresponding trig function is. So in order to find the inverse sign, you would press second and then sign inverse cosine second and then cosign or inverse tangent second and then tangent and then you'll just type your value in and come to an answer. Now, when doing this, you'll typically be in radiant mode unless otherwise specified. Now that we know how to evaluate any inverse trig function sine cosine or tangent. 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 inverse tangent of negative square root of three. Now when working with inverse trick functions, we know that we can also think of this as OK, the tangent of what angle is equal to negative root three. And we want to find the angle for which that is true. Now, looking at our unit circle here, we know that when working with the inverse tangent, our value has to be in between negative pi over two and pi over two, which is actually the exact same interval that we dealt with when working with the inverse sign. Now again, here we can get even more specific because all of my tangent values in quadrant one are going to be positive. Whereas all of my tangent values in quadrant four are going to be negative. So if I'm taking the inverse tangent of a positive value, I know my solution has to be an angle in quadrant one. Whereas if I'm taking the inverse tangent of a negative value, my answer has to be an angle in quadrant four. Now here, since I'm taking the inverse tangent of the negative square root of three. I am looking for an angle in this bottom quadrant quadrant four. So for which of these angles is the tangent negative square root of three. Well, for the for negative pi over three, since this is a reference angle to pi over three, I know that the tangent of this value is going to be negative square root of three. So that represents my solution negative pi over three. So this is my solution. The inverse tangent of negative root three is negative pi over three. And we are good to go here. Thanks so much for watching and I'll see you in the next one.

Evaluate the expression.

$\tan^{-1}0$

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

Evaluate the expression.

$\tan^{-1}1$

• A. $0$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{2}$
• D. $-\frac{\pi}{4}$

Evaluate the expression.

$\tan^{-1}\left(-\frac{\sqrt3}{3}\right)$

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

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.

$\tan^{-1}\left(5\right)$

• A. – 5
• B. – 3.38
• C. 1.37
• D. 78.69

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.

$\sin^{-1}\left(-\frac13\right)$

• A. 0.34
• B. – 0.34
• C. – 0.33
• D. – 19.47

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.

$\cos^{-1}\left(\frac14\right)$

• A. 75.52
• B. 1.82
• C. 1.32
• D. 0.97