Inverse Sine, Cosine, & Tangent - Online Tutor, Practice Problems & Exam Prep

Inverse trigonometric functions, such as inverse sine, cosine, and tangent, allow us to find angles corresponding to specific values. For inverse cosine, the output angle is restricted to the interval [0, π], while inverse sine and tangent output angles are limited to [-π/2, π/2]. Evaluating expressions like $\cos \left(\theta \right)=1/2,$ involves identifying angles on the unit circle that meet these criteria, ensuring a clear understanding of their respective domains and ranges.

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 inverse 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 that you can think of this, and this will work for any inverse trig function, is for the inverse cosine of one half, I can also think of this as, okay, 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 1, for my angle π/3, this has a corresponding cosine value of one half. So this π/3 represents an angle that gives me my corresponding value. Now, on my unit circle, as I continue around, I also know that in quadrant 4, all of my cosine values are positive here as well. So for my angle of 5π/3, that also has a cosine value of positive one half. So 5π/3 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. 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 one to one function, if I were to reflect this entire cosine graph over that line y equals x, 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 0 and pi. So if I 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 0 and pi. So when faced with evaluating an inverse cosine expression, our solution can only be angles within the interval 0 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 0 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 5π/3 is right down here in quadrant 4, but this cannot be a solution because it is not within my specified interval. So that leaves me with my final answer of π/3, which is up here in quadrant 1 between 0 and pi. So my final answer here is that the inverse cosine of one half is π/3. 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 0 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 over 2. Now, whenever working with an inverse trigonometric function, remember that we can also think of this as OK, the cosine of what angle is equal to negative root 3 over 2, and we want to find the angle for which that is true. Now, when working with the inverse cosine, we know that our angles can only be between 0 and π. But we can actually get even more specific here because we know that all our cosine values are going to be positive in quadrant 1, and all our cosine values in quadrant 2 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 1. 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 2. 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 over 2? Well, I know that the cosine of 5π over 6 is equal to just that so that represents my solution. My angle here is 5π over 6, 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 = 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 −π2 and π2. So flipping that, reflecting it over the line y = x, I end up with a graph that looks something like this for my inverse sine 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. The angles that we get back out have to be between −π2 and π2. So now with this interval in mind, we can go ahead and evaluate inverse sine expressions with no problem.

So let's come down here to our first example, which asks us to evaluate the expression, the inverse sine of $\frac{1}{2}$. Now remember when working with inverse trig 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, okay, the sine of what angle θ gives me a value of $\frac{1}{2}$. 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 negative one and positive one, which $\frac{1}{2}$ definitely is. And the angle that we get back out for our answer has to be between −π2 and π2. So let's come down to our unit circle here. Now on our unit circle, we want to find an angle θ for which our sine value is $\frac{1}{2}$. And looking at my unit circle here, I see that for my angle of π6, I end up with a sine value of $\frac{1}{2}$. So this represents the solution to my expression, π6. Now, you may be tempted here to come over to 5π6 and think, Okay, isn't the sine of 5π6 also equal to $\frac{1}{2}$? And it is, but the problem is that this is not within our specified interval from −π2 to π2. So I actually don't care about that half of my unit circle at all when dealing with the inverse sine. I only want to consider angles on this half of my unit circle between −π2 and π2. So this solution, π6, is my final solution here, and it's my only solution to the expression inverse sine of $\frac{1}{2}$.

Let's come to our next example here, which is the inverse sine of negative root 2 over 2. Now remember, we can also think of this as, Okay, the sine of what angle θ gives me a value of negative root 2 over 2. And coming over here to my unit circle, I see that down here in quadrant 4, for negative π4, I know that my sine value is going to be negative root 2 over 2. But something else that you might have noticed here is that all of these values in quadrant 4 are negative. And on a typical unit circle, this negative π4 would actually be 7π4. So what's going on here? Well, the problem is is that 7π4 is not within my specified interval of −π2 to π2. So whenever dealing with the inverse sine function, we consider this quadrant 4 as being measured clockwise from 0 rather than counterclockwise so that all of our values are within that interval, −π2 to π2. So here my solution is negative π4 within that specified interval. So the inverse sine of negative root 2 over 2 is negative π4. So now that we know how to evaluate inverse sine 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 sine of negative one. Now remember, when working with inverse trig functions, we can also think of this as, "Okay, the sine of what angle is equal to negative 1?" and we're looking for the angle for which that is true.

When working with the inverse sine specifically, we know that our solution can only be angles between negative pi over 2 and positive pi over 2, but we can get even more specific here. Because all of my sine values in the first quadrant, quadrant 1, from 0 to pi over 2, are going to be positive. Whereas in the fourth quadrant, from negative pi over 2 to 0, all of these sine values are going to be negative.

So if I'm taking the inverse sine of a positive value, I know that my solution has to be in quadrant 1. Whereas if I'm taking the inverse sine of a negative value, like we are here, the inverse sine of negative one, we know that our solution has to be in quadrant 4 between negative pi over 2 and 0.

So here, since we're taking the inverse sine of negative one, I already know that my solution has to be down here in this fourth quadrant. So, where is the sine equal to negative one? Well, for negative pi over 2, I know that my sine or my y value is equal to negative one. So here, my angle and my solution to this expression is negative pi over 2, and we are done here. Thanks for watching, and I'll see you in the next one.

− π 1 2

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 in 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, I, of course, need to reflect this over the line y = 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 2 and positive pi over 2. So reflecting this over the line y = 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 negative infinity and infinity. But the angles that we get back out are restricted. They must be in between negative pi over 2 and positive pi over 2.

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 3. Now, remember, for any inverse trig functions, we can also think of this as, okay, the tangent of what angle theta gives me a value of root 3. Now looking at our unit circle here, the specific interval that I am working with is between negative pi over 2 and positive pi over 2. So I really just don't care what's going on in this half of my graph at all. I am only looking for an angle for which the tangent is the square root of 3 on this half of my graph. And it's actually the same exact interval we already saw when working with the inverse sine. So let's go ahead and find our angle for which the tangent is the square root of 3. Now looking in this first quadrant here, I see that for pi over 3 my tangent is the root is root 3. So that tells me that my answer here is pi over 3. 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, okay, the tangent of what angle theta gives me a value of negative 1, and we want to find that angle within our specified interval. So in my interval from negative pi over 2 to positive pi over 2, where is my tangent negative 1? Well, I know that for my angle negative pi over 4, my tangent is going to end up being negative 1. So this gives me my final answer. The inverse tangent of negative one is negative pi over 4, and we're done here.

Now, as you continue to work through problems, you may have to evaluate some inverse trig functions using a calculator, either because you're explicitly asked to 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 sine, you would press the second button and then sine; for inverse cosine, the second button and then cosine; or for inverse tangent, the second button 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 radian 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 3. Now, when working with inverse trig functions, we know that we can also think of this as, okay, the tangent of what angle is equal to negative root 3, 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 between negative pi over 2 and pi over 2, which is actually the exact same interval that we dealt with when working with the inverse sine. Now again here, we could get even more specific because all of my tangent values in quadrant 1 are going to be positive whereas all of my tangent values in quadrant 4 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 1. Whereas if I'm taking the inverse tangent of a negative value, my answer has to be an angle in quadrant 4. Now here, since I'm taking the inverse tangent of the negative square root of 3, I am looking for an angle in this bottom quadrant, quadrant 4. So for which of these angles is the tangent negative square root of 3? Well, for the for negative pi over 3, since this is a reference angle to pi over 3, I know that the tangent of this value is going to be negative square root of 3. So that represents my solution, negative pi over 3. π/3. So this is my solution, the inverse tangent of negative root 3 is −π/3, 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

Here’s what students ask on this topic:

What is the inverse cosine of 1/2?

The inverse cosine of 1/2 is the angle whose cosine value is 1/2. Using the unit circle, we find that cos(π/3) = 1/2. Since the range of the inverse cosine function is [0, π], we only consider angles within this interval. Therefore, the inverse cosine of 1/2 is π/3.

How do you evaluate the inverse sine of -√2/2?

To evaluate the inverse sine of -√2/2, we need to find the angle whose sine value is -√2/2. On the unit circle, sin(-π/4) = -√2/2. The range of the inverse sine function is [-π/2, π/2], so we only consider angles within this interval. Therefore, the inverse sine of -√2/2 is -π/4.

What is the inverse tangent of √3?

The inverse tangent of √3 is the angle whose tangent value is √3. Using the unit circle, we find that tan(π/3) = √3. The range of the inverse tangent function is [-π/2, π/2], so we only consider angles within this interval. Therefore, the inverse tangent of √3 is π/3.

How do you find the inverse cosine of -1?

To find the inverse cosine of -1, we need to determine the angle whose cosine value is -1. On the unit circle, cos(π) = -1. The range of the inverse cosine function is [0, π], so we only consider angles within this interval. Therefore, the inverse cosine of -1 is π.

What is the inverse sine of 1/2?

The inverse sine of 1/2 is the angle whose sine value is 1/2. Using the unit circle, we find that sin(π/6) = 1/2. The range of the inverse sine function is [-π/2, π/2], so we only consider angles within this interval. Therefore, the inverse sine of 1/2 is π/6.