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## Precalculus

### Course: Precalculus>Unit 2

Lesson 3: Inverse trigonometric functions

# Inverse trigonometric functions review

Review your knowledge of the inverse trigonometric functions, arcsin(x), arccos(x), & arctan(x).

## What are the inverse trigonometric functions?

$\mathrm{arcsin}\left(x\right)$, or ${\mathrm{sin}}^{-1}\left(x\right)$, is the inverse of $\mathrm{sin}\left(x\right)$.
$\mathrm{arccos}\left(x\right)$, or ${\mathrm{cos}}^{-1}\left(x\right)$, is the inverse of $\mathrm{cos}\left(x\right)$.
$\mathrm{arctan}\left(x\right)$, or ${\mathrm{tan}}^{-1}\left(x\right)$, is the inverse of $\mathrm{tan}\left(x\right)$.

## Range of the inverse trig functions

RadiansDegrees
$-\frac{\pi }{2}\le \mathrm{arcsin}\left(\theta \right)\le \frac{\pi }{2}$$-{90}^{\circ }\le \mathrm{arcsin}\left(\theta \right)\le {90}^{\circ }$
$0\le \mathrm{arccos}\left(\theta \right)\le \pi$${0}^{\circ }\le \mathrm{arccos}\left(\theta \right)\le {180}^{\circ }$
$-\frac{\pi }{2}<\mathrm{arctan}\left(\theta \right)<\frac{\pi }{2}$$-{90}^{\circ }<\mathrm{arctan}\left(\theta \right)<{90}^{\circ }$
The trigonometric functions aren't really invertible, because they have multiple inputs that have the same output. For example, $\mathrm{sin}\left(0\right)=\mathrm{sin}\left(\pi \right)=0$. So what should be ${\mathrm{sin}}^{-1}\left(0\right)$?
In order to define the inverse functions, we have to restrict the domain of the original functions to an interval where they are invertible. These domains determine the range of the inverse functions.
The value from the appropriate range that an inverse function returns is called the principal value of the function.
Want to learn more about arcsin(x)? Check out this video.
Want to learn more about arccos(x)? Check out this video.
Want to learn more about arctan(x)? Check out this video.

## Check your understanding

Problem 1
The sine value of all options is $0.98$. Which is the principal value of $\mathrm{arcsin}\left(0.98\right)$?
All measures are in radians.
Choose 1 answer:

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• Can the range of arctan also be from 0 to 180? If we divide the circle into quadrants where 0-90 is 1st, 90-180 2nd, 180-270 3rd and 270-360 4th quadrant than the 1st and 3rd have the same tan and the 2nd and 4th have the same tan. As I understand it range is determined so there is only one value for each tan and 1st and 2nd quadrant don't give the same tan.
(30 votes)
• Right, but if we define the range as [0, π], then we would actually have to exclude π/2 because tan π/2 is undefined. Also we would exclude either 0 or π because both of those angles have the same tangent. Therefore, we would have to write: (0, π/2) ∪ (π/2, π] or [0, π/2) ∪ (π/2, π). Since it is simpler to write (-π/2, π/2), this is what we use for the range of arctan.
(77 votes)
• I don't understand this particular unit
(41 votes)
• Fr they could atleast go into detail the steps they use to get the principal value using an example but instead we are to fill in the blanks
(9 votes)
• I've gone through Algebra I & II and half of Geometry and all the Trig up until here, and I've understood everything, but this makes zero sense to me. I get the domains of each function, but how should we know how many radians cos−1(0.32)? Or how can we know what tan−1(−21) is and whether it's inside the domain of pi/2 and -pi/2? And why are the radians not in terms of pi? What's 1.62 radians? Are we supposed to assume there are roughly 6.28 radians in a circle and estimate whether each of the answers would be inside of the domain of the function?
(15 votes)
• A lot of questions here! Let me address them one at a time.

"how should we know how many radians cos−1(0.32)"

We can't, at least not without a calculator. You can approximate the value using something known as a Taylor Polynomial, but at this stage, punch it into a calculator.

" Or how can we know what tan−1(−21) is and whether it's inside the domain of pi/2 and -pi/2"

Pretty much the same thing. Use a calculator. As for the "range" (not domain. The domain of arctan(x) is all reals), find the answer first, which comes to -1.5232. Now, the range of arctan(x) is (-1.57,1.57). Clearly, our answer falls in this interval. That's how you check for it.

"And why are the radians not in terms of pi? What's 1.62 radians?"

Thing is, they are in terms of pi. For example, if my question was "what is the value of arccos(0)?", usually you'd say pi/2 radians. But, one can also say 1.57 radians, and they'd be right. You'll mostly deal with radians having pi in them (because it helps in scaling a big fraction to a manageable one) but if you see a decimal, you'd probably want to use a calculator.

"Are we supposed to assume there are roughly 6.28 radians in a circle and estimate whether each of the answers would be inside of the domain of the function?"

Pretty much, actually. Knowing the decimal values of the multiples of pi isn't bad. So, if you know that the range of a function is (-1.57,1.57) and your answer comes to 1.5, you can be sure that you're right. But from my experience, in higher classes, you'll rarely see values of radians in decimals, as fractions of pi are much easier to deal with.
(31 votes)
• How do you find the sin(arctan(2)) without using a calculator? 2 is not a value on the unit circle, so I'm not sure how to do this problem by hand. Another problem like this would be cos(arcsin(1/4)). Is there a way to do these types of problems using right triangles? If so, how?
Thank you!
(8 votes)
• sin(arctan 2)
Let's take a look at what this question is really asking for. First, we are taking the arctangent of 2. That is, we are finding some angle 𝜃 such that tan 𝜃 = 2. Then, after finding that angle, we are taking the sine of that angle! In other words, if we have an angle 𝜃 such that tan 𝜃 = 2, we must find sin 𝜃.

This is possible to do with a right triangle. If 𝜃 is an angle in a right triangle such that the opposite side is 2 and the adjacent side is 1, then tan 𝜃 = 2. Then, the sine of this angle is 2/ℎ such that ℎ is the hypotenuse of the triangle. By the Pythagorean theorem, the hypotenuse is √5. Thus, sin 𝜃 = 2/√5 and we have:
sin(arctan 2) = 2/√5
You can use a similar argument for the second question. You could have also solved this using the Pythagorean trig identities (which are basically condensed versions of the process we used above).
(17 votes)
• How did you guys figure out the trigonometric functions?
By using calculaters?
(3 votes)
• All the answers for these excercises are done using elimination of options while thinking about the restricted domain of the functions
(12 votes)
• All I can figure out is what quadrant arcsin (0.98) is in. I don't understand where the 1.37 comes from.
(3 votes)
• 1.37 is the angle in radians (in degrees it is approximately 78.52º) in which its sine is equal to 0.98 and, in fact, it is also the first angle that has sine of 0.98 if you follow the trigonometric circle counterclockwise from 0 radians (0º) to 2pi radians (360º). Therefore, it is the principal value of arcsin(0.98), which can also be written as sin^-1(0.98).
(10 votes)
• Are there inverse trigonometric functions for the reciprocal trigonometric functions (secant,cosecant, and cotangent)?
(3 votes)
• Yes, there are! Here are all of the inverse trigonometric functions:

Name   commonly seen as
arcsine ----------- arcsin or sin⁻¹
arccosine -------- arccos or cos⁻¹
arctangent ------ arctan or tan⁻¹
arccosecant ----- arccsc or csc⁻¹  ←  inverse reciprocal
arcsecant -------- arcsec or sec⁻¹  ←  inverse reciprocal
arccotangent ---- arccot or cot⁻¹  ←  inverse reciprocal

Learn more about all of the inverse trigonometric functions:
https://en.wikipedia.org/wiki/Inverse_trigonometric_functions
(5 votes)
• why are they call arc? I looked it up on wiki and cant find more other source :(
(1 vote)
• I believe it is because the inverse trig functions give the measure of the angle, which if it's in radians and in the unit circle, is also the length of the arc made by that angle.
(8 votes)
• What do you do when the angle isn't within the domain? Do you have to reflect it?
(1 vote)
• You subtract the amount from the domain and work with what is left.

To use an simple example: imagine you were given a function f(x) = x + 1 and told that x must be more than 0 and less than 10. Now imagine you were given x = 1985 and told to plug it into that function. you would subtract 10 from 1985 until you got a number that fit the domain. In this case that would be five (AKA the remainder after dividing by ten) and you would get f(x) = 5 + 1 = 6

I know that seems really complicated. It comes from a field of mathematics called modular arithmetic which is often used in cryptography (that's the field of codes and ciphers). It isn't as complicated as it sounds but is very hard to explain without diagrams. Khan academy has a section on modular arithmetic under the cryptography section in computer science.
(7 votes)
• what's the difference between an inverse function and a reciprocal function?
(2 votes)
• Lets start with the words inverse and reciprocal without dealing with functions first. So an inverse sort of cancels another operation, so you can have an additive inverse (3 and -3), a multiplicative inverse (3 and 1/3), or even the root/power inverse. A reciprocal is basically the same as the multiplicative inverse. When you add the word function, an inverse function undoes another function so f(g(x))=g(f(x))=x. So inverse functions are related to some original function, but inverse relationships do not always have to be a function. A reciprocal function has a constant in the numerator and an expression usually with a variable in the denominator. It is not dependent on some other function, but you could find the inverse of a reciprocal function.
(5 votes)