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# Inverse trigonometric functions review

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

## What are the inverse trigonometric functions?

\arcsin, left parenthesis, x, right parenthesis, or sine, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, is the inverse of sine, left parenthesis, x, right parenthesis.
\arccos, left parenthesis, x, right parenthesis, or cosine, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, is the inverse of cosine, left parenthesis, x, right parenthesis.
\arctan, left parenthesis, x, right parenthesis, or tangent, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, is the inverse of tangent, left parenthesis, x, right parenthesis.

## Range of the inverse trig functions

minus, start fraction, pi, divided by, 2, end fraction, is less than or equal to, \arcsin, left parenthesis, theta, right parenthesis, is less than or equal to, start fraction, pi, divided by, 2, end fractionminus, 90, degrees, is less than or equal to, \arcsin, left parenthesis, theta, right parenthesis, is less than or equal to, 90, degrees
0, is less than or equal to, \arccos, left parenthesis, theta, right parenthesis, is less than or equal to, pi0, degrees, is less than or equal to, \arccos, left parenthesis, theta, right parenthesis, is less than or equal to, 180, degrees
minus, start fraction, pi, divided by, 2, end fraction, is less than, \arctan, left parenthesis, theta, right parenthesis, is less than, start fraction, pi, divided by, 2, end fractionminus, 90, degrees, is less than, \arctan, left parenthesis, theta, right parenthesis, is less than, 90, degrees
The trigonometric functions aren't really invertible, because they have multiple inputs that have the same output. For example, sine, left parenthesis, 0, right parenthesis, equals, sine, left parenthesis, pi, right parenthesis, equals, 0. So what should be sine, start superscript, minus, 1, end superscript, left parenthesis, 0, right parenthesis?
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.

Problem 1
The sine value of all options is 0, point, 98. Which is the principal value of \arcsin, left parenthesis, 0, point, 98, right parenthesis?

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. • The range of arctan cannot be from 0 to 180 because the 90 degrees angle is undefined and plus as you said the tans will be different in the first and second and in the third and fourth. tan(theta) can have infinite values and that is why several angles can function for many of the values of tan, therefore in order to have an inverse function, you need to restrict the values to the 1st and 4th quadrant in order to avoid 2 angles giving the same tan and so avoid making the function uninvertible. Plus I think you kinda answered your own question. Hopefully, this helps.
• 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! • 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).
• All I can figure out is what quadrant arcsin (0.98) is in. I don't understand where the 1.37 comes from. • 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).
• Hello.

For example, I know how to solve for arcsine(45 degrees), arcsin(3pi/4), and arcsin(square root of 3 / 2). But when it comes to solving for arcsin(0.98), I have no idea how to deal with the number 0.98 and I'm totally confused about how the process works. When I read the explanation down below the exercise 1, it didn't tell me how to get the range of [-pi/2, pi/2]. How would you know the range when you are dealing with the number 0.98? And can somebody explain how the process works to get the answer?

Best regards,
Alex • I think you are confusing sine and arcsine. The domain of arcsin(x) is [-1, 1], and neither 45º nor 3π/4 is in that set. However, sin(45º) and sin(3π/4) are both √2/2.

At this level, the only way to compute arcsin(0.98) is with a calculator. In calculus, you'll develop techniques to compute it by hand, but that's beyond the scope of precalculus.

Remember what happens to a functions graph when inverted: it's reflected across the line y=x. If we do this to sin(x), we get a wave going up the y-axis. But this isn't a function. So instead of reflecting the entirety of sin(x), we just reflect the part in [-π/2, π/2]. This smaller piece passes the vertical line test, and it's the largest possible piece that does so.

So because the domain of sin(x) is [-π/2, π/2], this is also the range of arcsin(x).
• Are there inverse trigonometric functions for the reciprocal trigonometric functions (secant,cosecant, and cotangent)? • why can I writearcsin(sin (4pi/3)) into arcsin(sin (-pi/3))?
(1 vote) • 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.   