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# The Inverse Sine Function - Problem 2

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

I’m talking about the inverse sine function and I’ve got the inverse sine function graphed here along with the restricted sine function which inverse sine function is the inverse of and I want to talk a little bit about their domains for a second. Remember that we had to restrict the domain of the sine function to between negative pi over 2 and pi over 2 so they would become one so that we can invert it.

Well the domain of the restricted sine function becomes the range of the inverse sine function. The domain of the inverse sine function between -1 and 1. So we need to remember the domains of these two functions when we’re writing down these inverse identities.

Now note that these are inverses of one another and so the inverse identities are going to hold. You remember those are f of f, inverse of x equals x and f inverse of f of x equals x. Now these identities are only true if x is within the domain of the inside function. In this case in the domain of f inverse, in this case in the domain of f. Let’s see how that applies to these two guys.

Well the first one would becomes sine of inverse sine of x equals x but in order for this to be true, x has to be in the domain of inverse sine. And the domain of inverse sine is between -1 and 1. So this is not true, if x is not in that interval. Secondly inverse sine of sine of x equals x as long as x is in the domain of the restricted sine function so it’s got to be between negative pi over 2 and pi over 2.

Let’s take a look at an example. The problem asks us to evaluate inverse sine of sine of pi over 4. Now here pi over 4 is in the domain of the restricted sine function, is between negative pi over 2 and pi over 2. This answer will be pi over 4. But here the identity doesn’t work, because this number is not between negative pi over 2 and pi over 2. If we want to evaluate this, let’s work from the inside out.

We’ll get inverse sine of what? The sine of 5 I over 6 is ½. And the we have to figure out what angle has a sine of ½ that’s between negative pi 2 over and pi over 2 and it’s pi over 6. So inverse sine of sine of 5pi over 6 is pi over 6.

Finally sine of inverse sine of 2, it’s not 2 because 2 is not in the domain of inverse sine. Domain of inverse sine is between -1 and 1, in fact inverse sine is not defined here so this is undefined. When you’re using the inverse identities always remember the domain of the inside function tells you whether the identity is true.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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