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# The Inverse Sine Function - Problem 1

###### 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 want to talk more about the inverse sine function. Remember that the inverse sine function is the inverse of the restricted sine function. Not the whole sine function. This is a little piece of the sine function going from negative pi over 2 of pi over 2. Of course the regular sine function continues in both directions but we needed to restrict the sine function to a one to one piece.

Do you remember how to invert the graph of a function? You basically switch the coordinates x and y so for example on the inverse sine function, (0,0) would stay put. Pi over 2,1 becomes 1,pi over 2, about there and -pi over 2,-1 becomes -1,-pi over 2. I want to try to draw the inverse graph.

Imagine the reflection of this graph around the line y equals x, so something like this. Here we go. This is a graph of the inverse sine function. Remember y equals inverse sine of x has as its domain the numbers between -1 and one and has its range, remember that it's output will always be an angle between negative pi over 2 and pi over 2. That’s the range.

Let’s take a look at the formal definition of inverse sine, I have it written up here. Y equals inverse of x means x equals sine y but only if y is between -pi over 2 and pi over 2. That’s from the domain restriction. But the way to think of the inverse is that y which is the inverse sine value, is the number in the interval between negative pi over 2 and pi over 2 whose sine is x.

So for example, if we want to evaluate these guys here, I want to find this inverse sine, its value will be the number between negative pi over 2 and pi over 2 whose sine is ½. Now I happen to know that sine of pi over 6 is ½ and pi over 6 is in the interval negative pi over 2 to pi over 2. The inverse sine of ½ would be pi over 6. You got to be careful. The sine of 5pi over 6 is also ½. But the inverse sine of ½ is not 5pi over 6 because it’s not in the interval negative pi over 2 to pi over 2.

Let’s try another example. Inverse sine of -1. I need to find a value between negative pi over 2 and pi over 2 whose sine is -1. And if you recall sine of negative pi over 2 is -1 and of course negative pi over 2 is in this interval because it’s one of the endpoints. It’s the left endpoint of the interval therefore inverse sine of -1 equals negative pi over 2.

When you’re dealing with functions and their inverses, inputs and outputs switch, that’s the whole point of functions is that they’re sort of un-doers of each other. And the thing to always remember for the trig functions the input is an angle and the output is a number. For inverse trig functions think of the input as a number and the output as an angle.

The way to think about inverse sine is to remember that you can always rewrite an inverse sine equation in the form a of a sine equation and think of the inverse sine as the number in the interval between negative pi over 2 and pi over 2 whose sine is x.

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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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