 ###### 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.

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The Inverse Sine Function - Problem 2

# 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.

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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.