Sin and inverse sin multiplied

[binalbean]

Homework Statement

Evaluate the following:

Sin^-1(sin(19*pi/7) in terms of pi

The Attempt at a Solution

i can't see any other way apart from the functions cancel to give 19pi/7 but checking on wolfram gives 2*pi/7, can anybody explain or help?

Thanks

 

[I like Serena]

Homework Statement

Evaluate the following:

Sin^-1(sin(19*pi/7) in terms of pi

The Attempt at a Solution

i can't see any other way apart from the functions cancel to give 19pi/7 but checking on wolfram gives 2*pi/7, can anybody explain or help?

Thanks

Welcome to PF, binalbean!

The inverse sine is defined to yield a result between -pi/2 and +pi/2.
So they do not just cancel.

The typical method is to draw a unit circle, find where the angle 19pi/7 is, and find the angle that has the same y coordinate but which is between -pi/2 and +pi/2.

 

[Mark44]

The equation f^-1(f(x)) = x is true only if x is in the domain of f, and f(x) is in the domain of f^-1.

For example, ln(e^-1) = -1, but e^ln(-1) is undefined. Here -1 is not in the domain of the natural log function.

 

[LCKurtz]

And a nit-pick. The Sin^-1 and sin aren't "multiplied". That is a composition of the two functions, or "function of a function".

 

[binalbean]

Thanks very much for your help! :D

 

[I like Serena]

Thanks very much for your help! :D

You're welcome. :)

The equation f^-1(f(x)) = x is true only if x is in the domain of f, and f(x) is in the domain of f^-1.

This problem is an example that these conditions are not sufficient.
x also has to be in the image of f^-1.

Btw, it's weird if f(x) is not in the domain of f^-1.