# Simplifying?

• September 20th 2010, 06:06 PM
DannyMath
Simplifying?
I'm a little rusty on my trig and I'm faced with this:

Simplify cos(sin^-1(x))

the -1 just means the inverse.

I've looked the answer up and I know it's sqrt(1-x^2) but I don't remember how to get that. Any help is appreciated.
• September 20th 2010, 06:24 PM
Prove It
Remember that $\cos{\theta} = \sqrt{1 - \sin^2{\theta}}$ (from the Pythagorean Identity).

So $\cos{(\sin^{-1}{x})} = \sqrt{1 - [\sin{(\sin^{-1}{x})}]^2}$

$= \sqrt{1 - x^2}$.
• September 20th 2010, 06:45 PM
DannyMath
Ok I understand it until the sin(sin^-1(x)). How does that turn into x specifically?
• September 20th 2010, 06:59 PM
mr fantastic
Quote:

Originally Posted by DannyMath
Ok I understand it until the sin(sin^-1(x)). How does that turn into x specifically?

By definition, $f(f^{-1}(x)) = f^{-1}(f(x)) = x$.
• September 20th 2010, 07:06 PM
DannyMath
Oh yes of course, thank you for the clarification! :)