# How do you prove cos(sin^-1x)=sqrt(1-x^2)?

Nov 15, 2016

For the conventional inverse sine, the answer is $\sqrt{1 - {x}^{2}}$. See explanation.

#### Explanation:

Let $a = {\sin}^{- 1} x \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ ( for the conventional inverse ),

wherein cosine is positive. Then, sin a = x.

The given expression is cos a = sqrt(1-sin^2a)=sqrt(1-x^2)#, for the

conventional ${\sin}^{- 1} x$.

The general value of $a \in \left(- \infty , \infty\right)$, and so, might be in any

quadrant, for an arbitrary $x \in \left[- 1 , 1\right]$. The cosine might be >=< 0.

So, here, the answer is $\pm \sqrt{1 - {x}^{2}}$