# Why $\frac{1}{\cos(\sin^{-1}(x))}=\frac{1}{\sqrt{1-x^2}}$?

This is the inverse function of sin. Why is $\cos(\sin^{-1}x)=\sqrt{1-x^2}$?

Thanks a lot.

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$\cos\,x=\pm\sqrt{1-\sin^2 x}$... – J. M. Nov 18 '11 at 13:13

Draw a right angled triangle a marked angle $\theta$. We can scale the triangle to set the hypotenuse equal to 1. Label the side opposite $\theta$ by some length $x$, then the adjacent side has length $\sqrt{1-x^2}.$
$\phi = \sin^{-1} x$ denotes the angle such that $\sin \phi = x$. So in the triangle, $\sin^{-1} (x) = \theta.$
Then $\displaystyle \cos \theta = \frac{ \text{Adjacent} }{ \text{Hypotenuse} } = \frac{ \sqrt{1-x^2} }{1} = \sqrt{1-x^2}$ so $$\cos ( \sin^{-1} (x) ) = \sqrt{1-x^2}.$$
Can you give me a hint about how to get the $tan^{-1}$? – Andrew Nov 18 '11 at 14:08
Did you mean $\tan(\sin^{-1}(x))$? If $\sin^{-1}(x)=\theta, \tan(\sin^{-1}(x))=\frac{x}{\sqrt{1-x^2}}$ by the same argument. If you mean $\cos(\tan^{-1}(x))$, draw a triangle with opposite side $x$, adjacent side $1$, hypotenuse $\sqrt{1+x^2}$, so the result is $\cos(\tan^{-1}(x))=\frac{1}{\sqrt{1+x^2}}$ This all applies to angles in the first quadrant, where all signs are positive-otherwise you need to worry about minus signs. – Ross Millikan Nov 18 '11 at 15:08