I need help with proving $\cos(\arcsin(x)) = \sqrt{1-x^2}$

Question

As I've said I need help with this particular trigonometric identity:

$$\cos(\arcsin(x)) = \sqrt{1-x^2}$$

I've tried finding a proof for it but I had no luck.

Any help would be much appreciated. Thanks in advance!

Answer 1

We have $\cos^2(\arcsin(x))+x^2=\cos^2(\arcsin(x))+\sin^2(\arcsin(x))=1$, hence

$\cos^2(\arcsin(x))=1-x^2$.

Can you proceed ?

Answer 2

Given a right triangle with hypotenuse $1$, angle $\theta$, opposite leg $x$ and adjacent leg $y$, we have $$ x^2 + y^2 = 1\\ \arcsin(x) = \theta\\ \cos(\theta) = y $$ If we put these together in the right order, we get your identity.

Answer 3

Geometrically, for $|x|\leq 1,$ in $\Bbb R^2 $ let $O=(0,0), $ $ A=(\sqrt {1-x^2}\;, x), $ and $ B=(\sqrt {1-x^2},x). $ Then $\angle AOB$ (evaluated as negative if $x<0$) is $\arcsin x.$ And $|\angle AOB|\leq \pi /2.$ So $\cos \angle AOB= OB/OA=OB=\sqrt {1-x^2}\;.$

Answer 4

If you set $\alpha=\arcsin x$, then $x=\sin\alpha$ and $-\pi/2\le\alpha\le\pi/2$.

Therefore the proposed identity becomes $$ \cos\alpha=\sqrt{1-\sin^2\alpha} $$ which is true because $\cos\alpha\ge0$ when $-\pi/2\le\alpha\le\pi/2$ and we know that $\cos^2\alpha=1-\sin^2\alpha$.

^{Be careful in spelling out the condition $\alpha\in[-\pi/2,\pi/2]$, because a claim such as $\cos\alpha=\sqrt{1-\sin^2\alpha}$ is false in general.}

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The analytic proof considering $f(x)=\sqrt{1-x^2}-\cos\arcsin x$ seems a bit more complicated, but here it is: $$ f'(x)=-\frac{x}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-x^2}}\sin\arcsin x=0 $$ for $-1<x<1$. By continuity, the function $f$ is constant on $[-1,1]$. Since $f(0)=0$ we are done.