# How do you simplify cos(sin^-1x)?

Aug 11, 2017

After using $u = A r c \sin \left(x\right)$ or $x = \sin u$ transforms,

$\cos \left(\arcsin x\right)$

=$\cos u$

=$\sqrt{1 - {\left(\sin u\right)}^{2}}$

= $\sqrt{1 - {x}^{2}}$

Aug 11, 2017

$\cos \theta = \sqrt{1 - {x}^{2}}$

#### Explanation:

$\cos \left({\sin}^{-} 1 x\right)$

Let   ${\sin}^{-} 1 x = \theta$

Then, $x = \sin \theta$

Now put the value for $x$ in $\cos \left({\sin}^{-} 1 x\right)$

$\implies \cos \left({\sin}^{-} 1 \left(\sin \theta\right)\right)$

So the equation becomes,

$\implies \cos \theta$

We know that ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

$\implies {\cos}^{2} \theta = 1 - {\sin}^{2} \theta$

$\implies \cos \theta = \sqrt{1 - {\sin}^{2} \theta}$

We have already found that $x = \sin \theta ,$ then ${x}^{2} = {\sin}^{2} \theta .$ Now put ${x}^{2}$ in the place for ${\sin}^{2} \theta .$

$\cos \theta = \sqrt{1 - {x}^{2}}$