Use the identity $\cos^2x+ \sin^2x=1$ to show that $\cos(\arcsin x)=\sqrt{1-x^2}$

Question

Use the identity $\cos^2x+ \sin^2x=1$ to show that $\cos(\arcsin x)=\sqrt{1-x^2}$

How do I begin? I have no idea where to start...

Answer 1

Let $\displaystyle\arcsin x=\theta\implies\sin\theta= x$

and $\displaystyle-\frac\pi2\le \theta\le \frac\pi2$ (Principal value)

$\displaystyle\cos(\arcsin x)=\cos\theta=+\sqrt{1-\sin^2\theta}$

Answer 2

Hint: put $\arcsin(x)=y$, so $\sin(y)=x$. Now use the formula to express $\cos(y)$ in terms of $x$.

Answer 3

We know that $\cos^2 x + \sin^2 x = 1$. Subtract $\sin^2x$ from both sides to get $$\cos^2 x = 1 - \sin^2 x.$$ Take the square root to get $$\cos x = \pm \sqrt{1 - \sin^2 x}.$$ Now let $x = \arcsin y$, and you have your formula :) (since $\sin^2(\arcsin y))=y^2$).