How to show that $\cos(\sin^{-1}(x))$ is $\sqrt{1-x^2}$?

Question

How to show that $\cos(\sin^{-1}(x))$ is $\sqrt{1-x^2}$?

I remember having to draw a triangle, but I'm not sure anymore.

The function $\sin()$ takes a ratio, $x/1$ in this case, and returns an angle $\theta = \sin(x/1)$. Draw a triangle that models this, and then say what $\cos(\theta)$ is. — Mike Pierce, Aug 04 '18 at 01:33

score 4 · Answer 1 · edited Aug 04 '18 at 02:22

4

$\cos(\sin^{-1}(x)) = \cos(\sin^{-1}(x/1))$

Since sin is opposite hypotenuse we know $x$ is the opposite side and the hypotenuse is 1. To get cos we would need the adjacent side of the right triangle, so it equals $\sqrt{1^2 - x^2}$

edited Aug 04 '18 at 02:22

Simply Beautiful Art

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answered Aug 04 '18 at 01:44

Bas

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score 3 · Accepted Answer · answered Aug 04 '18 at 01:48

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

Let $\sin^{-1}x=\theta$ $$\implies\sin\theta=x\\\sin^2\theta=x^2\\1-\cos^2\theta=x^2\\\cos^2\theta=1-x^2\\\cos\theta=\pm\sqrt{1-x^2}\\\theta=\cos^{-1}\pm\sqrt{1-x^2}$$ Now, plug in $\theta=\cos^{-1}\pm\sqrt{1-x^2}$ in $\cos(\sin^{-1}x)$ and we get$$\pm\sqrt{1-x^2}$$But note that $\sin^{-1}x$ is in $\left[\dfrac{-\pi}{2},\dfrac{\pi}{2}\right]$ so $\cos(\sin^{-1}x)\ge0$

Therefore, $\cos(\sin^{-1}x)=\sqrt{1-x^2}$

How to show that $\cos(\sin^{-1}(x))$ is $\sqrt{1-x^2}$?

2 Answers2