Question #13738

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Question #13738

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Answer 1 · 2018-02-09T15:43:16

$sin (2 {sin}^{- 1} (x)) = 2 x \sqrt{1 - x^{2}}$ $\sin(2\sin^{-1}(x))=2x\sqrt{1-x^2}$ .

Explanation:

This requires a little care. The best approach is to define $θ = {sin}^{- 1} (x)$ $\theta=\sin^{-1}(x)$ so that \sin \theta=x#. Then we can put in the double angle formula

$sin (2 θ) = 2 sin θ cos θ$ $\sin(2\theta)=2\sin \theta \cos \theta$

We have $sin θ = x$ $\sin \theta=x$ and from the Pythagorean identity, $cos θ = \pm \sqrt{1 - x^{2}}$ $\cos \theta=\pm\sqrt{1-x^2}$ . But also, for the inverse sine function, $- (\frac{π}{2}) \leq θ \leq (\frac{π}{2})$ $-(\pi/2)\le \theta \le (\pi/2)$ so $cos θ \geq 0$ $\cos \theta \ge 0$ . So then

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

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Question #13738

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