# What is sin(2*sin^-1(x))?

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#### Explanation

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#### Explanation:

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Feb 9, 2018

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

#### Explanation:

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

$\setminus \sin \left(2 \setminus \theta\right) = 2 \setminus \sin \setminus \theta \setminus \cos \setminus \theta$

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

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

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