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

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Feb 9, 2018

Answer:

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

Explanation:

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

#\sin(2\theta)=2\sin \theta \cos \theta#

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

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

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