Prove that $\sin(2\sin^{-1}(\alpha)) = 2\alpha \sqrt{-\alpha^2+1}$.
I was doing a trigonometric substitution problem in Calculus and came across this and wanted to know the proof of it.
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This should be helpful. – Git Gud Dec 18 '15 at 22:33
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Hint: Let $\alpha = \sin \phi$, then $$ 2\alpha\sqrt{1-\alpha^2} = 2\sin\phi|\cos\phi| = \sin2\phi \text{ or } -\sin2\phi $$
Henricus V.
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It should be $2\sin \phi|\cos \phi|$. Though you can let $\phi\in[-\pi/2,\pi/2]$. – user236182 Dec 18 '15 at 22:30
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There exists a $\phi\in[-\pi/2,\pi/2]$ such that $\alpha =\sin \phi$. Then you'll get $\sin 2\phi$. – user236182 Dec 18 '15 at 22:34
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Hint: use the double angle formula, $\sin{2 p} = 2 \sin{p} \cos{p}$ and then use the fact that $\sin^2 p + \cos^2 p = 1$.
Dmoreno
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Double angular formel: $sin(2p)=2*cos(p)*sin(p)=2*\sqrt{1-sin^2(p)}*sin(p)$
Let $\;p=sin^{-1}(\alpha)\quad$ and you get your result
XPenguen
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It's only true if $p\in[-\pi/2+2\pi k,\pi/2+2\pi k],, k\in\mathbb Z$. $, \sqrt{1-\sin^2 x}=|\cos x|$. – user236182 Dec 18 '15 at 22:48