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richyw
could someone give me a refresher on what to do with somthing like this \[\sin\left( 2\sin^{-1}\left( \frac{1}{2}\right)\right)\]
Start from inside out: 1/2 = sin 30° -> 2arcsin1/2 = 2 * 30 = ... => sin (.. ) = ...
thanks, I made a typo though. I meant to say \[\sin\left( 2\sin^{-1}\left( \frac{x}{2}\right)\right)\]
\[y=\sin\left( 2\sin^{-1}\left( \frac{x}{2}\right)\right)\]\[\sin^{-1}y=2\sin^{-1}\left(\frac{x}{2}\right)\]
sin2(theta) = sin(theta)*cos(theta)
Use the half angle formula: sin2a = 2 sina cosa with a = arcsin (x/2)
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so x/2 * (sqrt(4-x^2))/2
thanks! totally forgot how useful those triangle diagrams are!