Question

could someone give me a refresher on what to do with somthing like this

\[\sin\left( 2\sin^{-1}\left( \frac{1}{2}\right)\right)\]

Answer 1

Start from inside out:
1/2 = sin 30°
-> 2arcsin1/2 = 2 * 30 = ...
=> sin (.. ) = ...

Answer 2

thanks, I made a typo though. I meant to say \[\sin\left( 2\sin^{-1}\left( \frac{x}{2}\right)\right)\]

Answer 3

\[y=\sin\left( 2\sin^{-1}\left( \frac{x}{2}\right)\right)\]\[\sin^{-1}y=2\sin^{-1}\left(\frac{x}{2}\right)\]

Answer 4

nope. super lost!

Answer 5

sin2(theta) = sin(theta)*cos(theta)

Answer 6

Use the half angle formula:
sin2a = 2 sina cosa
with a = arcsin (x/2)

Answer 7

so x/2 * (sqrt(4-x^2))/2

Answer 8

thanks! totally forgot how useful those triangle diagrams are!