# How do you solve and find the value of sin(2sin^-1(1/2))?

Dec 21, 2016

$\pm \frac{\sqrt{3}}{2}$

#### Explanation:

$\sin \left(2 {\sin}^{-} 1 \left(\frac{1}{2}\right)\right) - \to \sin \left(2 \arcsin \left(\frac{1}{2}\right)\right)$
Trig table and unit circle give:
$\sin x = \frac{1}{2}$ --> arc $x = \frac{\pi}{6}$ and arc $x = \frac{5 \pi}{6}$

a. $\sin 2 x = \sin \left(2 \frac{\pi}{6}\right) = \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
b. $\sin 2 x = \sin \left(2 \frac{5 \pi}{6}\right) = \sin \left(\frac{5 \pi}{3}\right) = - \sin \left(\frac{\pi}{3}\right) = - \frac{\sqrt{3}}{2}$
Finally:
$\sin \left(2 {\sin}^{-} 1 \left(\frac{1}{2}\right)\right) = \pm \frac{\sqrt{3}}{2}$