# How do you find the exact value for sin^-1 (cos (2pi/3))?

Mar 29, 2018

${\sin}^{-} 1 \left(\cos \left(2 \frac{\pi}{3}\right)\right) = 7 \frac{\pi}{6} , 11 \frac{\pi}{6}$
Among which the first positive solution happens to be

${\sin}^{-} 1 \left(\cos \left(2 \frac{\pi}{3}\right)\right) = 7 \frac{\pi}{6}$

#### Explanation:

sin^-1(cos(2pi/3))=?

$2 \frac{\pi}{3} = \pi - \frac{\pi}{3}$

$\cos \left(2 \frac{\pi}{3}\right) = \cos \left(\pi - \frac{\pi}{3}\right)$

$\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$

$\cos \left(\pi - \frac{\pi}{3}\right) = \cos \pi \cos \left(\frac{\pi}{3}\right) + \sin \pi \sin \left(\frac{\pi}{3}\right)$

$\cos \pi = - 1$

$\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$

$\sin \pi = 0$

$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

$\cos \left(\pi - \frac{\pi}{3}\right) = - 1 \times \frac{1}{2} + 0 \times \frac{\sqrt{3}}{2}$

$\cos \left(2 \frac{\pi}{3}\right) = - \frac{1}{2}$

${\sin}^{-} 1 \left(\cos \left(2 \frac{\pi}{3}\right)\right) = {\sin}^{-} 1 \left(- \frac{1}{2}\right)$

Let

$u = {\sin}^{-} 1 \left(- \frac{1}{2}\right)$

$\sin u = - \frac{1}{2}$

$\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$
sine ratio is negative in 3rd and 4th quadrants

For angles u<2pi,

$\sin \left(\pi + \frac{\pi}{6}\right) = - \sin \left(\frac{\pi}{6}\right)$->

$u = \pi + \frac{\pi}{6} = 7 \frac{\pi}{6}$

$\sin \left(2 \pi - \frac{\pi}{6}\right) = - \sin \left(\frac{\pi}{6}\right)$->

$u = 2 \pi - \frac{\pi}{6} = 11 \frac{\pi}{6}$

Hence,
Exact value of

${\sin}^{-} 1 \left(\cos \left(2 \frac{\pi}{3}\right)\right) = 7 \frac{\pi}{6} , 11 \frac{\pi}{6}$