# How do you do simplify sin^-1(cos ((5pi)/6)) ?

Nov 1, 2015

${\sin}^{- 1} \left(\cos \left(\frac{5 \pi}{6}\right)\right) = - \frac{\pi}{3}$

#### Explanation:

Since $\cos \left(\pi - x\right) = - \cos \left(x\right)$,
$\cos \left(\frac{5 \pi}{6}\right) = - \cos \left(\frac{\pi}{6}\right)$.

$\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ can be seen from the geometry of a 30,60,90 degree triangle, which is half of an equilateral triangle. Therefore,

$\cos \left(\frac{5 \pi}{6}\right) = - \frac{\sqrt{3}}{2}$.

Since ${\sin}^{- 1} \left(- x\right) = - {\sin}^{- 1} \left(x\right)$, and ${\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$ (which can be seen from the same 30,60,90 degree triangle),

${\sin}^{- 1} \left(\cos \left(\frac{5 \pi}{6}\right)\right) = {\sin}^{- 1} \left(- \frac{\sqrt{3}}{2}\right)$
$= - {\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right) = - \frac{\pi}{3}$