How do you prove Sin^-1(-1/2)= -pi/6?

Oct 2, 2016

See the Explanation.

Explanation:

The Defn. of ${\sin}^{-} 1$ fun. is :

${\sin}^{-} 1 x = \theta , | x | \le 1 \iff \sin \theta = x , \theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] .$

So, we have to find a $\theta ,$ which satisfies the conds., namely,

$\left(1\right) : - \frac{\pi}{2} \le \theta \le \frac{\pi}{2} , \text{ & } \left(2\right) : \sin \theta = - \frac{1}{2.}$

Knowing that, $\sin \left(- \frac{\pi}{6}\right) = - \sin \left(\frac{\pi}{6}\right) = - \frac{1}{2} ,$ and, noting

that, $- \frac{\pi}{6} \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$, we have,

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