# How do you evaluate arcsin -1/2?

##### 1 Answer
May 29, 2015

Consider an equilateral triangle with sides of length 2.
Each of the internal angles will be $\frac{\pi}{3}$ (i.e. ${60}^{o}$).

Now split the triangle into two mirror image right angled triangles.
The shortest side of each will have length $1$, and the smallest angle opposite it will be $\frac{\pi}{6}$ (i.e. ${30}^{o}$).

Then by definition $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$ - the length of the shortest side divided by the length of the hypotenuse.

Now $\sin \left(- \theta\right) = - \sin \left(\theta\right)$, so

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

The range of $\arcsin$ is $- \frac{\pi}{2} \le \theta \le \frac{\pi}{2}$.

$- \frac{\pi}{6}$ lies in this range so $\arcsin \left(- \frac{1}{2}\right) = - \frac{\pi}{6}$