# How do you evaluate sin^-1(-sqrt3/2) without a calculator?

-60°
A right-angled triangle with hypotenuse $1$ and one side $\frac{\sqrt{3}}{2}$ has the third side of length $\frac{1}{4}$ by Pythagoras. Draw a unit circle centred on the origin $O$ and sketch a regular hexagon in it, with vertices $\left(1 , 0\right)$, $\left(\frac{1}{2} , \frac{\sqrt{3}}{2}\right)$, $\left(- \frac{1}{2} , \frac{\sqrt{3}}{2}\right)$, $\left(- 1 , 0\right)$, $\left(- \frac{1}{2} , - \frac{\sqrt{3}}{2}\right)$, $\left(\frac{1}{2} , - \frac{\sqrt{3}}{2}\right)$. Let $P$ the last vertex in this list and let $N$ be the foot of the perpendicular from $P$ to the $x$-axis. Then, disregarding signs, angle $\hat{P O N} = {\sin}^{-} 1 \left(\frac{\sqrt{3}}{2}\right)$. But clearly $\hat{P O N}$=360°/6 by symmetry.
Applying the definition of "principal value" of the inverse sin function (which requires the angle to be within the inclusive range +-90°) you get -60°#.