Question

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

Answer 1

`-60°`

Explanation:

A right-angled triangle with hypotenuse `1` and one side `sqrt(3)/2` has the third side of length `1/4` by Pythagoras. Draw a unit circle centred on the origin `O` and sketch a regular hexagon in it, with vertices `(1,0)`, `(1/2,sqrt(3)/2)`, `(-1/2, sqrt(3)/2)`, `(-1,0)`, `(-1/2,-sqrt(3)/2)`, `(1/2,-sqrt(3)/2)`. 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{PON}=sin^-1(sqrt(3)/2)`. But clearly `hat{PON}`=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°`.

There are many other equivalent ways of visualizing this result.