Sine of Right Angle

Theorem

$\sin 90^\circ = \sin \dfrac \pi 2 = 1$

where $\sin$ denotes the sine function.

Proof

A direct implementation of Sine of Half-Integer Multiple of Pi:

$\forall n \in \Z: \sin \left({n + \dfrac 1 2}\right) \pi = \left({-1}\right)^n$

In this case, $n = 0$ and so:

$\sin \dfrac 1 2 \pi = \left({-1}\right)^0 = 1$

$\blacksquare$