Sine of 45 Degrees/Proof 1

Theorem

$\sin 45 \degrees = \sin \dfrac \pi 4 = \dfrac {\sqrt 2} 2$

Proof

Let $ABCD$ be a square of side $r$.

By definition, each angle of $\triangle ABCD$ is equal to $90 \degrees$.

Let $AC$ be a diagonal of $ABCD$.

As $\triangle ABC$ is a right angled triangle, it follows from Pythagoras's Theorem that $AC = \sqrt 2 A B$.

As $AC$ is a bisector of $\angle DAB$ it follows that $\angle CAB = 45 \degrees$.

So by definition of sine function:

$\map \sin {\angle CAB} = \dfrac r {r \sqrt 2} = \dfrac {\sqrt 2} 2$

$\blacksquare$