Sine of 45 Degrees/Proof 1
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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:
- $\map \sin {\angle CAB} = \dfrac r {r \sqrt 2} = \dfrac {\sqrt 2} 2$
$\blacksquare$