Sine of 45 Degrees/Proof 5

Theorem

 * $\sin 45^\circ = \sin \dfrac \pi 4 = \dfrac {\sqrt 2} 2$

where $\sin$ denotes the sine function.

Proof
The negative solution is rejected because $45^\circ$ is an acute angle and Sine of Acute Angle is Positive.

Therefore:
 * $\sin 45^\circ = \dfrac 1 {\sqrt 2} = \dfrac {\sqrt 2} 2$