Sine of 45 Degrees/Proof 3

Theorem

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

$\ds \sin 45 \degrees$

$=$

$\ds \map \sin {3 \times 15 \degrees}$

$\ds $

$=$

$\ds 3 \sin 15 \degrees - 4 \sin^3 15 \degrees$

$\ds $

$=$

$\ds 3 \paren {\frac {\sqrt 6 - \sqrt 2} 4} - 4 \paren {\frac {\sqrt 6 - \sqrt 2} 4}^3$

$\ds $

$=$

$\ds \frac {3 \sqrt 6 - 3 \sqrt 2} 4 - \frac {6 \sqrt 6 - 18 \sqrt 2 + 6 \sqrt 6 - 2 \sqrt 2} {16}$

$\ds $

$=$

$\ds \frac {3 \sqrt 6 - 3 \sqrt 2} 4 - \frac {12 \sqrt 6 - 20\sqrt 2} {16}$

$\ds $

$=$

$\ds \frac {3 \sqrt 6 - 3 \sqrt 2} 4 - \frac {3 \sqrt 6 - 5 \sqrt 2} 4$

$\ds $

$=$

$\ds \frac {2 \sqrt 2} 4$

$\ds $

$=$

$\ds \frac {\sqrt 2} 2$

$\blacksquare$