Sine of 15 Degrees/Proof 2

Theorem

$\sin 15 \degrees = \sin \dfrac \pi {12} = \dfrac {\sqrt 6 - \sqrt 2} 4$

Proof

\(\ds \sin 15 \degrees\)

\(=\)

\(\ds \map \sin {45 \degrees - 30 \degrees}\)

\(\ds \)

\(=\)

\(\ds \sin 45 \degrees \cos 30 \degrees - \cos 45 \degrees \sin 30 \degrees\)

Sine of Difference

\(\ds \)

\(=\)

\(\ds \paren {\frac {\sqrt 2} 2} \paren {\frac {\sqrt 3} 2} - \paren {\frac {\sqrt 2} 2} \paren {\dfrac 1 2}\)

Sine of $45 \degrees$, Cosine of $30 \degrees$, Cosine of $45 \degrees$, Sine of $30 \degrees$

\(\ds \)

\(=\)

\(\ds \frac {\sqrt 6} 4 - \frac {\sqrt 2} 4\)

\(\ds \)

\(=\)

\(\ds \frac {\sqrt 6 - \sqrt 2} 4\)

$\blacksquare$