Sine of 45 Degrees/Proof 4

Theorem

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

Proof

\(\ds \sin 45 \degrees\)

\(=\)

\(\ds \map \sin {60 \degrees - 15 \degrees}\)

\(\ds \)

\(=\)

\(\ds \sin 60 \degrees \cos 15 \degrees - \cos 60 \degrees \sin 15 \degrees\)

Sine of Difference

\(\ds \)

\(=\)

\(\ds \paren {\frac {\sqrt 3} 2} \paren {\frac {\sqrt 6 + \sqrt 2} 4} - \paren {\frac 1 2} \paren {\frac {\sqrt 6 - \sqrt 2} 4}\)

Sine of $60 \degrees$, Cosine of $15 \degrees$, Cosine of $60 \degrees$, Sine of $15 \degrees$

\(\ds \)

\(=\)

\(\ds \frac 1 8 \paren {\sqrt 3 \paren {\sqrt 6 + \sqrt 2} - \paren {\sqrt 6 - \sqrt 2} }\)

simplifying

\(\ds \)

\(=\)

\(\ds \frac 1 8 \paren {3 \sqrt 2 + \sqrt 6 - \sqrt 6 + \sqrt 2}\)

simplifying

\(\ds \)

\(=\)

\(\ds \frac 1 8 \paren {3 \sqrt 2 + \sqrt 2}\)

simplifying

\(\ds \)

\(=\)

\(\ds \frac 1 8 \paren {4 \sqrt 2}\)

simplifying

\(\ds \)

\(=\)

\(\ds \frac {\sqrt 2} 2\)

$\blacksquare$