Sine of 15 Degrees/Proof 1
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Theorem
- $\sin 15 \degrees = \sin \dfrac \pi {12} = \dfrac {\sqrt 6 - \sqrt 2} 4$
Proof
\(\ds \sin 15 \degrees\) | \(=\) | \(\ds \sin \frac {30 \degrees} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\frac {1 - \cos 30 \degrees} 2}\) | Half Angle Formula for Sine: $\theta$ is in Quadrant I | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\frac {1 - \frac {\sqrt 3} 2} 2}\) | Cosine of $30 \degrees$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\frac {2 - \sqrt 3} 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\frac {8 - 4 \sqrt 3} {16} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\frac {6 + 2 - 2 \sqrt 2 \sqrt 6} {16} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\frac {\paren {\sqrt 6 - \sqrt 2}^2} {16} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt 6 - \sqrt 2} 4\) | positive because $\theta$ is in the first quadrant |
$\blacksquare$