Cosine of 15 Degrees/Proof 1
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Theorem
- $\cos 15 \degrees = \cos \dfrac \pi {12} = \dfrac {\sqrt 6 + \sqrt 2} 4$
Proof
\(\ds \cos 15 \degrees\) | \(=\) | \(\ds \cos \frac {30 \degrees} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\frac {1 + \cos 30 \degrees} 2}\) | Half Angle Formula for Cosine: $\theta$ 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 |
$\blacksquare$