Cosine of 15 Degrees

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Theorem

$\cos 15 \degrees = \cos \dfrac \pi {12} = \dfrac {\sqrt 6 + \sqrt 2} 4$

where $\cos$ denotes the cosine.


Proof 1

\(\displaystyle \cos 15^\circ\) \(=\) \(\displaystyle \cos \frac {30 \degrees} 2\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {\frac {1 + \cos 30 \degrees} 2}\) Half Angle Formula for Cosine: $\theta$ is in Quadrant I
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {\frac {1 + \frac {\sqrt 3} 2} 2}\) Cosine of $30 \degrees$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {\frac {2 + \sqrt 3} 4}\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {\frac {8 + 4 \sqrt 3} {16} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {\frac {6 + 2 + 2 \sqrt 2 \sqrt 6} {16} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {\frac {\paren {\sqrt 6 + \sqrt 2}^2} {16} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt 6 + \sqrt 2} 4\) positive because $\theta$ is in the first quadrant

$\blacksquare$


Proof 2

\(\displaystyle \cos 15 \degrees\) \(=\) \(\displaystyle \map \cos {45 \degrees - 30 \degrees}\)
\(\displaystyle \) \(=\) \(\displaystyle \cos 45 \degrees \cos 30 \degrees + \sin 45 \degrees \sin 30 \degrees\) Cosine of Difference
\(\displaystyle \) \(=\) \(\displaystyle \paren {\frac {\sqrt 2} 2} \paren {\frac {\sqrt 3} 2} + \paren {\frac {\sqrt 2} 2} \paren {\dfrac 1 2}\) Cosine of $45 \degrees$, Cosine of $30 \degrees$, Sine of $45 \degrees$, Sine of $30 \degrees$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt 6} 4 + \frac {\sqrt 2} 4\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt 6 + \sqrt 2} 4\)

$\blacksquare$


Sources