Cosine of 15 Degrees

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 $30 \degrees$ $\displaystyle$ $=$ $\displaystyle \frac {\sqrt 6} 4 + \frac {\sqrt 2} 4$ $\displaystyle$ $=$ $\displaystyle \frac {\sqrt 6 + \sqrt 2} 4$

$\blacksquare$