..Originally Posted by Another user. Dont ask.
You can use this: $\displaystyle \sin\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}$ and $\displaystyle \cos\frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}$
$\displaystyle \sin\frac{\pi}{16}=\sqrt{\frac{1-\cos\frac{\pi}{8}}{2}}$
$\displaystyle \cos\frac{\pi}{8}=\sqrt{\frac{1+\cos\frac{\pi}{4}} {2}}$