<s>[INST] What is a fast way to evaluate $\cos(\pi/4 + 2\pi/3)$ and $\cos(\pi/4 + 4\pi/3)$ (maybe using inspection) Given 
$\cos(\pi/4 + 2\pi/3)$ 
and  
$\cos(\pi/4 + 4\pi/3)$
I wish to show that these values are given by $(\pm(\sqrt{6}) - \sqrt{2})/4$.
What is a quick way to arrive at these exact solutions (perhaps even by inspection)?
I am currently using the addition of cosine formula, but it is too slow. I'm thinking perhaps a graphic method can be used? 
 [/INST] Is it really that slow?
$\cos (\frac{\pi}{4} + \frac{2\pi}{3}) =\cos \frac{\pi}{4}\cos \frac{2\pi}{3} - \sin \frac{\pi}{4}\sin\frac{2\pi}{3} = (\frac {\sqrt{2}}{2})(\frac {-1}2) - (\frac {\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) = \frac {-\sqrt 2 - \sqrt 6}{4}$
That seems pretty straight forward to me.
$\cos (\frac{\pi}{4} + \frac{4\pi}{3})$ isn't any more difficult.
If you want to use symmetry arguments... 
$\cos (\frac{\pi}{12}) = \cos (\frac{\pi}{3}-\frac{\pi}{4}) = \frac {\sqrt 6 + \sqrt 2}{4}\\
\sin (\frac{\pi}{12}) = \sin (\frac{\pi}{3}-\frac{\pi}{4}) = \frac {\sqrt 6 - \sqrt 2}{4}$
(which is every bit as much work as done in the first line.)
Then you can say:
$\cos (\frac{\pi}{4} + \frac{2\pi}{3}) = \cos (\frac{11\pi}{12}) = \cos(\pi - \frac {\pi}{12}) = -\cos \frac {\pi}{12}$
$\cos (\frac{\pi}{4} + \frac{4\pi}{3}) = \cos (\frac{19\pi}{12}) = \cos(2\pi - \frac {5\pi}{12}) = \cos \frac {5\pi}{12} = \cos(\frac {\pi}{2} - \frac {\pi}{12}) =\sin \frac{\pi}{12}  $
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