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? 
 A: 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}  $
A: Use the fact that when angles correspond to equally spaced points around a circle, their sines add up to zero and their cosines do the same.  Thus
$\cos(\frac{\pi}{4})+\cos(\frac{\pi}{4}+\frac{2\pi}{3})+\cos(\frac{\pi}{4}+\frac{4\pi}{3})=0$
Put $\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$, then:
$\cos(\frac{\pi}{4}+\frac{2\pi}{3})=a-\frac{\sqrt{2}}{4}$
$\cos(\frac{\pi}{4}+\frac{4\pi}{3})=-a-\frac{\sqrt{2}}{4}$
To find $a$, simply subtract and use the appropriate sum-product relation:
$2a=\cos(\frac{\pi}{4}+\frac{2\pi}{3})-\cos(\frac{\pi}{4}+\frac{4\pi}{3})=2\sin(\frac{(\frac{\pi}{4}+\frac{2\pi}{3})+(\frac{\pi}{4}+\frac{4\pi}{3})}{2})\sin(\frac{(\frac{\pi}{4}+\frac{4\pi}{3})-(\frac{\pi}{4}+\frac{2\pi}{3})}{2})=2\sin(\frac{5\pi}{4})\sin(\frac{\pi}{3})$
Then $\sin(\frac{5\pi}{4})=(-\sqrt{2})/2, \sin(\frac{\pi}{3})=(\sqrt{3})/2$ and we get $a=(-\sqrt{6})/4$.
A: Let $\,a=\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}=\dfrac{1+i}{\sqrt{2}}\,$ and  $\,b=\cos \dfrac{2\pi}{3} + i \sin \dfrac{2\pi}{3}=\dfrac{-1+i\sqrt{3}}{2}\,$, so:
$$
ab = \dfrac{(1+i)(-1+i\sqrt{3})}{2 \sqrt{2}}=\dfrac{\left(-1-\sqrt{3}\right)+i\left(-1+\sqrt{3}\right)}{2 \sqrt{2}}
$$
Then $\,\cos\left(\dfrac{\pi}{4} + \dfrac{2 \pi}{3}\right) = \operatorname{Re}(ab) = \dfrac{-1-\sqrt{3}}{2\sqrt{2}}\,$.
