Ugly summation

My solution:

$=4\sin (\frac{11}{12}\pi)+4\cos (\frac{23}{12}\pi)+2\sqrt{2}\tan (\frac{19}{12}\pi)$ $=4\sin (\pi-\frac{\pi}{12})+4\cos (2\pi-\frac{\pi}{12})+2\sqrt{2}\tan (2\pi-\frac{5\pi}{12})$ $=4\sin (\frac{\pi}{12})+4\cos (\frac{\pi}{12})-2\sqrt{2}\tan (\frac{5\pi}{12}) ...\space ...\space ...(i)$

Now $\sin\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}}$

so $\sin\frac{\pi}{12}=\sqrt{\frac{1-\cos\frac{5\pi}{6}}{2}} = \frac{\sqrt{2+\sqrt{3}}}{2}$ we know $\sin\theta=\sqrt{1-\cos^{2}\theta}$ so $\cos\frac{\pi}{12}=\frac{\sqrt{2-\sqrt{3}}}{2}$

by putting these values in (i) , squaring and some calculations we have answer $\boxed{32}$