TITLE: How to find the value of $\cos\left (\frac{\pi }{28} \right )-\cos\left (\frac{3\pi }{28} \right )+\sin\left ( \frac{5\pi }{28} \right )$?
QUESTION [5 upvotes]: Once , I do a problem  $ \sum_{n=1}^{14} \cos\left ( \frac{n^{2}\pi }{14} \right )$
when angle modular by $28$  we've got  
$2\left ( \cos\left (\frac{\pi }{14}  \right ) - \cos\left (\frac{2\pi }{14}  \right ) + \cos\left (\frac{3\pi }{14}  \right ) + \cos\left (\frac{4\pi }{14}  \right ) - \cos\left (\frac{5\pi }{14}  \right ) - \cos\left (\frac{6\pi }{14}  \right ) \right ) + 1$
and then using identity $\cos\left ( a \right ) - \cos\left ( b \right )$ and  $\cos\left ( a \right ) + \cos\left ( b \right )$ 
leads to   $2\sqrt{2} \left (\cos\left ( \frac{\pi }{28} \right ) -\cos\left ( \frac{3\pi }{28} \right )+\sin\left ( \frac{5\pi }{28} \right )\right ) + 1$ 
the final answer is $\sqrt{7}$  but I don't know how to compute $\left ( \cos\left ( \frac{\pi }{28} \right ) - \cos\left ( \frac{3\pi }{28} \right )+\sin\left ( \frac{5\pi }{28} \right )\right )$ by hands.
I appreciate for your helps.

REPLY [2 votes]: We need to prove that:
$$2\sqrt2\left(\cos\frac{\pi }{28}-\cos\frac{3\pi }{28}+\sin\frac{5\pi }{28}\right)+1=\sqrt7$$ or
$$8\left(\cos\frac{\pi }{28}-\cos\frac{3\pi }{28}+\cos\frac{9\pi }{28}\right)^2=8-2\sqrt7$$ or
$$\left(4-4\left(\cos\frac{\pi }{28}-\cos\frac{3\pi }{28}+\cos\frac{9\pi }{28}\right)^2\right)^2=7.$$
Indeed,
$$\left(4-4\left(\cos\frac{\pi }{28}-\cos\frac{3\pi }{28}+\cos\frac{9\pi }{28}\right)^2\right)^2=$$
$$=\left(4-4\left(\cos^2\frac{\pi }{28}+\cos^2\frac{3\pi }{28}+\cos^2\frac{9\pi }{28}-2\cos\tfrac{\pi}{28}\cos\tfrac{3\pi}{28}+2\cos\tfrac{\pi}{28}\cos\tfrac{9\pi}{28}-2\cos\tfrac{3\pi}{28}\cos\tfrac{9\pi}{28}\right)\right)^2=$$
$$=\left(4-2\left(3+\cos\tfrac{\pi }{14}+\cos\tfrac{3\pi }{14}+\cos\tfrac{9\pi }{14}-2\cos\tfrac{\pi}{7}-2\cos\tfrac{\pi}{14}+2\cos\tfrac{5\pi}{14}+2\cos\tfrac{2\pi}{7}-2\cos\tfrac{3\pi}{7}-2\cos\tfrac{3\pi}{14}\right)\right)^2=$$
$$=\left(2+2\left(-\cos\tfrac{\pi }{14}-\cos\tfrac{3\pi }{14}+\cos\tfrac{5\pi }{14}-2\cos\tfrac{\pi}{7}+2\cos\tfrac{2\pi}{7}-2\cos\tfrac{3\pi}{7}\right)\right)^2=$$
$$=\left(2+2\left(-\cos\tfrac{\pi }{14}-\cos\tfrac{3\pi }{14}+\cos\tfrac{5\pi }{14}+\frac{-2\sin\tfrac{\pi}{7}\cos\tfrac{\pi}{7}+2\sin\tfrac{\pi}{7}\cos\tfrac{2\pi}{7}-2\sin\tfrac{\pi}{7}\cos\tfrac{3\pi}{7}}{\sin\tfrac{\pi}{7}}\right)\right)^2=$$
$$=\left(2+2\left(-\cos\tfrac{\pi }{14}-\cos\tfrac{3\pi }{14}+\cos\tfrac{5\pi }{14}+\frac{-\sin\tfrac{2\pi}{7}+\sin\frac{3\pi}{7}-\sin\tfrac{\pi}{7}-\sin\frac{4\pi}{7}+\sin\tfrac{2\pi}{7}}{\sin\frac{\pi}{7}}\right)\right)^2=$$
$$=4\left(-\cos\frac{\pi }{14}-\cos\frac{3\pi }{14}+\cos\frac{5\pi }{14}\right)^2=$$
$$=4\left(\cos^2\frac{\pi }{14}+\cos^2\frac{3\pi }{14}+\cos^2\frac{5\pi }{14}+2\cos\frac{\pi }{14}\cos\frac{3\pi }{14}-2\cos\frac{\pi }{14}\cos\frac{5\pi }{14}-2\cos\frac{3\pi }{14}\cos\frac{5\pi }{14}\right)=$$
$$=2\left(3+\cos\tfrac{\pi }{7}+\cos\tfrac{3\pi }{7}+\cos\tfrac{5\pi }{7}+2\cos\tfrac{2\pi }{7}+2\cos\tfrac{\pi }{7}-2\cos\tfrac{3\pi }{7}-2\cos\tfrac{2\pi }{7}-2\cos\tfrac{4\pi }{7}-2\cos\tfrac{\pi }{7}\right)=$$
$$=2\left(3+\cos\tfrac{\pi }{7}+\cos\tfrac{3\pi }{7}+\cos\tfrac{5\pi }{7}\right)=6+\frac{2\sin\frac{\pi}{7}\cos\tfrac{\pi }{7}+2\sin\frac{\pi}{7}\cos\tfrac{3\pi }{7}+2\sin\frac{\pi}{7}\cos\frac{5\pi }{7}}{\sin\frac{\pi}{7}}=$$
$$=6+\frac{\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}-\sin\frac{2\pi }{7}+\sin\frac{6\pi}{7}-\sin\frac{4\pi }{7}}{\sin\frac{\pi}{7}}=7$$ and since easy to see that 
$$0<\cos\frac{\pi }{28}-\cos\frac{3\pi }{28}+\sin\frac{5\pi }{28}<1,$$ we are done!