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

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.

Answer 1

Your sum can be written in terms of complex exponentials as

$$ 1+\exp(\pi i/14) + \exp(3 \pi i/14) + \exp(4 \pi i/14) + \exp(8 \pi i/14) + \exp(9 \pi i/14) + \exp(12 \pi i/14) + \exp(16 \pi i/14) + \exp(19 \pi i/14) + \exp(20 \pi i/14) + \exp(24 \pi i/14) + \exp(25 \pi i/14) + \exp(27 \pi i/14) $$ If $w = e^{i\pi/14}$, this is $$Q(w) = {w}^{27}+{w}^{25}+{w}^{24}+{w}^{20}+{w}^{19}+{w}^{16}+{w}^{12}+{w}^{9} +{w}^{8}+{w}^{4}+{w}^{3}+w+1 $$ The claim is that $Q(w)^2 - 7 = 0$ which must mean that $Q(w)^2-7$ is divisible by the minimal polynomial of $w$. Since $w$ is a primitive $28$'th root of $1$, that minimal polynomial is the $28$'th cyclotomic polynomial $C_{28}(w)$, namely ${w}^{12}-{w}^{10}+{w}^{8}-{w}^{6}+{w}^{4}-{w}^{2}+1$. And indeed it turns out that $(Q(w)^2-7)/C_{28}(w)$ is a polynomial in $w$ of degree $42$ (but I'd hate to have to verify that by hand).

Answer 2

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!