Finding the value of $\prod^{6}_{r=0}\cos\bigg(\frac{\pi}{21}+\frac{r\pi}{7}\bigg)$

Question

Find value of $$\prod^{6}_{r=0}\cos\left(\frac{\pi}{21}+\frac{r\pi}{7}\right)$$

What I try:

$$\cos \frac{\pi}{21}\cdot \cos \frac{4\pi}{21}\cdot \cos\frac{7\pi}{21}\cdot \cos \frac{10\pi}{21}\cdot \cos \frac{13\pi}{21}\cdot \cos \frac{16\pi}{21}\cdot \cos \frac{19\pi}{21}$$

$\displaystyle \cos \frac{19\pi}{21} = -\cos \frac{2\pi}{21}$ and $\displaystyle \cos \frac{13\pi}{21}=-\cos \frac{8\pi}{21}$

How do I solve it? Help me, please.

Answer 1

$$\cos \frac{\pi}{21}\cdot \cos \frac{4\pi}{21}\cdot \cos\frac{7\pi}{21}\cdot \cos \frac{10\pi}{21}\cdot \cos \frac{13\pi}{21}\cdot \cos \frac{16\pi}{21}\cdot \cos \frac{19\pi}{21}=$$ $$=\cos \frac{\pi}{21}\cdot \cos \frac{4\pi}{21}\cdot\frac{1}{2}\cdot \cos \frac{10\pi}{21}\cdot \left(-\cos \frac{8\pi}{21}\right)\cdot \cos \frac{16\pi}{21}\cdot \left(-\cos \frac{2\pi}{21}\right)=$$ $$=\frac{32\sin\frac{\pi}{21} \cos\frac{\pi}{21}\cdot \cos \frac{2\pi}{21}\cdot \cos\frac{4\pi}{21}\cdot \cos \frac{8\pi}{21}\cdot \cos \frac{16\pi}{21}\cdot \cos \frac{10\pi}{21}}{64\sin\frac{\pi}{21}}=$$ $$=\frac{\sin\frac{32\pi}{21} \cdot \cos \frac{10\pi}{21}}{64\sin\frac{\pi}{21}}=-\frac{2\sin\frac{10\pi}{21} \cdot \cos \frac{10\pi}{21}}{128\sin\frac{\pi}{21}}=-\frac{\sin\frac{20\pi}{21} }{128\sin\frac{\pi}{21}}=-\frac{1}{128}.$$

Answer 2

I believe this is how the problem came into being

Use $\cos(\pi-x)=-\cos x,$

$P=-\prod_{r=0}^6\cos(\pi/21+2r\pi/7),$

Observe that $\cos7(x+2r\pi/7)=\cos7x,$

Now if $\cos(2n+1)y=\cos(2n+1)x$ which is $2^{2n}\cos^{2n+1}x+\cdots+(-1)^n(2n+1)\cos x$

So, the roots of $$2^{2n}\cos^{2n+1}x+\cdots+(-1)^n(2n+1)\cos x-\cos(2n+1)y=0$$ are $\cos(2r\pi/(2n+1)+y)$ where $0\le r\le2n$

So, the product of the roots will be $$\dfrac{\cos7y}{2^{2n}}$$

Here $7y=\pi/3$ and $2n+1=7\iff n=?$