I wrote about the book here.
It's not for nothing that I wrote somewhere that THIS BOOK IS FULL OF SCRATCHES. (Maybe that's why the author withdrew Part I from sale)
We are now debating Problem 62***, page 157 (solved on pages 516-519).
" 62***. Calculate $P_n\; Q_n\;and\; R_n$, where $n\in \mathbb{N},\;n\geqslant 3,\;a\in\mathbb{R}^*\;:$
$$a)\;\;\;P_n=\prod_{k=0}^{n-1} \cos \left (a+\frac{k\pi}{n}\right )\;;$$
$$b)\;\;Q_n=\prod_{k=0}^{n-1}\sin \left (a+\frac{k\pi}{n}\right )\;;$$
$$c)\;\;R_n=\prod_{k=0}^{n-1}\tan \left (a+\frac{k\pi}{n}\right )."$$
ANSWER
$a)\;\;P_n=$
$b)\;\;Q_n=\frac{\sin na}{2^{n-1}}$
$c)\;\;R_n=$
Solution CiP
We will start with b). A solution that, at some point, requires a square root extraction is found in TURTOIU Fanică - Probleme de Trigonometrie, Ed. Tehnică, București, 1979 (pages 86-87, Problem 2.38 - received by the kindness of Liviu PODGORNEI). This is a personal attempt.
Let $z_k=\cos\left (a+\frac{k\pi}{n}\right )+\imath \sin \left (a+\frac{k\pi}{n}\right )\;,\;\;0\leqslant k\leqslant n-1.$ We have
$|z_k|=1=z_k\cdot \bar z_k$ so $\bar z_k=\cos \left (a+\frac{k\pi}{n}\right )-\imath \sin \left (a+\frac{k\pi}{n}\right )=\frac{1}{z_k}.$ In addition
$z_k^{2n}=\cos 2n \left (a+\frac{k\pi}{n}\right )+\imath \sin 2n\left (a+\frac{k\pi}{n}\right )=\cos 2na+\imath \sin 2na \tag{1}$
Then
$\sin \left (a+\frac{k\pi}{n}\right )=\frac{z_k-\bar z_k}{2\imath}=\frac{z_k-\frac{1}{z_k}}{2\imath}=\frac{z_k^2-1}{2\imath z_k} \tag{2}$
From (1) it follows that the equation
$z^{2n}=\cos 2na+\imath \sin 2na \tag{3}$
has the roots $\pm z_k\;,\;\;k\in \{0,\;1,\dots ,\;n-1 \}.$ Denoting
$f(z)=z^{2n}-(\cos na +\imath \sin na) \tag{4}$
we have $f(z)=(z-z_0)(z-z_1)\dots (z-z_{n-1})(z+z_0)(z+z_1)\dots (z+z_{n-1})\;$, or
$$f(z)=\prod_{k=0}^{n-1}(z^2-z_k^2) \tag{5}$$
From here we go further $$\prod_{k=0}^{n-1}(z_k^2-1)=(-1)^n\prod_{k=0}^{n-1}(1-z_k^2)=(-1)^nf(1)\overset{(4)}{=}(-1)^n(1-\cos 2na-\sin 2na)=$$
$$=(-1)^n(2\sin^2na-2\imath \sin na \cos na)=(-1)^n(-2\imath^2\sin^2 na-2\imath \sin na \cos na)$$
so
$$\prod_{k=0}^{n-1}(z_k^2-1)=-2\imath (-1)^n\sin na(\cos na+\imath \sin na) \tag{6}$$
On the other hand
$$\prod_{k=0}^{n-1}z_k=\prod_{k=0}^{n-1}\left [(\cos a+\imath \sin a)\left (\cos \frac{k\pi}{n}+\imath \sin\frac{k\pi}{n}\right )\right]=$$
$$=(\cos a+\imath \sin a)^n\prod_{k=0}^{n-1}\left (\cos \frac{k\pi}{n}+\imath \sin \frac{k\pi}{n}\right )=$$
$$=(\cos na+\imath \sin na)\left [\cos \left (\sum_{k=0}^{n-1}k\right )\frac{\pi}{n}+\imath \sin \left (\sum_{k=0}^{n-1}k\right )\frac{\pi}{n}\right ]=$$
$=(\cos na+\imath \sin na)\left [\cos (n-1)\frac{\pi}{2}+\imath \sin (n-1)\frac{\pi}{2}\right ]$
because $\sum_{k=0}^{n-1}k=\frac{(n-1)n}{2}.$ Analyzing the cases $n=4m,\;4m+1,\;4m+2,\;4m+3$ it is immediately seen that $\cos (n-1)\frac{\pi}{2}+\imath \sin (n-1)\frac{\pi}{2}=-\imath^{n+1}$ , and then the previous calculation gives us
$$\prod_{k=0}^{n-1}z_k=-\imath^{n+1}(\cos na+\imath\sin na) \tag{7}$$
Now we can finish
$$Q_n=\prod_{k=0}^{n-1}\sin \left (a+\frac{k\pi}{n}\right )\overset{(2)}{=}\frac{\prod_{k=0}^{n-1}(z_k^2-1)}{2^n\imath^n\prod_{k=0}^{n-1}z_k}\overset{(6)}{\underset{(7)}{=}}\frac{-2\imath(-1)^n\sin na(\cos na+\imath\sin na)}{2^n\imath^n(-\imath^{n+1})(\cos na+\imath \sin na)}=$$
$=\frac{\imath \cdot (-1)^n\sin na}{2^{n-1}\imath^{2n}\cdot \imath}=\frac{(-1)^n \sin na}{2^{n-1}(-1)^n}=\frac{\sin na}{2^{n-1}}.$ We got the answer of b).
Remark CiP On page 518 the following stupid answer is given :
$$Q_n=\begin{cases}\frac{(-1)^n\cdot \sin na}{2^{n-1}}\;\;\;if\;\;n-even\\\frac{(-1)^{n+1}\cdot \sin na}{2^{n-1}}\;\;\;if\;\;n-odd\end{cases} \tag{Q}$$
............................................................................................................................
(in construction)
