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
$a)\;\;P_{2n}=\frac{(-1)^n\sin 2na}{2^{2n-1}}\;\;;\;\;P_{2n+1}=\frac{(-1)^n\cos (2n+1)a}{2^{2n}}$
$b)\;\;Q_n=\frac{\sin na}{2^{n-1}}$
$c)\;\;R_{2n}=-1\;\;;\;\;R_{2n+1}=(-1)^n\tan (2n+1)a$
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 :
$$Q_n=\begin{cases}\frac{(-1)^n\cdot \sin na}{2^{n-1}}\;\;\;if\;\;n-even\\\frac{(-1)^{n+1}\cdot \sin na}{2^{n-1}}\;\;\;if\;\;n-odd\end{cases} \tag{Q}$$
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a) With the same notations as in point b), first we have
$\cos\left (a+\frac{k\pi}{n}\right )=\frac{z_k+\bar z_k}{2}=\frac{z_k^2+1}{2z_k} \tag{8}$
Then, starting from (5) we see that
$$f(\imath)=\prod_{k=0}^{n-1}(-1-z_k^2)=(-1)^n\prod_{k=0}^{n-1}(z_k^2+1)$$
so
$$\prod_{k=0}^{n-1}(z_k^2+1)=(-1)^nf(\imath) \tag{9}$$
Then we can write
$$P_n=\prod_{k=0}^{n-1}\cos \left (a+\frac{k\pi}{n}\right )\overset{(8)}{=}\frac{\prod_{k=0}^{n-1}(z_k^2+1)}{2^n\prod_{k=0}^{n-1}z_k}\overset{(7)}{\underset{(9)}{=}}\frac{(-1)^nf(\imath)}{2^n(-\imath^{n+1})(\cos na+\imath \sin na)} \tag{10}$$
But $f(\imath)\underset{(4)}{=}(-1)^n-\cos 2na-\imath \sin 2na$ and we will treat the $n-even$ and $n-odd$ cases separately.
$\fbox{$n$-even}$
$f(\imath)=1-\cos na-\imath \sin na=2\sin^2 na-2\imath \sin na \cos na=$
$=-2\imath^2\sin^2 na-2\imath \sin na \cos na=-2\imath \sin na(\cos na+\imath \sin na)$
In (10) also appears $-\imath^{n+1}=-\imath^n\cdot \imath=(-\imath)(-1)^{\frac{n}{2}}$ and then we continue like this
$P_n=\frac{-2\imath \sin na(\cos na+\imath \sin na)}{2^n(-\imath)(-1)^{\frac{n}{2}}(\cos na+\imath \sin na)}=\frac{(-1)^{\frac{n}{2}}\sin na}{2^{n-1}}.$
$\fbox{$n$-odd}$
$(-1)^nf(\imath)=-(-1-\cos 2na-\imath \sin 2na)=2\cos^2 na+2\imath \sin na \cos na=$
$=2\cos na(\cos na+\imath \sin na)$
and because $-\imath^{n+1}=-(-1)^{\frac{n+1}{2}}=-(-1)^{\frac{n-1}{2}+1}=-(-1)^{\frac{n-1}{2}}(-1)=(-1)^{\frac{n-1}{2}}$
we have $P_n\underset{(10)}{=}\frac{2\cos na(\cos na+\imath \sin na)}{2^n(-1)^{\frac{n-1}{2}}(\cos na+\imath \sin na)}=\frac{(-1)^{\frac{n-1}{2}}\cos na}{2^{n-1}}.$ We got the answer of a).
Remark CiP Although in the solution I used COPILOT, who first told me incorrectly that the minus sign always appears in the formula, then came back and corrected the mistake, we must admit that in the book, on page 517, the correct answer is given. Marius DRĂGAN and Sorin RĂDULESCU are cited as the authors of the solution.
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c) $R_{2n}=\frac{Q_{2n}}{P_{2n}}=\frac{\frac{\sin 2na}{2^{2n-1}}}{\frac{(-1)^n\sin 2na}{2^{2n-1}}}=(-1)^n\;;$
$R_{2n+1}=\frac{Q_{2n+1}}{P_{2n+1}}=\frac{\frac{\sin (2n+1)a}{2^{2n}}}{\frac{(-1)^n\cos (2n+1)a}{2^{2n}}}=(-1)^n\tan(2n+1)a.$
$\blacksquare\;\blacksquare\;\blacksquare$
