Product of cosines

For $n$ natural number, and $a = \frac{2\pi}{2n+1}$ find the value of $$\cos a . \cos 2a . \cos 3a . \dots . \cos na$$

Solution

We first prove the following claim: for any $n$ positive odd number, $$\sin (nx) = \sin x P_n (\cos x)$$ $$\cos (nx) = \cos x Q_n (\cos x)$$ where $P_n(t)$ and $Q_n(t)$ are polynomials in the form of $2^{n-1} t^{n-1} + ...$ (In other words, they have degree $n-1$ and the leading coefficient $2^{n-1}$

Proof by induction, evident for $n=1$ and $n=3$. Inductive step: $$\sin (n+2)x = \sin nx \cos 2x + \cos nx \sin 2x = \sin x P_n (\cos x) (2 \cos^2 x - 1) + \cos x Q_n (\cos x) .2\sin x \cos x$$ $$= \sin x ( P_n (\cos x) (2 \cos^2 x - 1) + 2\cos^2 x Q_n (\cos x) ) = \sin x P_{n+2} (\cos x)$$ where $P_{n+2}$ has degree $n+2$ and the leading coefficient $2. 2^{n-1} + 2.2^{n-1} = 2^{n+1}$

Likewise: $$\cos (n+2) x = \cos nx \cos 2x - \sin nx \sin 2x = \cos x Q_n(\cos x)(2 \cos^2 x - 1) - \sin x P_n (\cos x) . 2 \sin x \cos x$$ $$= \cos x ( Q_n(\cos x)(2 \cos^2 x - 1) - 2 P_n (\cos x) (1 - \cos^2 x)) = \cos x Q_{n+2} (\cos x)$$ like before, the polynomial $Q_{n+2}$ has degree $n+2$ and leading coefficient $2^{n+1}$

So now, observe that $x = 0,a,2a, \dots, 2na$ are all solutions of the equation $$\cos (2n+1)x = 1 = \cos x Q_{2n+1}(\cos x)$$ Therefore $t = \cos 0, \cos a, \dots, \cos 2na$ are all roots of the polynomial $$S(t) = tQ_{2n+1}(t) - 1$$ Because $S(t)$ has degree $2n+1$, then $\cos 0, \dots \cos 2na$ are ALL of the roots, and the product of all those roots is $\frac{1}{2^{2n}}$ (because $S(t)$ has leading coefficient $2^{2n}$

Now, $\cos a = \cos 2n a, \cos 2a = \cos (n-1) a$ and so on. So: $$\cos 0 . \cos a .\dots. \cos 2na = 1 . (\cos a. \dots . \cos na)^2 = \frac{1}{2^{2n}}$$ So: $$\cos a . \dots . \cos na = \frac{1}{2^n}$$