Find the value of : $\cos x \cos 2x...\cos 999x$ given that $x=\frac {2\pi}{1999}$

Given $x=\frac {2\pi}{1999}$

Find the value of $$\cos x \cos 2x \cos 3x ...\cos 999x$$

So I tried expanding $\sin {2000x}=2\sin 1000x \cos 1000x$

Then rewriting $\cos 1000x= \cos {(999x+x)}$

No luck so far.

Solution 1:

Two approaches for evaluating:

$$\prod_{k=0}^{n-1}\cos\left(\frac{2\pi k}n\right)$$

when $n$ is odd. This is the square of the value you are looking for, so we'll have to figure out the sign of the actual product from $k=1$ to $k=\frac{n-1}{2}$.

Complex number approach

The first approach is to write: $z=e^{2\pi i/n}$ then seek the value of $$\prod_{k=0}^{n-1}\cos\left(\frac{2\pi k}n\right)=\prod_{k=0}^{n-1} \frac{1}{2}\left(z^k+\frac{1}{z^k}\right)=\frac{1}{2^{n}}\prod_k z^{-k} \prod (z^{2k}+1)$$ But the $z^{2k}$ are just exactly the roots of $x^{n}-1$, since $n$ is odd, so $z^{2k}+1$ are just the roots of $(z-1)^n -1$, and their product must be $2$, since $(z-1)^n-1$ is a polynomial of odd degree $n$ with value $-2$ at $z=0$.

We also see that $\prod_{k=0}^{n} z^{-k} = z^{-n(n-1)/2} = 1$ since $n$ is odd so $n\mid n(n-1)/2$ and $z^n=1$.

So $$\prod_{k=0}^{n-1} \cos \frac{2\pi k}{n} = \frac{1}{2^{n-1}}$$

Chebyshev polynomial approach

If you know Chebyshev polynomials, take:

$$\cos(nx)=T_n(\cos(x))$$

where $T_n$ is the Chebyshev polynomial of the first kind.

Then the product of the roots of $T_n(x)-1$ is the product:

$$\prod_{k=0}^{n-1} \cos\left(\frac{2\pi k}{n}\right)$$

This is true because, although there are repeated values here, there are repeated roots of $T_n(x)-1$ because where $k\neq 0$, the values $x_k=\cos\left(\frac{2\pi k}{n}\right)$ are local maximums.

The product of the roots of $T_n(x)-1$ is $\frac{1}{2^{n-1}}$ when $n$ is odd, since the lead coefficient if $T_n$ is $2^{n-1}$ and the constant coefficient is $-1$ .

Determining the sign

We have, via either technique, that $$\left(\prod_{k=0}^{(n-1)/2} \cos\left(\frac{2\pi k}n\right)\right)^2=\frac{1}{2^{n-1}}$$

So the product we are seeking is:

$$\frac{\pm 1}{2^{(n-1)/2}}$$

The sign will be determined by how many negative terms there are:

$$\frac{\pi}{2}<\frac{2\pi k}{n}<\pi$$

or:

$$\frac{n}{2}<2k<n$$

For $n=1999$, this means $1000\leq 2k < 1999$ or $500\leq k\leq 999$ so there are $500$ negative terms, so the product is positive.

For more general odd $n$, the equation is:

$$\frac{n+1}{4}\leq k< \frac{n+1}{2}$$

So the number of negative terms is $$\frac{n+1}{2}-\left\lceil\frac{n+1}{4}\right\rceil= \begin{cases}\frac{n-1}{4}&n\equiv 1\pmod 4\\ \frac{n+1}{4}&n\equiv 3\pmod 4 \end{cases} $$

Thus the final answer can be seen to be:

$$\frac{(-1)^{\lceil(n-1)/4\rceil}}{2^{(n-1)/2}}$$

Solution 2:

If $\cos(2n+1)x=1,(2n+1)x=2m\pi$ where $m$ is any integer

$x=\dfrac{2m\pi}{2n+1}$ where $m=0,1,2,\cdots,2n$

But $\cos(2n+1)x=2^{2n}\cos x+\cdots+(-1)^n\cos x$

So, the roots of $2^{2n}\cos^{2n+1}x+\cdots+(-1)^n\cos x-1=0$ are $\cos\dfrac{2m\pi}{2n+1}$ where $m=0,1,2,\cdots,2n$

$\implies\prod_{m=0}^{2n}\cos\dfrac{2m\pi}{2n+1}=\dfrac1{2^{2n}}$

For $m=0,\cos\dfrac{2m\pi}{2n+1}=1$ $\implies\prod_{m=1}^{2n}\cos\dfrac{2m\pi}{2n+1}=\dfrac1{2^{2n}}$

Again, $\cos\dfrac{(2n+1-m)2\pi}{2n+1}=\cos\left(2\pi-\dfrac{2m\pi}{2n+1}\right)=\cos\dfrac{2m\pi}{2n+1}$

$\implies\prod_{m=1}^{2n}\cos\dfrac{2m\pi}{2n+1}=\left(\prod_{m=1}^n\cos\dfrac{2m\pi}{2n+1}\right)^2$

Consequently, $\prod_{m=1}^n\cos\dfrac{2m\pi}{2n+1}=\pm\dfrac1{2^n}$

Now $\cos\dfrac{2m\pi}{2n+1}<0$ if $\dfrac{3\pi}2>\dfrac{2m\pi}{2n+1}>\dfrac\pi2\iff\dfrac{3(2n+1)}4>m>\dfrac{2n+1}4$

But $\dfrac{3(2n+1)}4>n$

So, there are $n-\left\lceil\dfrac{2n+1}4\right\rceil$ negative roots, which will dictate the sign of $\prod_{m=1}^n\cos\dfrac{2m\pi}{2n+1}$