The Cosine Product!

Relevant wiki: Chebyshev Polynomials - Application to Polynomial Interpolation

Let $I = f\left( \dfrac{2\pi}{1999} \right)$ . Since $I$ represents the product of all cosine of positive angles less than $\pi$ , then $I > 0$ .

By recalling the identity $\cos(\pi - x) = - \cos x$ , we can rewrite $I$ as

$\begin{aligned} I& =& \left [ - \cos \left( \dfrac{1997\pi}{1999} \right) \right ] \left [ - \cos \left( \dfrac{1995\pi}{1999} \right) \right ] \left [ - \cos \left( \dfrac{1993\pi}{1999} \right) \right ] \cdots \left [ - \cos \left( \dfrac{\pi}{1999} \right) \right ] \\ &=& (-1)^{999} \cos \left( \dfrac{\pi}{1999} \right) \cos \left( \dfrac{3\pi}{1999} \right) \cos \left( \dfrac{5\pi}{1999} \right) \cdots \cos \left( \dfrac{\pi}{1999} \right) \end{aligned}$

Thus, $\displaystyle I^2 = - \prod_{n=1}^{1998} \cos \left( \dfrac{n\pi}{1999} \right) = \prod_{n=1}^{1999} \cos \left( \dfrac{n\pi}{1999} \right)$ .

Note that $\left \{ \cos \left( \dfrac{n\pi}{1999}\right) \right \}_{n=1}^{n=1999}$ are roots of the equation $\cos(1999y) = 0$ .

By Chebyshev polynomials , we may rewrite $\cos(1999y)$ as a polynomial of $\cos y$ , namely,

$\cos(1999y) = 2^{1998} \cos^{1999} (y ) - \cos y (\cdots) - 1 \; .$

By Vieta's formula , $\displaystyle \prod_{n=1}^{1999} \cos \left( \dfrac{n\pi}{1999} \right)$ can be expressed as the product of roots of the equation, $2^{1998} z^{1999} - \cdots - 1 = 0$ . That is,

$I^2 = \prod_{n=1}^{1999} \cos \left( \dfrac{n\pi}{1999} \right) = \dfrac{ 1}{2^{1998}} \quad \Rightarrow \quad I = \dfrac1{2^{999}} \quad \Rightarrow \quad k = \boxed{999} \; .$

Let $P:=f(2\pi/1999)$ . It is well-known that $\prod_{k=1}^{n-1}\cos(k\pi/n)=\frac{\sin(n\pi/2)}{2^{n-1}}.$ In particular, $\prod_{k=1}^{1998}\cos(k\pi/1999)=\frac1{2^{1998}}.$ Thus, since $\begin{aligned} \prod_{k=1}^{1998}\cos(k\pi/1999)&=\prod_{k=1}^{999}\cos(2k\pi/1999)\prod_{k=1}^{999}\cos((2k-1)\pi/1999) =P \prod_{k=1}^{999}\cos(\pi-(2k-1)\pi/1999)\\ &=P \prod_{k=1}^{999}\cos((2000-2k)\pi/1999) =P \prod_{k=1}^{999}\cos(2k\pi/1999) =P^2, \end{aligned}$

we deduce that $P=\sqrt{1/2^{1998}}$ . Therefore, $k=999$ .