Chebyshev plus one

 A puzzle inspired by a recent problem from the JEE 2026 advanced math paper.

Find the value of 

$$ \prod_{k=0}^{4} \left(1 - \sqrt{2} \cos \left(\frac{(2k+1)\pi}{11}\right)\right)^2 $$

Scroll down for a solution.

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The actual value is $$ \frac{(\sqrt{2} - 1)^2}{32}$$

We will prove the following identity:

$\textbf{Proposition:}$

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

Setting $\theta = \frac{\pi}{4}$ and $n = 5$ in the above solves the puzzle.

So let's try to prove the above identity.

$\textbf{Proof:}$

We first make the following claim:

$\textbf{Lemma:}$ There is a polynomial $T$ of degree $2n+1$ and leading coefficient $2^{2n}$ such that $T(\cos \theta) = \cos (2n+1)\theta$

$\textbf{Proof of Lemma:}$ This is easily proved using complex numbers and DeMoivre's formula. So we will skip it.

Now $\cos (2n+1)\theta = -1$ for $\theta = \frac{(2k+1)\pi}{2n+1}$, $k = 0 \dots n$

Differentiating $T(\cos \theta) = \cos (2n+1)\theta$ we get

$$T'(\cos \theta) \sin \theta = (2n+1) \sin (2n+1)\theta$$

Thus for $k = 0, \dots n-1$, we have that $T'\left(\cos \frac{(2k+1)\pi}{2n+1}\right) = 0$

Thus the $\cos \frac{(2k+1)\pi}{2n+1}$ are double roots of $T(x) + 1 = 0$, for $k = 0 \dots n-1$. Since the leading coefficient is $2^{2n}$, we have the factorization of $T + 1$

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

The $x+1$ factor corresponds to $k = n$, and the rest correspond to the double roots for $k=0, \dots, n-1$, totaling $2n+1$ in number, which is the degree of $T$.

Setting $x = \cos \theta$ gives us the identity we wanted to prove.