P 12 , JEEE 2026

Method 1: Since $ 1 - 2 \cos{2 \theta} = 3 - 4 \cos^2 \theta = \frac{3 \cos{\theta} - 4 \cos^3{\theta}}{\cos{\theta}}= \frac{- \cos{3 \theta}}{\cos{\theta}}. $
The problem can be rephrased as follows.

The above conversion is due to the fact that $ \cos{(\pi- \theta)} = - \cos{\theta}. $ This becomes a telescopic product.

Method:2

If you are an explorer, then follow this method. This method by, virtue of being non-routine, is long but you will learn a lot. I had suggested this method to Dr. K D Joshi and he was glad to include it on his blog and give the credit. Consider the equation $Z^{11}=-1$. The roots of the equation are as follows. $Z_{k}= \cos{(\frac{(2k+1)\pi}{11})} + i \sin{(\frac{(2k+1)\pi}{11})} $, where $ k \in {0,1,2, \dots, 10}$. For $k=5$,the root $Z_{5} = -1.$ The equation $ Z^{11}+1=0 $ can be factorised as:

Therefore, the roots of equation $ Z^{10}-Z^{9}+ \dots +Z^2-Z+1 = 0 $ are $ Z_{k}= \cos{(\frac{(2k+1)\pi}{11})} + i \sin{(\frac{(2k+1)\pi}{11})} $, where $ k \in {0,1,2,3,4,6,7,8,9, 10}$. Notice that $\Re(Z_{k})= \Re(Z_{11-k})$. So, the equation

has ten roots, of which every pair has the same real part. Observe that

so each conjugate pair determines the same value of

Hence the ten roots of equation $(A)$ give rise to exactly five distinct values of $x$, namely

Since none of the roots of equation $(A)$ is zero, we may divide by $Z^5$ to obtain

Set $ Z+\frac{1}{Z} = x. $ Since $Z = e^{i\theta } \implies Z+\frac{1}{Z} = 2 \cos{\theta} = x. $ Define $Z^n+ \frac{1}{Z^n}= S_{n} \implies S_{0}= 2, S_{1}= x $. It is obvious that $S_{n}$ satisfies the recurrence

Substituting these results into $(B)$, we get

For every root $Z$ of equation $(A)$, equation $(B)$ is satisfied. Since each quantity

is a polynomial in

every corresponding value of $x$ satisfies

As this polynomial is of degree five, these are precisely all of its roots. Hence,

Put $x=1$ to get $\alpha$.