JEE Advanced 2026 · Paper 1, Mathematics question 12

Reduce the angles modulo $2\pi$: $\frac{27\pi}{11}\to\frac{5\pi}{11}$, and $\cos\frac{81\pi}{11}=\cos\frac{15\pi}{11}=\cos\frac{7\pi}{11}$. So
$$\alpha=\prod_{k\in\{1,3,5,7,9\}}\left(1-2\cos\frac{k\pi}{11}\right).$$

Identity. $\sin\frac{3\theta}2=\sin\frac\theta2\,(1+2\cos\theta)$. With $1-2\cos\theta=1+2\cos(\pi-\theta)$ and $\pi-\frac{k\pi}{11}=\frac{m\pi}{11}$, $m=10,8,6,4,2$:
$$\alpha=\prod_{j=1}^{5}\frac{\sin(3j\pi/11)}{\sin(j\pi/11)}.$$

Numerator. For $j=1,\dots,5$, $3j\pi/11$ gives $\sin\frac{3\pi}{11}$, $\sin\frac{6\pi}{11}=\sin\frac{5\pi}{11}$, $\sin\frac{9\pi}{11}=\sin\frac{2\pi}{11}$, $\sin\frac{12\pi}{11}=-\sin\frac{\pi}{11}$, $\sin\frac{15\pi}{11}=-\sin\frac{4\pi}{11}$. The two signs cancel, so the numerator equals the denominator and $\alpha=1$.

So $5-\alpha^2=4$.