JEE Advanced 2026 Paper 1, Mathematics Section 3 Q4: Trigonometric Ratios And Identities

Set $\theta =\pi /11$, so the product is $\prod _{k=0}^4\left(1-2\cos (3^k\theta )\right)$.

Key telescoping identity: Using $1-2\cos \varphi =3-4{\cos }^2(\varphi /2)=-\frac{4{\cos }^3(\varphi /2)-3\cos (\varphi /2)}{\cos (\varphi /2)}=-\frac{\cos (3\varphi /2)}{\cos (\varphi /2)}$

Applying this to each factor with ${\varphi }_k=3^k\theta$:

$1-2\cos (3^k\theta )=-\frac{\cos (3^{k+1}\theta /2)}{\cos (3^k\theta /2)}$

Telescoping:

$\alpha =\prod _{k=0}^4\left(-\frac{\cos (3^{k+1}\theta /2)}{\cos (3^k\theta /2)}\right)=(-1{)}^5\cdot \frac{\cos (3^5\theta /2)}{\cos (\theta /2)}=-\frac{\cos (243\pi /22)}{\cos (\pi /22)}$

Reducing the angle: $\frac{243\pi }{22}=11\pi +\frac{\pi }{22}$, so $\cos (243\pi /22)=\cos (11\pi +\pi /22)=-\cos (\pi /22)$.

Therefore $\alpha =-\frac{-\cos (\pi /22)}{\cos (\pi /22)}=1$.

Hence $5-{\alpha }^2=5-1=4$.