JEE Advanced 2019 · Paper 2, Mathematics question 3

Closed form. Let $\theta=\frac\pi{n+2}$. Using $\sin A\sin(A+\theta)=\frac12[\cos\theta-\cos(2A+\theta)]$ and $\sin^2A=\frac12(1-\cos2A)$, the cosine sums over these equally spaced angles simplify. The result is
$$f(n)=\cos\frac\pi{n+2}.$$
Check at $n=0$: the numerator is $\sin\frac\pi2\sin\pi=0$ and $\cos\frac\pi2=0$. ✓

The options.

• (A) $f(4)=\cos\frac\pi6=\frac{\sqrt3}2$. True.
• (B) $f(n)\to\cos0=1$, not $\frac12$. False.
• (C) $\alpha=\tan\frac\pi8=\sqrt2-1$, and $\alpha^2+2\alpha-1=(3-2\sqrt2)+(2\sqrt2-2)-1=0$. True.
• (D) $\cos^{-1}f(5)=\frac\pi7$, so $\sin\left(7\cdot\frac\pi7\right)=\sin\pi=0$. True.

(A), (C), (D)