For non-negative integers n, let \[f(n)=\frac{\sum_{k=0}^{n} \sin \left(\frac{k+1}{n+2} \pi\right) \sin \left(\frac{k+2}{n+2} \pi\right)}{\sum_{k=0}^{n} \sin ^{2}\left(\frac{k+1}{n+2} \pi\right)}\] Assuming $\cos ^{-1} \mathrm{x}$ takes values in $[0, \pi]$, which of the following options is/are correct ?

(A) $\sin \left(7 \cos ^{-1} \mathrm{f}(5)\right)=0$; (B) $f(4)=\frac{\sqrt{3}}{2}$; (D) If $\alpha=\tan \left(\cos ^{-1} \mathrm{f}(6)\right)$, then $\alpha^{2}+2 \alpha-1=0$

(C) $\lim _{\mathrm{n} \rightarrow \infty} \mathrm{f}(\mathrm{n})=\frac{1}{2}$

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Trigonometric Ratios and Identities

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For non-negative integers n, let

f (n) = \frac{\sum _{k = 0}^{n} sin ( \frac{k + 1}{n + 2} π ) sin ( \frac{k + 2}{n + 2} π )}{\sum _{k = 0}^{n} sin ^{2} ( \frac{k + 1}{n + 2} π )}

Assuming

cos^{- 1} x

takes values in

[0, π]

, which of the following options is/are correct ?

(A) $sin (7 cos^{- 1} f (5)) = 0$ ✓ Correct
(B) $f (4) = \frac{3}{2}$ ✓ Correct
(C) $lim_{n \to \infty} f (n) = \frac{1}{2}$
(D)If $α = tan (cos^{- 1} f (6))$ , then $α^{2} + 2 α - 1 = 0$ ✓ Correct

Answer: (A), (B), (D)

Solution

Using product-to-sum formulas and telescoping sums, we can show that

f (n) = cos (\frac{π}{n + 2})

. (A)

f (5) = cos \frac{π}{7}

, so

sin (7 cos^{- 1} cos \frac{π}{7}) = sin π = 0

. (B)

f (4) = cos \frac{π}{6} = \frac{3}{2}

. (C)

lim_{n \to \infty} cos (\frac{π}{n + 2}) = 1 \neq = \frac{1}{2}

. (D)

f (6) = cos \frac{π}{8}

, so

α = tan \frac{π}{8} = 2 - 1

. Then

α + 1 = 2

, squaring:

α^{2} + 2 α + 1 = 2

, giving

α^{2} + 2 α - 1 = 0

. Options (A), (B), and (D) are correct.

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Trigonometric Ratios and Identities

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