For non-negative integers n , let f ( n ) = k = 0 n sin ( k + 1 n + 2 ) sin ( k + 2 n + 2 ) k = 0 n sin 2 ( k + 1 n + 2 ) . Assuming cos - 1 x takes values in [ 0 , ] , which of the following options is/are correct? [2019]

Question

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? [2019]

Smart Achievers · Accepted Answer

(2, 3, 4)

$f (n) = \frac{\sum_{k = 0}^{n} 2 \sin (\frac{k + 1}{n + 2} π) \sin (\frac{k + 2}{n + 2} π)}{\sum_{k = 0}^{n} 2 \sin^{2} (\frac{k + 1}{n + 2}) π},$

where $n$ is a non-negative integer.

$= \frac{\sum_{k = 0}^{n} [\cos (\frac{π}{n + 2}) - \cos (\frac{(2 k + 3) π}{n + 2})]}{\sum_{k = 0}^{n} [1 - \cos (\frac{2 (k + 1) π}{n + 2})]}$

$= \frac{(n + 1) \cos (\frac{π}{n + 2}) - [\cos (\frac{3 π}{n + 2}) + \cos (\frac{5 π}{n + 2}) + \dots + \cos (\frac{(2 n + 3) π}{n + 2})]}{n + 1 - [\cos (\frac{2 π}{n + 2}) + \cos (\frac{4 π}{n + 2}) + \dots + \cos (\frac{2 (n + 1) π}{n + 2})]}$

$= \frac{(n + 1) \cos (\frac{π}{n + 2}) - \frac{\sin (\frac{(n + 1) π}{n + 2})}{\sin (\frac{π}{n + 2})} \cos (\frac{(2 n + 6) π}{2 (n + 2)})}{n + 1 - \frac{\sin (\frac{(n + 1) π}{n + 2})}{\sin (\frac{π}{n + 2})} \cos (\frac{(2 n + 4) π}{2 (n + 2)})}$

$= \frac{(n + 1) \cos (\frac{π}{n + 2}) + \cos (\frac{π}{n + 2})}{(n + 1) + 1} = \frac{(n + 2) \cos (\frac{π}{n + 2})}{n + 2}$

$∴$ $f (n) = \cos (\frac{π}{n + 2})$

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

$∴$ $Option (1) is incorrect$

$f (4) = \cos (\frac{π}{4 + 2}) = \cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}$

$∴$ $Option (2) is correct$

$If α = \tan (\cos^{- 1} f (6)),$

$= \tan (\cos^{- 1} (\cos \frac{π}{8})) = \tan \frac{π}{8}$

Now, $\tan \frac{π}{4} = 1 = \frac{2 \tan \frac{π}{8}}{1 - \tan^{2} \frac{π}{8}}$

$\Rightarrow \frac{2 α}{1 - α^{2}} = 1 \Rightarrow α^{2} + 2 α - 1 = 0$

$∴$ $Option (3) is correct$

$\sin (7 \cos^{- 1} f (5)) = \sin (7 \cos^{- 1} (\cos \frac{π}{7})) = \sin (7 \times \frac{π}{7})$

$= \sin π = 0$

$∴$ $Option (4) is correct$

Solution

Q.
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 [2019]

1 $\lim_{n \to \infty} f (n) = \frac{1}{2}$

2 $f (4) = \frac{\sqrt{3}}{2}$

3 If $α = \tan (\cos^{- 1} f (6)),$ then $α^{2} + 2 α - 1 = 0$

4 $\sin (7 \cos^{- 1} f (5)) = 0$

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Solution

Q. For non-negative integers n, let f(n)=∑k=0nsin(k+1n+2π)sin(k+2n+2π)∑k=0nsin2(k+1n+2π). Assuming cos-1x takes values in [0,π], which of the following options is/are correct [2019]

1 limn→∞f(n)=12

2 f(4)=32

3 If α=tan(cos-1f(6)), then α2+2α-1=0