For non-negative integers n, letAssuming cos–1x takes values in [0, π], which of the following options is/are correct?

Engineering

Mathematics

Trigonometric Limits

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^–1x takes values in [0, π], which of the following options is/are correct?

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

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

sin (7cos^–1 f(5)) = 0

If α = tan (cos^–1 f(6)), then α²+2α–1 = 0

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

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

$f (n)= \frac{(n + 1) \cos (\frac{π}{n + 2}) - (\frac{\sin (\frac{(n + 1) π}{n + 2})}{\sin (\frac{π}{n + 2})} \cdot \cos (\frac{n + 3}{n + 2}) π)}{(n + 1) - (\frac{\sin (\frac{(n + 1) π}{n + 2})}{\sin (\frac{π}{n + 2})} \cdot \cos (\frac{2 (n + 2) π}{2 (n + 2)}))}$

$f (n) = \frac{(n + 1) \cos (\frac{π}{n + 2}) + \cos (\frac{π}{n + 2})}{(n + 1) + 1} \Rightarrow g (x) = \cos (\frac{π}{n + 2})$

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

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

(D) $α = \tan (\cos^{- 1} \cos \frac{π}{8}) = \sqrt{2} - 1 \Rightarrow α + 1 = \sqrt{2}$

⇒ α² + 2α + 1 = 0.