Question

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

• A. f(4) = $\frac{\sqrt{3}}{2}$
• B. $lim_{n\to\infty} f\left(n\right) = \frac{1}{2}$
• C. If $\alpha = tan \left(cos^{-1}f\left(6\right)\right),$ then $\alpha^{2} +2\alpha -1 = 0$
• D. $sin \left(7 \,cos^{-1} f\left(5\right)\right) = 0$

Answer 1

Answer: (A), (C), (D)

$f \left(n\right)=\frac{\displaystyle \sum_{k=1}^n\left(cos\left(\frac{\pi}{n+2}\right)-cos\left(\frac{2k+3}{n+2}\right)\pi\right)}{\displaystyle \sum_{k=1}^n\left(1-cos\left(\frac{2k+2}{n+2}\right)\pi\right)}$
$f \left(n\right)=\frac{\left(n+1\right)cos\left(\frac{\pi}{n+2}\right)-\left(\displaystyle \sum_{k=1}^ncos\left(\frac{2k+3}{n+2}\right)\pi\right)}{\left(n+1\right)-\left(\displaystyle \sum_{k=1}^ncos\left(\frac{2k+2}{n+2}\right)\pi\right)}$
$f \left(n\right)=\frac{\left(n+1\right)cos \frac{\pi}{n+2}-\left(\frac{sin \left(\frac{\left(n+1\right)\pi}{n+2}\right)}{sin \left(\frac{\pi}{n+2}\right)}.cos \left(\frac{n+3}{n+2}\right)\pi\right)}{\left(n+1\right)-\left(\frac{sin\left(\frac{\left(n+1\right)\pi\pi}{n+2}\right)}{sin \left(\frac{\pi}{n+2}\right)}.cos \left(\frac{2\left(n+2\right)\pi}{2\left(n+2\right)}\right)\right)}$
$f \left(n\right)=\frac{\left(n+1\right)cos\left(\frac{\pi}{n+2}\right)+cos\left(\frac{\pi}{n+2}\right)}{\left(n+1\right)+1} \Rightarrow g\left(x\right)=cos\left(\frac{\pi}{n+2}\right)$
$\left(A\right) sin\left(7\,cos^{-1}\, \frac{\pi}{7}\right)=sin\,\pi=0$
$\left(B\right) f \left(4\right)=cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$
$\left(C\right)$ $\displaystyle \lim_{n \to \infty}$ $cos\left(\frac{\pi}{n+2}\right)=1$
$\left(D\right) \alpha=tan\left(cos^{-1}\,cos \frac{\pi}{8}\right)=\sqrt{2}-1 \Rightarrow \alpha+1 \sqrt{2}$
$\alpha^{2}+2\alpha-1=0$