Question

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

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

Answer 1

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

$f\left(n\right)=\frac{\sum _{k=1}^n\left(cos\left(\frac{\pi }{n+2}\right)-cos\left(\frac{2k+3}{n+2}\right)\pi \right)}{\sum _{k=1}^n\left(1-cos\left(\frac{2k+2}{n+2}\right)\pi \right)}$
$f\left(n\right)=\frac{\left(n+1\right)cos\left(\frac{\pi }{n+2}\right)-\left(\sum _{k=1}^{n}cos\left(\frac{2k+3}{n+2}\right)\pi \right)}{\left(n+1\right)-\left(\sum _{k=1}^{n}cos\left(\frac{2k+2}{n+2}\right)\pi \right)}$
$f\left(n\right)=\frac{\left(n+1\right)cos\frac{\pi }{n+2}-\left(\frac{sin\left(\frac{\left(n+1\right)\pi }{n+2}\right)}{sin\left(\frac{\pi }{n+2}\right)}.cos\left(\frac{n+3}{n+2}\right)\pi \right)}{\left(n+1\right)-\left(\frac{sin\left(\frac{\left(n+1\right)\pi \pi }{n+2}\right)}{sin\left(\frac{\pi }{n+2}\right)}.cos\left(\frac{2\left(n+2\right)\pi }{2\left(n+2\right)}\right)\right)}$
$f\left(n\right)=\frac{\left(n+1\right)cos\left(\frac{\pi }{n+2}\right)+cos\left(\frac{\pi }{n+2}\right)}{\left(n+1\right)+1}\Rightarrow g\left(x\right)=cos\left(\frac{\pi }{n+2}\right)$
$\left(A\right)sin\left(7co{s}^{-1}\frac{\pi }{7}\right)=sin\pi =0$
$\left(B\right)f\left(4\right)=cos\frac{\pi }{6}=\frac{\sqrt{3}}{2}$
$\left(C\right)$ $\underset{n\to \infty }{\lim }$ $cos\left(\frac{\pi }{n+2}\right)=1$
$\left(D\right)\alpha =tan\left(co{s}^{-1}cos\frac{\pi }{8}\right)=\sqrt{2}-1\Rightarrow \alpha +1\sqrt{2}$
${\alpha }^2+2\alpha -1=0$