Question

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

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

Answer 1

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

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

$f\left(n\right)=\frac{\sum _{k=0}^n[\cos \left(\frac{\pi }{n+2}\right)-\cos \left(\frac{2k+3}{n+2}\pi \right)]}{\sum _{k=0}^{n}1-\cos 2\left(\frac{k+1}{n+2}\pi \right)}$

$\Rightarrow f\left(n\right)=\frac{(n+1)\cos \left(\frac{\pi }{n+2}\right)-\sum _{k=0}^n\cos \left(\frac{2k+3}{n+2}\pi \right)}{(n+1)-\sum _{k=0}^n\cos 2\left(\frac{k+1}{n+2}\pi \right)}$

$\Rightarrow f\left(n\right)=\frac{(n+1)\cos \left(\frac{\pi }{n+2}\right)-[\cos \frac{3\pi }{n+2}+\cos \frac{5\pi }{n+2}+...+\cos \frac{2n+3}{n+2}\pi ]}{(n+1)-[\cos \frac{2\pi }{n+2}+\cos \frac{4\pi }{n+2}+...+\cos 2\frac{n+1}{n+2}\pi ]}$

$\Rightarrow f\left(n\right)=\frac{(n+1)\cos \left(\frac{\pi }{n+2}\right)-\left[\frac{\sin \frac{(n+1)\pi }{n+2}}{\sin \frac{\pi }{n+2}}\right]\cos \frac{(n+3)\pi }{(n+2)}}{(n+1)-\left[\frac{\sin \frac{(n+1)\pi }{n+2}}{\sin \frac{\pi }{n+2}}\right]\cos \frac{2(n+2)\pi }{2(n+2)}}$

$\Rightarrow f\left(n\right)=\frac{(n+1)\cos \left(\frac{\pi }{n+2}\right)-\left[\frac{\sin (\pi -\frac{\pi }{n+2})}{\sin \frac{\pi }{n+2}}\right]\cos (\pi +\frac{\pi }{(n+2)})}{(n+1)-\left[\frac{\sin (\pi -\frac{\pi }{n+2})}{\sin \frac{\pi }{n+2}}\right]\cos \pi }$

$\Rightarrow f\left(n\right)=\frac{(n+1)\cos \left(\frac{\pi }{n+2}\right)+\cos \frac{\pi }{n+2}}{(n+1)-\cos \pi }$
$f\left(n\right)=\frac{(n+2)\cos \left(\frac{\pi }{n+2}\right)}{(n+2)}=\cos \left(\frac{\pi }{n+2}\right)$

Option 1:
$f\left(4\right)=\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$

Option 2:
$\underset{n\to \infty }{lim}f\left(n\right)=\underset{n\to \infty }{lim}\cos \frac{\pi }{n+2}=\cos \left(0\right)\ne \frac{1}{2}$

Option 3:
$\text{If}\alpha =\tan \left({\cos }^{-1}\cos \frac{\pi }{8}\right)=\tan \frac{\pi }{8}=\sqrt{2}-1$
$\text{then}{\alpha }^2+2\alpha -1=(\sqrt{2}-1{)}^2+2(\sqrt{2}-1)-1=0$

Option 4:
$\sin \left(7{\cos }^{-1}\cos \frac{\pi }{7}\right)=\sin \pi =0$