For non-negative integers n , let f n = ∑ k = 0 n sin ⁡ k + 1 n + 2 π sin ⁡ k + 2 n +…

$\begin{array}{cc}f\left(n\right)=\frac{\sum _{k=0}^n\sin \left(\left(\frac{k+1}{n+2}\right)\pi \right)\sin \left(\left(\frac{k+2}{n+2}\right)\pi \right) }{\sum _{k=0}^n\mathrm{s}\mathrm{i}{\mathrm{n}}^2\left(\left(\frac{k+1}{n+2}\right)\pi \right)}& \end{array}$

$\left(\begin{array}{c}\because 2\sin C\sin D=\cos \left(C-D\right)-\cos \left(C+D\right)\\ \because 2{\sin }^2\theta =1-\cos 2\theta \end{array}\right)$
$f\left(n\right)=\frac{\sum _{k=0}^n\left(\cos \left(\frac{\pi }{n+2}\right)-\cos \left(\frac{2k+3}{n+2}\pi \right)\right)}{\sum _{k=0}^n\left(1-\cos \left(\left(\frac{2k+2}{n+2}\right)\pi \right)\right)}$
$f\left(n\right)=\frac{\left(n+1\right)\cos \left(\frac{\pi }{n+2}\right)-\sum _{k=0}^n\cos \left(\frac{2k+3}{n+2}\right)\pi }{\left(n+1\right)-\sum _{k=0}^n\cos \left(\frac{2k+2}{n+2}\right)\pi }$
Now $\cos \alpha +\cos \left(\alpha +\beta \right)+\dots +\cos \left(\alpha +\left(n-1\right)\beta \right)=\frac{\sin \left(\frac{n\beta }{2}\right)}{\sin \left(\frac{\beta }{2}\right)}\cos \left(\alpha +\frac{\left(n-1\right)\beta }{2}\right)$
$f\left(n\right)=\frac{\left(n+1\right)\cos \left(\frac{\pi }{n+2}\right)-\frac{\sin \left(\frac{n+1}{n+2}\pi \right)}{\sin \left(\frac{\pi }{n+2}\right)}\cos \left(\frac{n+3}{n+2}\pi \right)}{\left(n+1\right)-\frac{\sin \left(\frac{n+1}{n+2}\right)\pi }{\sin \left(\frac{\pi }{n+2}\right)}\cos \left(\pi \right)}$
In numerator cosine sum $\alpha =\frac{3\pi }{n+2}$
$\beta =\frac{2\pi }{n+2}$
Number of terms $=\left(n+1\right)$
In denominator cosine sum $\alpha =\frac{2\pi }{n+2}$
$\beta =\frac{2\pi }{n+2}$
Number of terms $=\left(n+1\right)$
Now $\sin \left(\frac{n+1}{n+2}\pi \right)=\sin \left(\pi -\frac{\pi }{n+2}\right)=\sin \frac{\pi }{n+2}$
$\cos \left(\frac{n+3}{n+2}\pi \right)=\cos \left(\pi +\frac{\pi }{n+2}\right)=-\cos \frac{\pi }{n+2}$
$f\left(n\right)=\frac{\left(n+1\right)\cos \left(\frac{\pi }{n+2}\right)+\cos \left(\frac{\pi }{n+2}\right)}{\left(n+1\right)-\left(-1\right)}$
$f\left(n\right)=\frac{\cos \left(\frac{\pi }{n+2}\right)\left(n+2\right)}{\left(n+2\right)}$
$f\left(n\right)=\cos \left(\frac{\pi }{n+2}\right)$
Hence, $\left(A\right) f\left(4\right)=\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$
$\left(3\right) \underset{n+\infty }{\lim }f\left(n\right)=\cos \left(0\right)=1$
$\left(4\right) f\left(6\right)=\cos \left(\frac{\pi }{8}\right):$ then $\alpha =\tan \left(\frac{\pi }{8}\right)=\sqrt{2}-1$
$\alpha =\sqrt{2}\left(5\right)=\cos \left(\frac{\pi }{7}\right)$
$\sin \left(7{\cos }^{-1}\left(\cos \frac{\pi }{7}\right)\right)=\sin \pi =0$