Question

For non-negative integers $n$, let
$f\left(n\right)=\frac{{\displaystyle \sum _{k=0}^n}\sin \left({\displaystyle \frac{\mathrm{k}+1}{\mathrm{n}+2}}\mathrm{\pi }\right)\sin \left({\displaystyle \frac{k+2}{n+2}}\mathrm{\pi }\right)}{{\displaystyle \sum _{k=0}^n}{\sin }^2\left({\displaystyle \frac{k+1}{n+2\mathrm{\pi }}}\right)}$

Assuming ${\cos }^{-1}x$ takes value in $[0,\pi ]$, which of the following options is/are correct?

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

Answer 1

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

$\sin (7{\cos }^{-1}f(5\left)\right)=0$

Explanation for the correct option:

Step 1. Evaluating the given function:

$\begin{array}{rcl}f\left(n\right)& =& \frac{{\displaystyle \sum _{k=0}^n}\sin \left({\displaystyle \frac{\mathrm{k}+1}{\mathrm{n}+2}}\mathrm{\pi }\right)\sin \left({\displaystyle \frac{k+2}{n+2}}\mathrm{\pi }\right)}{{\displaystyle \sum _{k=0}^n}{\sin }^2\left({\displaystyle \frac{k+1}{n+2\mathrm{\pi }}}\right)}\\ & =& \frac{{\displaystyle \sum _{k=0}^n\left[\cos \left(\frac{\mathrm{\pi }}{\mathrm{n}+2}\right)-\cos \left(\frac{2k+3}{n+2}\mathrm{\pi }\right)\right]}}{{\displaystyle \sum _{k=0}^n\left[1-\sin \left(\frac{2k+2}{n+2\mathrm{\pi }}\mathrm{\pi }\right)\right]}}\end{array}$

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

$\Rightarrow$$\begin{array}{rcl}f\left(n\right)& =& \frac{\left(n+1\right)\cos \left({\displaystyle \frac{\mathrm{\pi }}{n+2}}\right)+\cos \left({\displaystyle \frac{\mathrm{\pi }}{n+2}}\right)}{\left(n+1\right)+1}\\ & =& \cos \left(\frac{\mathrm{\pi }}{n+2}\right)\end{array}$

Step 2. Consider Option A to check if its true

(A) $\alpha =\tan \left({\cos }^{-1}f\right(6\left)\right)$

$$\begin{gathered} =\tan \left({\cos }^{-1}\left(\cos \frac{\mathrm{\pi }}{8}\right)\right) \\ =\tan \left(\frac{\mathrm{\pi }}{8}\right) \end{gathered}$$

$\Rightarrow$${\alpha }^2+2\alpha –1={\tan }^2\left(\frac{\mathrm{\pi }}{8}\right)+2\tan \left(\frac{\mathrm{\pi }}{8}\right)-1$

$\Rightarrow$ $\tan 2\left(\frac{\mathrm{\pi }}{8}\right)=\frac{2\tan \left(\frac{\mathrm{\pi }}{8}\right)}{1-{\tan }^2\left(\frac{\mathrm{\pi }}{8}\right)}$

$\Rightarrow$ $1=\frac{2\alpha }{1-{\alpha }^2}$

$\Rightarrow$${\alpha }^2+2\alpha –1=0$

$\therefore$Therefore, option ‘A’ is Correct.

Step 3. Consider Option C to check if it is true

(c) $\begin{array}{rcl}f\left(4\right)& =& \cos \left(\frac{\mathrm{\pi }}{6}\right)\\ & =& \frac{\sqrt{3}}{2}\end{array}$

$\therefore$Therefore, option ‘C’ is Correct.

Step 4. Consider Option D to check if its true

(D) $\begin{array}{rcl}\sin (7{\cos }^{-1}f(5\left)\right)& =& \sin \left(7{\cos }^{-1}\left(\cos \frac{\mathrm{\pi }}{7}\right)\right)\\ & =& \sin \left(7\times \frac{\mathrm{\pi }}{7}\right)\\ & =& \sin \mathrm{\pi }\\ & =& 0\end{array}$

$\therefore$Therefore, option ‘D’ is Correct.

Explanation for the incorrect option:

Consider Option B to check if its true

(B)$\begin{array}{rcl}\underset{n\to \infty }{\lim }f\left(n\right)& =& \underset{n\to \infty }{\lim }\cos \left(\frac{\mathrm{\pi }}{n+2}\right)\\ & =& \underset{\frac{1}{n}\to 0}{\lim }\cos \left(\frac{{\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}}{1+{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}}\right)\\ & =& 1\end{array}$

$\therefore$Therefore, option ‘B’ is Incorrect.

Hence, Option ‘A’, ‘C’ and ‘D’ is Correct.