Question

The positive integer value of $n>3$ satisfying the equation $\frac{1}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{n}}\right)}=\frac{1}{\sin \left({\displaystyle \frac{2\mathrm{\pi }}{n}}\right)}+\frac{1}{\sin \left({\displaystyle \frac{3\mathrm{\pi }}{n}}\right)}$ is

• A. $8$
• B. $6$
• C. $5$
• D. $7$

Answer 1

The correct option is D

$7$

Explanation for the correct option:

Find the positive integer value of $n$.

An equation $\frac{1}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{n}}\right)}=\frac{1}{\sin \left({\displaystyle \frac{2\mathrm{\pi }}{n}}\right)}+\frac{1}{\sin \left({\displaystyle \frac{3\mathrm{\pi }}{n}}\right)}$ is given.

Solve the given equation as follows:

$$\begin{gathered} \frac{1}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{n}}\right)}-\frac{1}{\sin \left({\displaystyle \frac{3\mathrm{\pi }}{n}}\right)}=\frac{1}{\sin \left({\displaystyle \frac{2\mathrm{\pi }}{n}}\right)} \\ \Rightarrow \frac{\sin \left(\frac{3\mathrm{\pi }}{n}\right)-\sin \left(\frac{\mathrm{\pi }}{n}\right)}{\sin \left(\frac{3\mathrm{\pi }}{n}\right)·\sin \left(\frac{\mathrm{\pi }}{n}\right)}=\frac{1}{\sin \left({\displaystyle \frac{2\mathrm{\pi }}{n}}\right)} \end{gathered}$$

Use $\mathit{s}\mathit{i}\mathit{n}\mathit{A}\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\mathit{B}\mathbf{=}\mathbf{2}\mathit{c}\mathit{o}\mathit{s}\left(\frac{A+B}{2}\right)\mathit{s}\mathit{i}\mathit{n}\left(\frac{A-B}{2}\right)$.

$$\begin{gathered} \frac{2\cos \left(\frac{2\mathrm{\pi }}{n}\right)\sin \left(\frac{\mathrm{\pi }}{n}\right)}{\sin \left(\frac{3\mathrm{\pi }}{n}\right)·\sin \left(\frac{\mathrm{\pi }}{n}\right)}=\frac{1}{\sin \left({\displaystyle \frac{2\mathrm{\pi }}{n}}\right)} \\ \Rightarrow \frac{2\cos \left(\frac{2\mathrm{\pi }}{n}\right)}{\sin \left(\frac{3\mathrm{\pi }}{n}\right)}=\frac{1}{\sin \left({\displaystyle \frac{2\mathrm{\pi }}{n}}\right)} \\ \Rightarrow 2\cos \left(\frac{2\mathrm{\pi }}{n}\right)\sin \left(\frac{2\mathrm{\pi }}{n}\right)=\sin \left(\frac{3\mathrm{\pi }}{n}\right) \end{gathered}$$

Since, $\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathbf{2}\mathit{\theta }\mathbf{\right)}\mathbf{=}\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\left(\theta \right)\mathit{c}\mathit{o}\mathit{s}\left(\theta \right)$.

$$\begin{gathered} \sin \left(\frac{4\mathrm{\pi }}{n}\right)=\sin \left(\frac{3\mathrm{\pi }}{n}\right) \\ \Rightarrow \sin \left(\mathrm{\pi }-\frac{4\mathrm{\pi }}{n}\right)=\sin \left(\frac{3\mathrm{\pi }}{n}\right) \end{gathered}$$

Therefore,

$$\begin{gathered} \mathrm{\pi }-\frac{4\mathrm{\pi }}{\mathrm{n}}=\frac{3\mathrm{\pi }}{\mathrm{n}} \\ \Rightarrow \frac{7\mathrm{\pi }}{\mathrm{n}}=\mathrm{\pi } \\ \Rightarrow \mathrm{n}=7 \end{gathered}$$

Therefore, The positive integer value of $n>3$ satisfying the equation $\frac{1}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{n}}\right)}=\frac{1}{\sin \left({\displaystyle \frac{2\mathrm{\pi }}{n}}\right)}+\frac{1}{\sin \left({\displaystyle \frac{3\mathrm{\pi }}{n}}\right)}$ is $7$.

Hence, option $D$ is the correct answer.