Positive integer value of n>3 satisfying equation frac{1}{sin left(frac{π}{n}right)}=frac{1}{sin

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Trigonometrical Equations

hard

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

$2$

$6$

$7$

$8$

(IIT-2011)

Solution

We have

$\frac{1}{\sin \frac{\pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}+\frac{1}{\sin \frac{3 \pi}{n}}$

$\Rightarrow \frac{1}{\sin \frac{\pi}{n}}-\frac{1}{\sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}$

$\Rightarrow \frac{\sin \frac{3 \pi}{n}-\sin \frac{\pi}{n}}{\sin \left(\frac{\pi}{n}\right) \cdot \sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}$

$\Rightarrow \frac{2 \cos \frac{2 \pi}{n} \sin \frac{\pi}{n}}{\sin \frac{\pi}{n} \sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}$

$\Rightarrow 2 \cos \frac{2 \pi}{n} \sin \frac{2 \pi}{n}=\sin \frac{3 \pi}{n}$

$\Rightarrow \sin \frac{4 \pi}{n}+\sin (0)=\sin \frac{3 \pi}{n}$

$\Rightarrow \frac{4 \pi}{n}+\frac{3 \pi}{n}=\pi$

$\Rightarrow 7 \pi= n \pi$

$n =7$

Hence, this is the answer.

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