Trigonometrical Equations 3 Question 11
12. The positive integer value of $n>3$ satisfying the equation $\frac{1}{\sin \frac{\pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}+\frac{1}{\sin \frac{3 \pi}{n}}$ is ……
(2011)
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Answer:
Correct Answer: 12. 7
Solution:
- Given, $n>3 \in$ Integer
$$ \begin{aligned} & \text { and } \frac{1}{\sin \frac{\pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}+\frac{1}{\sin \frac{3 \pi}{n}} \\ & \Rightarrow \quad \frac{1}{\sin \frac{\pi}{n}}-\frac{1}{\sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}} \end{aligned} $$
$$ \begin{array}{rlrl} & \Rightarrow \quad \frac{\sin \frac{3 \pi}{n}-\sin \frac{\pi}{n}}{\sin \frac{\pi}{n} \cdot \sin \frac{3 \pi}{n}} & =\frac{1}{\sin \frac{2 \pi}{n}} \\ \Rightarrow \quad & 2 \cos \frac{2 \pi}{n} \cdot \sin \frac{\pi}{n} & =\frac{\sin \frac{\pi}{n} \cdot \sin \frac{3 \pi}{n}}{\sin \frac{2 \pi}{n}} \\ \Rightarrow & \quad 2 \sin \frac{2 \pi}{n} \cdot \cos \frac{2 \pi}{n} & =\sin \frac{3 \pi}{n} \\ \Rightarrow & \sin \frac{4 \pi}{n} & =\sin \frac{3 \pi}{n} \\ \Rightarrow & & \frac{4 \pi}{n} & =\pi-\frac{3 \pi}{n} \\ \Rightarrow & & \frac{7 \pi}{n} & =\pi \Rightarrow n=7 \end{array} $$
