JEE Advanced · Mathematics · 8. Trigonometric Equations
The positive integer value of \(n>3\) satisfying the equation
\(\frac{1}{\sin \left(\frac{\pi}{n}\right)}=\frac{1}{\sin \left(\frac{2 \pi}{n}\right)}+\frac{1}{\sin \left(\frac{3 \pi}{n}\right)} \text { is }\)
- A 2
- B 3
- C 5
- D 7
Answer & Solution
Correct Answer
(D) 7
Step-by-step Solution
Detailed explanation
Given, \(n>3 \in\) Integer
\(\text { and } \frac{1}{\sin \left(\frac{\pi}{n}\right)}=\frac{1}{\sin \left(\frac{2 \pi}{n}\right)}+\frac{1}{\sin \left(\frac{3 \pi}{n}\right)} \)
\( \Rightarrow \frac{1}{\sin \frac{\pi}{n}}-\frac{1}{\sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}} \)
\( \Rightarrow \frac{\sin \frac{3 \pi}{n}-\sin \frac{\pi}{n}}{\sin \frac{\pi}{n} \cdot \sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}\)
\(\Rightarrow 2 \cos \left(\frac{2 \pi}{n}\right) \cdot \sin \frac{\pi}{n}=\frac{\sin \frac{\pi}{n} \cdot \sin \frac{3 \pi}{n}}{\sin \frac{2 \pi}{n}} \)
\( \Rightarrow 2 \sin \frac{2 \pi}{n} \cdot \cos \frac{2 \pi}{n}=\sin \frac{3 \pi}{n} \)
\( \Rightarrow \sin \frac{4 \pi}{n}=\sin \frac{3 \pi}{n} \)
\( \Rightarrow \frac{4 \pi}{n}=\pi-\frac{3 \pi}{n} \Rightarrow \frac{7 \pi}{n}=\pi \Rightarrow n=7\)
\(\text { and } \frac{1}{\sin \left(\frac{\pi}{n}\right)}=\frac{1}{\sin \left(\frac{2 \pi}{n}\right)}+\frac{1}{\sin \left(\frac{3 \pi}{n}\right)} \)
\( \Rightarrow \frac{1}{\sin \frac{\pi}{n}}-\frac{1}{\sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}} \)
\( \Rightarrow \frac{\sin \frac{3 \pi}{n}-\sin \frac{\pi}{n}}{\sin \frac{\pi}{n} \cdot \sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}\)
\(\Rightarrow 2 \cos \left(\frac{2 \pi}{n}\right) \cdot \sin \frac{\pi}{n}=\frac{\sin \frac{\pi}{n} \cdot \sin \frac{3 \pi}{n}}{\sin \frac{2 \pi}{n}} \)
\( \Rightarrow 2 \sin \frac{2 \pi}{n} \cdot \cos \frac{2 \pi}{n}=\sin \frac{3 \pi}{n} \)
\( \Rightarrow \sin \frac{4 \pi}{n}=\sin \frac{3 \pi}{n} \)
\( \Rightarrow \frac{4 \pi}{n}=\pi-\frac{3 \pi}{n} \Rightarrow \frac{7 \pi}{n}=\pi \Rightarrow n=7\)
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