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JEE Advanced 2011Paper 1
Exam
JEE Advanced
Year
2011
Paper
Paper 1
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Advanced 2011 · Paper 1 · Mathematics · Trigonometric Equations

The positive integer value of n 3 satisfying the equation (1)/(...

The positive integer value of n>3n \gt 3 satisfying the equation 1sin⁡(πn)=1sin⁡(2πn)+1sin⁡(3πn) is \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 }

Correct answer

7

Step-by-step solution

Given, n>3∈n \gt 3 \in Integer  and 1sin⁡(πn)=1sin⁡(2πn)+1sin⁡(3πn)⇒1sin⁡πn−1sin⁡3πn=1sin⁡2πn⇒sin⁡3πn−sin⁡πnsin⁡πn⋅sin⁡3πn=1sin⁡2πn\begin{aligned}& \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 \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}} \end{aligned}⇒2cos⁡(2πn)⋅sin⁡πn=sin⁡πn⋅sin⁡3πnsin⁡2πn⇒2sin⁡2πn⋅cos⁡2πn=sin⁡3πn⇒sin⁡4πn=sin⁡3πn⇒4πn=π−3πn⇒7πn=π\Rightarrown=7\begin{aligned}& \Rightarrow \quad 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 \quad 2 \sin \frac{2 \pi}{n} \cdot \cos \frac{2 \pi}{n}=\sin \frac{3 \pi}{n}\\ & \Rightarrow \quad \quad \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 \Rightarrown=7 \end{aligned}
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