The positive integer value of n 3 satisfyin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given equation: $\frac{1}{\sin(\frac{\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})} + \frac{1}{\sin(\frac{3\pi}{n})}$Rearranging the terms: $\frac{1}{

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Given equation: $\frac{1}{\sin(\frac{\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})} + \frac{1}{\sin(\frac{3\pi}{n})}$Rearranging the terms: $\frac{1}{\sin(\frac{\pi}{n})} - \frac{1}{\sin(\frac{3\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})}$Using the formula $\sin(A) - \sin(B) = 2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})$,we get:$\frac{\sin(\frac{3\pi}{n}) - \sin(\frac{\pi}{n})}{\sin(\frac{\pi}{n})\sin(\frac{3\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})}$$\frac{2\cos(\frac{2\pi}{n})\sin(\frac{\pi}{n})}{\sin(\frac{\pi}{n})\sin(\frac{3\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})}$$\frac{2\cos(\frac{2\pi}{n})}{\sin(\frac{3\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})}$$2\sin(\frac{2\pi}{n})\cos(\frac{2\pi}{n}) = \sin(\frac{3\pi}{n})$Using $\sin(2	heta) = 2\sin	heta\cos	heta$,we have:$\sin(\frac{4\pi}{n}) = \sin(\frac{3\pi}{n})$Since $\sin(A) = \sin(B)$ implies $A = \pi - B$ (for $A, B$ in the relevant range),we have:$\frac{4\pi}{n} = \pi - \frac{3\pi}{n}$$\frac{4\pi}{n} + \frac{3\pi}{n} = \pi$$\frac{7\pi}{n} = \pi$$n = 7$

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

Similar Questions

If $A = \{x \in [0, 2\pi] : \tan x - \tan^2 x > 0\}$ and $B = \{x \in [0, 2\pi] : |\sin x| < \frac{1}{2}\}$,then $A \cap B =$

If $0 < B < A < \frac{\pi}{4}$,$\cos^2 B - \sin^2 A = \frac{\sqrt{3}+1}{4\sqrt{2}}$ and $2 \cos A \cos B = \frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}$,then $\cos^2 \frac{4B}{3} - \sin^2 \frac{4A}{5} =$

If $|\sin x-\cos ^2 x| \geq|3-3 \sin x+\sin ^2 x|+4|\sin x-1|$,then $x=$

If $\sin A + \sin B = x$ and $\cos A + \cos B = y$,then $\sin(A + B) = $

Let $a$ and $b$ be non-negative real numbers. If $\sin x + a \cos x = b$,then $|a \sin x - \cos x| = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module