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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The positive integer value of $n>3$ satisfying the equation $\large\frac{1}{\sin\big(\Large\frac{\pi}{n}\big)}=\frac{1}{\sin\big(\Large\frac{2\pi}{n}\big)}+\frac{1}{\sin\big(\Large\frac{3\pi}{n}\big)}$ is

$(a)n=6\qquad(b)\;n=7\qquad(c)\;n=8\qquad(d)\;n=9$

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1 Answer

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Toolbox:
  • $sin x-siny=2cos\large\frac{x+y}{2}.$$sin\large\frac{x-y}{2}$
  • $2sinx.cosx=sin2x$
Given :
$\large\frac{1}{\sin\big(\Large\frac{\pi}{n}\big)}=\frac{1}{\sin\big(\Large\frac{2\pi}{n}\big)}+\frac{1}{\sin\big(\Large\frac{3\pi}{n}\big)}$
$\Rightarrow \large\frac{1}{\sin\pi/n}-\frac{1}{\sin \large\frac{3\pi}{n}}=\frac{1}{\sin\Large\frac{2\pi}{n}}$
$\Rightarrow \large\frac{\sin 3\pi/n-\sin\pi/n}{\sin\pi/n.\sin 3\pi/n}$$=\large\frac{1}{sin\large\frac{2\pi}{n}}$
$\Rightarrow\:\large\frac{2cos 2\pi/n.sin \pi/n}{sin\pi/n.sin3\pi/n}$$=\Rightarrow \large\frac{1}{\sin\large\frac{2\pi}{n}}$
$\Rightarrow 2\cos\big(\large\frac{2\pi}{n}\big).$$\sin\large\frac{2\pi}{n}$$=sin\frac{3\pi}{n}$
$\Rightarrow 2\sin \Large\frac{2\pi}{n}.$$\cos\Large\frac{2\pi}{n}=$$\sin\large\frac{3\pi}{n}$
$\Rightarrow \sin\large\frac{4\pi}{n}-$$\sin\large\frac{3\pi}{n}$
$\Rightarrow \large\frac{4\pi}{n}=$$\pi-\large\frac{3\pi}{n}$
$\Rightarrow \large\frac{7\pi}{n}=$$\pi$
$n=7$
Hence (b) is the correct answer.
answered Nov 18, 2013 by sreemathi.v
edited Mar 25, 2014 by rvidyagovindarajan_1
 

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