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Questions  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Q)

Solve the equation $\sin 9\theta=\sin \theta$

$\begin{array}{1 1}(A)\;\theta=\large\frac{n\pi}{4},\frac{(2n+1)\pi}{10}\\(B)\;\theta=\large\frac{n\pi}{6},\frac{(2n+1)\pi}{8}\\(C)\;\theta=\large\frac{n\pi}{2},\frac{(3n+1)\pi}{10}\\(D)\;\theta=\large\frac{\pi}{12},\frac{(2n+1)\pi}{12}\end{array} $

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A)
Toolbox:
  • $\sin a-\sin b=2\cos\large\frac{(a+b)}{2}$$\sin \large\frac{a-b}{2}$
Given :
$\sin 9\theta=\sin \theta$
$\sin 9\theta-\sin \theta=0$
$2\cos \large\frac{9\theta+\theta}{2}$$\sin \large\frac{9\theta-\theta}{2}$
$\cos 5\theta-\sin 4\theta=0$
$\cos 5\theta=0$
$5\theta=\large\frac{(2n+1)\pi}{2}$
$\theta=\large\frac{(2n+1)\pi}{10}$
$\sin 4\theta=0=\sin \pi$
$4\theta=n\pi$
$\theta=\large\frac{n\pi}{4}$
$\therefore \theta=\large\frac{n\pi}{4},\frac{(2n+1)\pi}{10}$
Where $n=0,\pm 1,\pm 2....$
Hence (A) is the correct answer.
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