Q140 / 200 · AP EAMCET 2001 · Mathematics

$\frac{\sin 5 \theta}{\sin \theta}$ is equal to

  • Correct answer
    A
    $16 \cos ^4 \theta-12 \cos ^2 \theta+1$
  • B
    $16 \cos ^4 \theta+12 \cos ^2 \theta+1$
  • C
    $16 \cos ^4 \theta-12 \cos ^2 \theta-1$
  • D
    $16 \cos ^4 \theta+12 \cos ^2 \theta-1$

Solution

$\begin{aligned} & \frac{\sin 5 \theta}{\sin \theta}=\frac{\sin (3 \theta+2 \theta)}{\sin \theta} \\ & =\frac{\sin 3 \theta \cos 2 \theta+\cos 3 \theta \sin 2 \theta}{\sin \theta} \\ & =\frac{\left(3 \sin \theta-4 \sin ^3 \theta\right)(\cos 2 \theta)+2 \cos 3 \theta \cdot \sin \theta \cos \theta}{\sin \theta} \\ & =\left(3-4 \sin ^2 \theta\right)\left(2 \cos ^2 \theta-1\right) \\ & \quad+2 \cos \theta\left(4 \cos ^3 \theta-3 \cos \theta\right) \\ & =\left[3-4\left(1-\cos ^2 \theta\right)\right]\left(2 \cos ^2 \theta-1\right)+8 \cos ^4 \theta-6 \cos ^2 \theta \\ & =\left[4 \cos ^2 \theta-1\right]\left(2 \cos ^2 \theta-1\right)+8 \cos ^4 \theta-6 \cos ^2 \theta \\ & =8 \cos ^4 \theta-6 \cos ^2 \theta+1+8 \cos ^4 \theta-6 \cos ^2 \theta+1 \\ & =16 \cos ^4 \theta-12 \cos ^2 \theta+1\end{aligned}$

Hence, the correct answer is option: 'A'.

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