MCQ
$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi = $
  • $\cos 2\theta $
  • B
    $cos 3\theta$
  • C
    $\sin 2\theta $
  • D
    $\sin 3\theta $

Answer

Correct option: A.
$\cos 2\theta $
a
(a) We have,$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi  $

Now, put $\theta = \phi = \frac{\pi }{4}$

$\cos 2\left( {\frac{\pi }{2}} \right) - 4\cos \left( {\frac{\pi }{2}} \right)\sin \left( {\frac{\pi }{4}} \right)\sin \left( {\frac{\pi }{4}} \right) + 2{\sin ^2}\left( {\frac{{2\pi }}{4}} \right) = 0$

Put $\theta = \phi = \pi /4$  in option $(a)$, 

then, $\cos 2\theta = \cos \pi /2 = 0$.

Hence option $(a)$ is correct.

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