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 $
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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