Cos 2(θ + phi ) - 4cos (θ + phi )sin θ sin phi + 2{sin ^2}phi =

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3.Trigonometrical Ratios, Functions and Identities

medium

$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi = $

$\cos 2\theta $

$cos 3\theta$

$\sin 2\theta $

$\sin 3\theta $

Solution

(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.

Standard 11

Mathematics

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medium

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$ \frac {sinAsin BsinC}{(sin 3A+ sin 3B+ sin 3C)}$ (wherever definied) is –

normal

$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi = $

Solution

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