- Home
- Standard 11
- Mathematics
3.Trigonometrical Ratios, Functions and Identities
medium
$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi = $
A
$\cos 2\theta $
B
$cos 3\theta$
C
$\sin 2\theta $
D
$\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
Similar Questions
normal
