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3.Trigonometrical Ratios, Functions and Identities
medium
If $\sin 6\theta = 32{\cos ^5}\theta \sin \theta - 32{\cos ^3}\theta \sin \theta + 3x,$ then $x = $
A
$\cos \theta $
B
$\cos 2\theta $
C
$\sin \theta $
D
$\sin 2\theta $
Solution
(d) $\sin 6\theta = 2\sin 3\theta \cos 3\theta $
$ = 2\,[3\sin \theta – 4{\sin ^3}\theta ]\,[4{\cos ^3}\theta – 3\cos \theta ]$
$=24 \sin \theta \cos \theta (\sin ^2 \theta + \cos ^2 \theta) -18\sin \theta \cos \theta -32 \sin ^2 \theta \cos^2 \theta$
$ = 32{\cos ^5}\theta \sin \theta – 32{\cos ^3}\theta \sin \theta + 3\sin 2\theta $
On comparing, $x = \sin 2\theta .$
Standard 11
Mathematics
