• # question_answer If $\sin 6\theta =32{{\cos }^{5}}\theta \sin \theta -32{{\cos }^{3}}\theta \sin \theta +3x,$ then $x=$ [EAMCET 2003] A) $\cos \theta$ B) $\cos 2\theta$ C) $\sin \theta$ D) $\sin 2\theta$

$\sin 6\theta =2\sin 3\theta \cos 3\theta$ $=2\,[3\sin \theta -4{{\sin }^{3}}\theta ]\,[4{{\cos }^{3}}\theta -3\cos \theta ]$ =24sinqcosq(sin2q+cos2q) -18sinqcosq - 32sin2qcos2q $=32{{\cos }^{5}}\theta \sin \theta -32{{\cos }^{3}}\theta \sin \theta +3\sin 2\theta$ On comparing, $x=\sin 2\theta .$