sin3θ = (A) 3sinθ + 4sin^3θ (B) 4sin^3θ − 3sinθ

Question

\(\sin 3 \theta=\)

(A) \(3 \sin \theta+4 \sin ^3 \theta\)

(B) \(4 \sin ^3 \theta-3 \sin \theta\)

(C) \(3 \sin \theta-4 \sin ^3 \theta\)

(D) \(4 \sin \theta-3 \sin ^3 \theta\)

Answer 1

Correct option is (C) \(3 \sin \theta-4 \sin ^3 \theta\)

\(\sin 3 \theta= \sin (2 \theta + \theta)\)

\(= \sin 2 \theta \cos \theta + \cos 2 \theta + \sin \theta\)

\(= 2(\sin \theta \cos \theta) \cos \theta + (\cos^2 \theta - \sin^2 \theta) \sin \theta\)

\(= 2 \sin \theta \cos^2 \theta + \cos^2\theta \sin\theta - \sin ^3 \theta\)

\(= 2 \sin \theta (1 - \sin^2 \theta) + (1 - \sin^2\theta) \sin \theta - \sin^3 \theta\)

\(= 2\sin\theta - 2\sin^3\theta + \sin \theta - \sin^3 \theta - \sin^3 \theta\)

\(= 3 \sin \theta - 4\sin^3 \theta\)