Question

Prove that:
sin 3θ = 3 sinθ + 4 sin³θ.

Answer 1

Answer:

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Answer 2

Correct question :-

Prove that:

sin 3θ = 3 sinθ - 4 sin³θ.

Solution :-

$\underline{\underline{\sf{LHS:-}}}$

$\sf{sin3\theta}\\\\\sf{=sin(\theta+2\theta)}\\\\\sf{[sin(A+B)=sinAcosB+cosAsinB]}\\\\\sf{= sin\theta cos2\theta+cos\theta sin2\theta}\\\\\sf{= sin\theta (1-2sin^2\theta)+cos\theta(2sin\theta cos\theta) }\\\\\sf{[As,\ cos2\theta=1-2sin^2\theta,\ and\ sin2\theta=2sin\theta cos\theta]}\\\\\sf{=sin\theta-2sin^3\theta+2sin\theta cos^2\theta}\\\\\sf{=sin\theta-2sin^3\theta+2sin\theta(1-sin^2\theta)}\\\\\sf{[As\ cos^2\theta=1-sin^2\theta]}\\\\\sf{=sin\theta-2sin^3\theta+2sin\theta-2sin^3\theta}$

$\sf{=3sin\theta-4sin^3\theta=\underline{\underline{RHS}}}$