Question

show that
$\sin 3θ = 3 \sin θ - {4}^{3} sinθ$
​

Answer 1

Given :

$\sin 3\theta = 3\sin\theta-4\sin^3\theta$

To Find :

$\text{Show that RHS=LHS }$

Explanation :

We can prove the given equations by using some basic trigonometric identities alike :-

$\cos^2 A = 1-\sin^2A$

$\sin (A+A)=\sin A \cos B + \cos A \sin B$

$\sin 2A = 2 \sin A \cos A$

$\cos 2A = \cos ^2 A - \sin ^2 A$

Solution :

$\sin 3\theta = \text {LHS}$

$\implies \sin (2\theta+\theta)$

$\implies \sin2\theta\cos \theta+\cos2\theta\sin\theta+(\cos^2\theta-\sin^2\theta)\sin\theta$

$(\because \sin 2A = 2 \sin A \cos A \:\:\&\:\:\cos 2A = \cos ^2 A - \sin ^2 A)$

$\implies 2\sin\theta\cos^2\theta+(1-\sin^2\theta-\sin^2\theta)\sin\theta$

$\implies 2\sin\theta\cos^2\theta+(1-2\sin^2\theta)\sin\theta$

$\implies 2\sin\theta\cos^2\theta+\sin\theta-2\sin^3\theta$

$\implies 2\sin\theta(1-\sin^2\theta)+\sin\theta-2\sin^2\theta$

$(\because \cos^2 A = 1-\sin^2A)$

$\implies 2\sin\theta-2\sin^3\theta+\sin\theta-2\sin^3\theta$

$\implies 3\sin\theta-4\sin^3\theta = \text{RHS}$

Hence Proved.

Answer 2

Answer:

Refer to the attachment hope it's helpful for