Question

Prove that $\sin 6\theta +\sin 4\theta -\sin 2\theta =4\cos \theta \cdot \sin 2\theta \cdot \cos 3\theta$

Answer 1

LHS :-

$\Rightarrow$ we know $sinx-siny=2cos\left(\frac{x+y}{2}\right)sin\left(\frac{x-y}{2}\right)$

$\Rightarrow sin6\theta +2cos3\theta sin\theta$

$\Rightarrow sin(3\theta +3\theta )+2cos3\theta sin\theta$

$\Rightarrow sin3\theta cos3\theta +sin3\theta cos3\theta +2cos3\theta sin\theta$

$\Rightarrow 2cos3\theta [sin3\theta +sin\theta ]$

$\Rightarrow 2cos3\theta [2sin2\theta .cos\theta ]$

$=4cos\theta .sin2\theta cos3\theta =RHS$