Question

Find the value of the expression $2\sin 3\theta \cos \theta -\sin 4\theta -\sin 2\theta $.

Answer 1

Hint: Use $\sin x+\sin y=2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$ to simplify the expression.
Alternatively, you can use $2\sin x\cos y=\sin \left( x+y \right)+\sin \left( x-y \right)$ to simplify the expression

Complete step by step answer:
We have
$2\sin 3\theta \cos \theta -\sin 4\theta -\sin 2\theta =2\sin 3\theta \cos \theta -\left( \sin 4\theta +\sin 2\theta \right)$
We know that $\sin x+\sin y=2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$
Put $x=4\theta $ and $y=2\theta $, we get
\[\sin 4\theta +\sin 2\theta =2\sin \left( \dfrac{4\theta +2\theta }{2} \right)\cos \left( \dfrac{4\theta -2\theta }{2} \right)=2\sin 3\theta \cos \theta \]
Hence, we have
$2\sin 3\theta \cos \theta -\sin 4\theta -\sin 2\theta =2\sin 3\theta \cos \theta -2\sin 3\theta \cos \theta =0$
Hence the expression identically goes to 0.

Note: Alternative Solution:
We know that $2\sin x\cos y=\sin \left( x+y \right)+\sin \left( x-y \right)$
Put $x=3\theta $ and $y=\theta $ , we get
$2\sin 3\theta \cos \theta =\sin \left( 3\theta +\theta \right)+\sin \left( 3\theta -\theta \right)=\sin 4\theta +\sin 2\theta $
Hence we have $2\sin 3\theta \cos \theta -\sin 4\theta -\sin 2\theta =\sin 4\theta +\sin 2\theta -\sin 4\theta -\sin 2\theta =0$
Hence the expression goes identically to 0.
We can remember the formulae involved in solving the question through this short Aid to memory:
[1] S+S = 2SC
[2] S-S = 2CS
[3] C+C= 2CC
[4] C-C=-2SS
Consider [3].
We can use it to remember two formulae
viz $\cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$ and \[2\cos x\cos y=\cos \left( x+y \right)+\cos \left( x-y \right)\].
Similarly, every equation shown above helps to memorise two equations.