## Sum of Sines Trigonometric Equation

We can solve the equation
$sin \theta + sin 4 \theta =sin 2 \theta n+ sin 3 \theta$

using the identities
$sin P +sin Q=2 sin (\frac{P+Q}{2}) cos (\frac{P+Q}{2})$
(1) and
$cos P-cos Q=2 sin (\frac{P+Q}{2})sin (\frac{P-Q}{2})$
(2).
using (1) on both sides of the equation gives
$2sin \frac{5 \theta}{2} cos \frac{3 \theta}{2} = 2 sin \frac{5 \theta}{2} cos \frac{\theta}{2}$

Subtracting
$2 sin \frac{5 \theta}{2} sin \frac{\theta}{2}$
from both sides and factorising gives
$2sin \frac{5 \theta}{2}( cos \frac{3 \theta}{2} - cos \frac{\theta}{2})=0$

Using identity (2) with the bracketed term gives
$2sin \frac{5 \theta}{2}(-2sin \theta sin \frac{\theta}{2})=0$

Hence
$\frac{5 \theta}{2} = n \rightarrow \theta = \frac{2n \pi}{5}$
or
$\theta = n \pi$
or
$\frac{\theta}{2} = n \pi \rightarrow \theta = 2n \pi$
.