Question

I REALLY NEED HELP: SEE ATTACHMENT

Answer 1

I think you have to use the sum to product formula.

Answer 2

\[\large \sin\alpha+\sin\beta=2\sin(\frac{\alpha +\beta}{2})\cos(\frac{\alpha -\beta}{2})\]

Answer 3

For the numerator, you have: \(\sin\theta+\sin(2\theta)\)
By applying the trig identity above you'll get:

\[\large \sin\theta+\sin(3\theta)=2\sin(\frac{\theta +3\theta}{2})\cos(\frac{\theta -3\theta}{2})\]

Answer 4

Simplify it\[
\large \sin\theta+\sin(3\theta)=2\sin(\frac{4\theta}{2})\cos(\frac{-2\theta}{2})=2\sin(2\theta)\cos(-\theta)\\\large=2\sin(2\theta)\cos(\theta)\]

Answer 5

Put everything back..

\[\large \frac{\sin\theta+\sin(3\theta)}{2\sin(2\theta)}=\frac{2\sin(2\theta)\cos(\theta)}{2\sin(2\theta)}=\cos\theta\]