From \(\displaystyle e^{i\theta}= cos(\theta)+ i sin(\theta)\), it immediately follows that \(\displaystyle e^{-i\theta}= cos(-\theta)+ i sin(-\theta)= cos(\theta)- i sin(\theta)\). Adding those, \(\displaystyle e^{i\theta}+ e^{-i\theta}= 2cos(\theta)\) or \(\displaystyle cos(\theta)= \frac{e^{i\theta}+ e^{-i\theta}}{2}\) and subtracting, \(\displaystyle e^{i\theta}- e^{-i\theta}= 2 sin(\theta)\) or \(\displaystyle sin(\theta)= \frac{e^{i\theta}- e^{-i\theta}}{2}\).
Replace the sine and cosines in what you want to prove with those.
