How to prove $1+\cos2\theta+\cos4\theta+\cos6\theta+\cos8\theta=\frac{(\cos4\theta)(\sin5\theta)}{\sin\theta} $?

Question

I need help to prove that the following is true:

$$1+\cos2\theta+\cos4\theta+\cos6\theta+\cos8\theta=\frac{(\cos4\theta)(\sin5 \theta)}{\sin\theta}$$

I realize that I must evaluate the real part of this, but whatever I get I am not quite sure how to get to the required expression. I have multiplied the numerator and denominator of the result of the geometric sum by the conjugate of $ e^{2i\theta} $ and still have no luck. $$\sum_{i=0}^4 e^{2ni\theta}$$

(Apologies for poor formatting)

Answer 1

According to the link posted as a possible duplicate and containing my answer in it, we have: $$1+\cos{\theta}+\cos{2\theta}+...+\cos{n\theta}=\frac{\cos{\frac{n\theta}{2}}\cdot\sin{\frac{(n+1)\theta}{2}}}{\sin{\frac{\theta}{2}}} \tag{1}$$ Now $$1 + \cos{2\theta} + \cos{4 \theta} + \cos{6 \theta} + \cos8{\theta}=\\ 1 + \cos{(1\cdot \color{red}{2\theta})} + \cos{(2\cdot \color{red}{2\theta})} + \cos{(3 \cdot \color{red}{2\theta})} + \cos{(4\cdot \color{red}{2\theta})}=\\ \frac{\cos{\frac{4\cdot\color{red}{2\theta}}{2}}\cdot\sin{\frac{(4+1)\cdot\color{red}{2\theta}}{2}}}{\sin{\frac{\color{red}{2\theta}}{2}}}= \frac{\cos{(4\color{red}{\theta})}\cdot\sin{(5\color{red}{\theta}})}{\sin{\color{red}{\theta}}}$$

Answer 2

It isn't true (try putting $\theta = \pi/2$). It should be $\sin5\theta$ on the RHS. To prove it, you could multiply the LHS by $2\sin\theta$ and use $2\sin\theta\cos2\theta = \sin3\theta - \sin\theta$, etc. Most of the terms cancel and you end up with $\sin\theta + \sin9\theta = 2\cos4\theta\sin5\theta$.