Note: I was originally trying essentially to prove the same thing as this Finalising proof of Lagrange's Trig Identity. However, I do not consider my question to be a duplicate. I am looking for someone to point out my mistake in my "proof," not to prove it for me.
My attempted "proof"
My goal was to show $$1+\cos\theta+\cdots+\cos n\theta= \frac12+\frac{\sin\frac{(2n+1)\theta}{2}} {2\sin\frac{\theta}{2}}$$ when $0<\theta<2 \pi$.
I see that $$\begin{align}LHS &= \Re\left(\frac {1-e^{i(n+1)\theta}} {1-e^{i\theta}}\right) \tag{1}\\[4pt] &= \frac {(\cos\theta-1)(\cos((n+1)\theta)-1)+\sin((n+1)\theta)\sin\theta} {(\cos\theta-1)^2+\sin^2\theta} \tag{2}\\[4pt] &=\frac{(\cos\theta-1)(\cos n\theta \cos \theta-\sin n\theta \sin\theta-1)+(\sin n\theta \cos\theta + \cos n\theta \sin\theta)\sin\theta} {2-2\cos\theta} \tag{3} \\[4pt] &= \frac {\cos n\theta-\cos\theta-\cos n\theta \cos\theta +\sin n\theta \sin\theta +1} {2-2\cos\theta} \tag{4}\\[4pt] &=\frac12+\frac12\cos n\theta+ \frac{\sin n\theta \sin\theta} {2-2\cos\theta} \tag{5} \end{align}$$
To finish proof, it suffices to show that $$\frac {\sin\frac{(2n+1)\theta}{2}} {2\sin\frac{\theta}{2}}=\frac12\cos n\theta+\frac{\sin n\theta \sin\theta} {2-2\cos\theta} \tag{6}$$
By our restriction on $\theta$, we must have $\sin\frac\theta2=\sqrt{\frac12(1-\cos \theta)}$.
We see that $\cos^2\frac\theta{2}=\frac12(1+\cos \theta)$.
Case 1: Suppose $\cos\frac{\theta}{2}=\sqrt{\frac12(1+cos \theta)}$. Then
$$\begin{align} \frac {\sin\frac{(2n+1)\theta}{2}} {2\sin\frac{\theta}{2}} &= \frac{\sin n\theta \cos\frac{\theta}{2}+\cos n\theta \sin\frac{\theta}{2}} {2\sin\frac{\theta}{2}} \tag{7} \\[4pt] &=\frac{\cos n\theta}{2}+\frac {\sin n \theta \sqrt {1+\cos \theta}} {2 \sqrt {1-\cos \theta}} \tag{8} \\[4pt] &=\frac12\cos n\theta + \frac{\sin n \theta \sin \theta} {2-2\cos \theta} \tag{9} \end{align}$$ as needed.
Case 2: Suppose $\cos\frac{\theta}{2}=-\sqrt{\frac12(1+\cos\theta)}$. Then, by similar logic, I obtained $$\frac{\sin\frac{(2n+1)\theta}{2}} {2\sin\frac{\theta}{2}}= \frac12 \cos n \theta- \frac{\sin n \theta \sin \theta} {2-2\cos \theta} \tag{10}$$ but this seems wrong to me.
Where's my mistake?
