Show $\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)}$ In working to prove that
$$1+\cos\theta+\cos(2\theta)+\dots+\cos(n\theta)=\frac{1}{2}+\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)} \tag{1}$$
I have shown
$$\begin{align}
1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\dots+\cos(n\theta) &=\Re\left(\frac{1-e^{(n+1)i\theta}}{1-e^{i\theta}}\right) \\[6pt]
&=\frac{1-\cos(\theta)+\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)} \\[6pt]
&=\frac{1}{2}+\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)} \tag{2}
\end{align}$$
but I am unsure how to proceed from here and get the last term of $(2)$ to match the last term of $(1)$:

$$\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)} \tag{3}$$

I have read this post "How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?", but I cannot seem to convert from the form their answer is in to my form.
If possible, I would like to avoid using too many identities as this is an exercise in my complex analysis book. 

To prove 
$$\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)}$$
examine the LHS
$$LHS = \frac{\cos(n\theta)-\cos(n\theta)\cos\theta+\sin(n\theta)\sin\theta}{2-2\cos\theta}$$
$$= \frac{\cos(n\theta)(1-\cos\theta)+\sin(n\theta)\sin\theta}{2(1-\cos\theta)}$$
Then, use $1-\cos\theta=2\sin^2\frac\theta2$ and $\sin\theta = 2\sin\frac\theta2 \sin\frac\theta2$ to simplify,
$$LHS= \frac{\cos(n\theta)\sin\frac\theta2+\sin(n\theta)\cos\frac\theta2}{2\sin\frac\theta2}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin\frac\theta2}=RHS$$
