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

Question

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.

Answer 1

$1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\dots+\cos(n\theta)=$

$\dfrac{2+e^{i\theta}+e^{-i\theta}+e^{2i\theta}+e^{-2i\theta}+e^{3i\theta}+e^{-3i\theta}\dots+e^{ni\theta}+e^{-ni\theta}}2=$

$\dfrac{1+e^{-ni\theta}+\dots+e^{-3i\theta}+e^{-2i\theta}+e^{-i\theta}+1+e^{i\theta}+e^{2i\theta}+e^{3i\theta}+\cdots+e^{ni\theta}}2=$

$\dfrac12$+$\dfrac{e^{-ni\theta}\left(\dfrac{1-e^{(2n+1)i\theta}}{1-e^{i\theta}}\right)}2=$

$\dfrac12$+$\dfrac{e^{-ni\theta}\left(\dfrac{e^{-i\theta/2}-e^{(2n+1/2)i\theta}}{e^{-i\theta/2}-e^{i\theta/2}}\right)}2=$

$\dfrac12$+$\dfrac{ \left(\dfrac{e^{-i(n+1/2)\theta}-e^{i(n+1/2)\theta}}{e^{-i\theta/2}-e^{i\theta/2}}\right)}2=$

$\dfrac12+\dfrac12\dfrac{ \sin\left((n+\frac12)\theta\right)}{\sin(\theta/2)}$

Answer 2

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,

$$LHS= \frac{\cos(n\theta)\sin\frac\theta2+\sin(n\theta)\cos\frac\theta2}{2\sin\frac\theta2}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin\frac\theta2}=RHS$$