Proving a result of $\text{Re}\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)$

Question

Show that for $0<\theta<2\pi$: $$\text{Re}\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)=\frac{1}{2}+\frac{\text{sin}\left(n+\frac{1}{2}\right)\theta}{2\text{sin}\frac{\theta}{2}}$$

I am stuck up to this point \begin{align} \text{Re}\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)&=\frac{1-\text{cos}\theta-\text{cos}(n\theta+\theta)+\text{cos}\theta\text{cos}(n\theta+\theta)}{2-2\text{cos}\theta} \\ &=\frac{1-\text{cos}\theta}{2-2\text{cos}\theta}+\frac{\text{cos}(n\theta+\theta)(\text{cos}\theta-1)}{2-2\text{cos}\theta} \\ &=\frac{1}{2}-\frac{\text{cos}(n\theta+\theta)}{2} \end{align}

I hint would be very helpful.

Answer 1

I don't quite obtain the result of your first line, but \begin{align} \operatorname{Re}\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right) & =\frac{1-\cos\theta-\cos(n+1)\theta+\cos n\theta }{2(1-\cos\theta)} \\[1ex] & = \frac12 + \frac{\cos n\theta-\cos(n+1)\theta}{2(1-\cos\theta)} \end{align} Now you can use the linearisation/factorisation formulæ:

• $1-\cos\theta=2\sin^2\dfrac\theta2$;
• $\cos p-\cos q=-2\sin\dfrac{p+q}2\,\sin\dfrac{p-q}2$, which yields here $$\cos n\theta-\cos(n+1)\theta=-2\sin\dfrac{(2n+1)\theta}2\,\sin\dfrac{-\theta}2=\sin\dfrac{(2n+1)\theta}2\,\sin\dfrac{\theta}2.$$

Answer 2

HINT

It seems useful refer to Lagrange's trigonometric identities

$$ \sum_{n=1}^N \cos (n\theta) = -\frac{1}{2}+\frac{\sin\left(\left(N+\frac{1}{2}\right)\theta\right)}{2\sin\left(\frac{\theta}{2}\right)}$$

observing that for $\theta \neq 2k\pi$

$$\text{Re}\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)=\text{Re}(1+e^{i\theta}+\ldots+ e^{in\theta})$$

Answer 3

$$\dfrac{1-e^{2mit}}{1-e^{2it}}=\dfrac{e^{imt}(e^{imt}-e^{imt})}{e^{it}(e^{it}-e^{-it})}$$

$$=e^{i(m-1)t}\cdot\dfrac{2i\sin mt}{2i\sin t}$$

$$=\dfrac{2\sin mt(\cos(m-1)t+i\sin(m-1)t)}{2\sin t}$$

Now $2\sin mt\cos(m-1)t=\sin(2m-1)t+\sin t$

Here $2t=\theta$ and $2mt=(n+1)\theta$