Proof of $\cos \theta+\cos 2\theta+\cos 3\theta+\cdots+\cos n\theta=\frac{\sin\frac12n\theta}{\sin\frac12\theta}\cos\frac12(n+1)\theta$

State the sum of the series $z+z^2+z^3+\cdots+z^n$, for $z\neq1$.

By letting $z=\cos\theta+i\sin\theta$, show that

$$\cos \theta+\cos 2\theta+\cos 3\theta+\cdots+\cos n\theta=\frac{\sin\frac12n\theta}{\sin\frac12\theta}\cos\frac12(n+1)\theta$$

Where $\sin\frac12\theta\neq0$.

I know the first part, The second part Im kind of stuck in showing that

My Attempt:

$$\begin{align}\Re{(\cos\theta+i\sin\theta)+(\cos\theta+i\sin\theta)^2+\ldots+(\cos\theta+i\sin\theta)^n}\end{align}$$

I realized that this is a Geometric Progression, so its in the form:

$a+ar+ar^2+....+ar^n$ , where $a=(\cos\theta+i\sin\theta)$ and $r=(\cos\theta+i\sin\theta)$

So I will apply the formula for the Sum of a G.P to my problem.

$$\begin{align}\Re \frac{(\cos\theta+i\sin\theta)(1-(\cos\theta+i\sin\theta)^n)}{1-(\cos\theta+i\sin\theta)}\end{align}$$

I applied the De Movire Theorem and simplified as follows:

$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta}{1-(\cos\theta+i\sin\theta)}\end{align}$$

$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta}{1-(\cos\theta+i\sin\theta)}\frac{(1+(cos\theta+isin\theta))}{(1+(cos\theta+isin\theta))}\end{align}$$

$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta+cos^2\theta+2icos\theta sin\theta-sin^2\theta-cos\theta cos(n+1)\theta- isin \theta cos(n+1)\theta+i cos\theta sin(n+1)\theta-sin\theta sin(n+1)\theta}{1-cos^2\theta-2i cos\theta sin\theta+sin^2\theta}\end{align}$$

$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta+cos (2\theta) +i sin (2\theta)- cos(n+1)\theta(cos\theta+isin \theta)+(i cos\theta -sin \theta) sin(n+1)\theta}{1-cos^2\theta-2i cos\theta sin\theta+sin^2\theta}\end{align}$$

$$\begin{align}\Re \frac{\cos\theta+i\sin\theta+cos (2\theta) +i sin (2\theta)- cos(n+1)\theta(1+cos\theta+isin \theta)+(1+i cos\theta -sin \theta) sin(n+1)\theta}{2sin^2\theta-2i cos\theta sin\theta}\end{align}$$

$$\begin{align}\Re \frac{[\cos\theta+i\sin\theta+cos (2\theta) +i sin (2\theta)- cos(n+1)\theta(1+cos\theta+isin \theta)+(1+i cos\theta -sin \theta) sin(n+1)\theta]}{2sin\theta(sin\theta-i cos\theta)} \frac{(sin\theta+i cos\theta)}{(sin\theta+i cos\theta)}\end{align}$$

$$\begin{align}\frac{[-cos(n+1)\theta(1+cos\theta+isin \theta)+(1+i cos\theta -sin \theta) sin(n+1)\theta]}{2sin\theta} \frac{(sin\theta+i cos\theta)}{1}\end{align}$$

$$\begin{align}\frac{[-cos(n+1)\theta(sin\theta+cos\theta sin\theta - cos\theta sin \theta)+(sin\theta-cos^2\theta -sin^2 \theta) sin(n+1)\theta]}{2sin\theta} \end{align}$$

$$\begin{align}\frac{[-cos(n+1)sin\theta+(sin\theta-1) sin(n+1)\theta]}{2sin\theta} \end{align}$$

I have no idea right now where I am taking this, I just dont know what the next step I should take. Please dont send me the solution (at least yet). Can anyone give me a hint (a little boost to my little mind) as to what I should do next, (make sure its a small hint, Don't give me the full next step) Just the help in order for me to construct the next step and carry on.

Solution 1:

Here are the main steps.

You may write $$ \begin{align} \sum_{k=1}^{n} \cos (k\theta)&=\Re \sum_{k=1}^{n} e^{ik\theta}\\\\ &=\Re\left( e^{i\theta}\frac{e^{in\theta}-1}{e^{i\theta}-1}\right)\\\\ &=\Re\left( e^{i\theta}\frac{e^{in\theta/2}\left(e^{in\theta/2}-e^{-in\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right)\\\\ &=\Re\left( e^{i\theta}\frac{e^{in\theta/2}\left(2i\sin(n\theta/2)\right)}{e^{i\theta/2}\left(2i\sin(\theta/2)\right)}\right)\\\\ &=\Re\left( e^{i(n+1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\Re\left( \left(\cos ((n+1)\theta/2)+i\sin ((n+1)\theta/2)\right)\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\frac{\sin(n\theta/2)}{\sin(\theta/2)}\cos ((n+1)\theta/2). \end{align} $$

Solution 2:

Using the appropriate prosthaphaeresis formula,

$$\begin{align} \sin{\frac{\theta}{2}}\cos{k\theta} &=\frac12\left[\sin{\left(\frac{\theta}{2}+k\theta\right)}+\sin{\left(\frac{\theta}{2}-k\theta\right)}\right]\\ &=\frac12\left[\sin{\left(\left(k+\frac{1}{2}\right)\theta\right)}-\sin{\left(\left(k-\frac{1}{2}\right)\theta\right)}\right]. \end{align}$$

In this form, summation of $k$ now transparently telescopes:

$$\begin{align} \sum_{k=1}^{n}\sin{\frac{\theta}{2}}\cos{k\theta} &=\sum_{k=1}^{n}\frac12\left[\sin{\left(\left(k+\frac{1}{2}\right)\theta\right)}-\sin{\left(\left(k-\frac{1}{2}\right)\theta\right)}\right]\\ &=\frac12\left[\sin{\left(\left(n+\frac{1}{2}\right)\theta\right)}-\sin{\left(\left(1-\frac{1}{2}\right)\theta\right)}\right]\\ &=\frac12\left[\sin{\left(\left(n+\frac{1}{2}\right)\theta\right)}-\sin{\left(\frac{\theta}{2}\right)}\right]\\ &=\sin{\left(\frac{n\theta}{2}\right)\cos{\left(\frac{(n+1)\theta}{2}\right)}}. \end{align}$$

Now divide through by $\sin\frac{\theta}{2}$.