Compute $\sum_{k=0}^n\cos(k\theta)\cos^k(\theta)$

Question

I want to compute $S=\sum_{k=0}^n\cos(k\theta)\cos^k(\theta)$ for real $\theta$ and positive integer $n$. I tried to solve this using Complex Numbers.

Let $C=\sum\limits_{k=0}^n\sin(k\theta)\cos^k\theta$

$\Rightarrow S+iC=\sum\limits_{k=0}^n\cos^k\theta\left[\cos(k\theta)+i\sin(k\theta)\right]$

$=\sum\limits_{k=0}^ne^{ik\theta}\cos^k(\theta)=\sum\limits_{k=0}^n\left[e^{i\theta}\cos\theta\right]^k$

using geometric progression formula $a=1,r=e^{i\theta}\cos\theta$

$\Rightarrow S+iC=\dfrac{e^{i(n+1)\theta}\cos^{n+1}\theta-1}{e^{i\theta}\cos\theta-1}$

At this point I tried many times to covert this into $a+ib$ form

this way i can let

$S=\sum_{k=0}^n\cos(k\theta)\cos^k(\theta)=a$

by Complex Numbers theorem

But seems like I am just stuck. Can someone help?

Answer 1

You got stuck at the point

$$S=\Re\left(\dfrac{e^{i(n+1)\theta}\cos^{n+1}\theta-1}{e^{i\theta}\cos\theta-1}\right)$$

Lets simplify this and ask about

$$\dfrac{e^{ix}a-1}{e^{iy}b-1}=\dfrac{e^{ix}a-1}{e^{iy}b-1}\cdot \frac{e^{-iy}b-1}{e^{-iy}b-1}=\frac{(e^{ix}a-1)(e^{-iy}b-1)}{b^2+1-2b\cos(y)}$$

The numerator here is

$$(e^{ix}a-1)(e^{-iy}b-1)=e^{i(x-y)}ab-e^{ix}a-e^{-iy}b+1$$

The real part of this is

$$\cos(x-y)ab-\cos(x)a-\cos(-y)b+1$$

Putting this all together we have

$$S=\frac{\cos(x-y)ab-\cos(x)a-\cos(y)b+1}{b^2+1-2b\cos(y)}$$

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

Answer 2

How do you convert any formula of the form $\frac{a+bi}{c+di}$ into $u+vi$ where $u,v$ are real? You multiply the numerator and the denominator by the complex conjugate of the denominator, $c-di.$

From your $S+iC,$ we multiplying top and bottom by the complex conjugate $e^{-i\theta}\cos\theta-1.$

You will get, in the denominator: $$\begin{align}(e^{i\theta}\cos\theta-1)(e^{-i\theta}\cos\theta-1)&=\cos^{2}\theta -\cos \theta\left(e^{i\theta}+e^{-i\theta}\right)+1\\&=1-\cos^2\theta\\&=\sin^2\theta.\end{align}$$

I'll leave the numerator to you.