How to prove $\sum^n_{k=0}\binom{n}k\cos\big((n-2k)\theta\big)=2^n\cos^n\theta$?

Question

Given that $n\in\mathbb{Z}$, for any $\theta\in\mathbb{R}$, prove that $$\sum^n_{k=0}\binom{n}k\cos\big((n-2k)\theta\big)=2^n\cos^n\theta\,.$$

I tried to finish the proof by Mathematical Induction. However, when I want to prove the case $P(k+1)$ is true, I am stucked.

I assumed that $P(m)$ is true for some positive integer $m$.

I don't know how to prove $2^m\cos^m\theta+\cos(m+2)\theta=2^{m+1}\cos^{m+1}\theta$

Can anyone help me? Thanks.

Answer 1

I don't follow your induction argument.

I see another route:

$$\sum_{k=0}^n \binom{n}{k} \cos{(n-2k) \theta} = \text{Re}\left\{\sum_{k=0}^n \binom{n}{k} e^{i(n-2k)\theta}\right\} = \text{Re}\{(e^{i\theta} + e^{-i\theta})^n\}$$

I have used binomial theorem in the last step.

Answer 2

If you want to use induction, note that, for any real (or complex) numbers $\phi$ and $\theta$, we have $$2\,\cos(\theta)\,\cos(\phi)=\cos(\phi+\theta)+\cos(\phi-\theta)\,.\tag{*}$$ Now, let $$f_n(\theta):=\sum_{k=0}^n\,\binom{n}{k}\,\cos\big((n-2k)\theta\big)$$ for each integer $n\geq 0$ and for each real (or complex) number $\theta$. We claim that $f_n(\theta)=2^n\,\cos^n(\theta)$. The basis step $n=0$ is trivial.

Now, for the inductive step, let $n\in\mathbb{Z}_{\geq 0}$ be such that $f_n(\theta)=2^n\,\cos^n(\theta)$. That is, $$2^n\,\cos(\theta)=\sum\limits_{k=0}^n\,\binom{n}{k}\,\cos\big((n-2k)\theta\big)\,.$$ Therefore, $$2^{n+1}\,\cos^{n+1}(\theta)=2\,\cos(\theta)\,\big(2^n\,\cos^n(\theta)\big)=2\,\cos(\theta)\,\left(\sum\limits_{k=0}^n\,\binom{n}{k}\,\cos\big((n-2k)\theta\big)\right)\,.$$ Therefore, $$2^{n+1}\,\cos^{n+1}(\theta)=\sum_{k=0}^n\,\binom{n}{k}\,2\,\cos(\theta)\,\cos\big((n-2k)\theta\big)\,.$$ From (*), we see that $$\begin{align}2\,\cos(\theta)\,\cos\big((n-2k)\theta\big)&=\cos\big((n-2k)\theta+\theta\big)+\cos\big((n-2k)\theta-\theta\big) \\ &=\cos\Big(\big((n+1)-2k\big)\theta\Big)+\cos\Big(\big((n+1)-2(k+1)\big)\theta\Big)\,.\end{align}$$ This shows that $$\begin{align}2^{n+1}\,\cos^{n+1}(\theta)&=\sum_{k=0}^n\,\binom{n}{k}\,\cos\Big(\big((n+1)-2k\big)\theta\Big)+\sum_{k=0}^n\,\binom{n}{k}\,\cos\Big(\big((n+1)-2(k+1)\big)\theta\Big) \\ &=\sum_{k=0}^{n+1}\,\binom{n}{k}\,\cos\Big(\big((n+1)-2k\big)\theta\Big)+\sum_{k=0}^{n+1}\,\binom{n}{k-1}\,\cos\Big(\big((n+1)-2k\big)\theta\Big)\,, \end{align}$$ where $\displaystyle\binom{n}{n+1}$ and $\displaystyle\binom{n}{-1}$ are conventionally defined to be $0$. Ergo, $$2^{n+1}\,\cos^{n+1}(\theta)=\sum_{k=0}^{n+1}\,\Biggl({{\binom{n}{k}+\binom{n}{k-1}}}\Biggr)\,\cos\Big(\big((n+1)-2k\big)\theta\Big)\,.$$ By Pascal's Rule, $$\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}$$ for all $k=0,1,2,\ldots,n+1$. It follows immediately that $$2^{n+1}\,\cos^{n+1}(\theta)=\sum_{k=0}^{n+1}\,\binom{n+1}{k}\,\cos\Big(\big((n+1)-2k\big)\theta\Big)=f_{n+1}(\theta)\,.$$ The proof is now complete.