Is there another method ? of this summation $\sum_{k=0}^{n}{n \choose k}\cos kx$ Calculate or simplify $$\sum_{k=0}^{n}{n \choose k}\cos (kx)$$
by using complex method.
Let $A=\sum_{k=0}^{n}{n \choose k}\cos(kx)$ and $B=\sum_{k=0}^{n}{n \choose k}\ i \sin(kx)$ , then
$$\begin{align*}
A+iB &= \sum_{k=0}^{n}{n \choose k}\cos(kx)+\sum_{k=0}^{n}{n \choose k}\ i \sin(kx) \\
&= \sum_{k=0}^{n}{n \choose k}\left(\cos(kx)+i \sin(kx)\right) \\
&=\sum_{k=0}^{n}{n \choose k}e^{ikx}\end{align*}$$
By Binomial coefficient $(x+y)^{n}=\sum_{k=0}^{n}{n \choose k}x^{n-k}y^{k}$, so
$$A+iB=(1+e^{ix})^{n},$$
then
$$Re(A+iB)=Re(1+e^{ix})^{n},$$
thus
$$A=Re\left \{ e^{\frac{nix}{2}} (e^{\frac{-ix}{2}}+e^{\frac{ix}{2}})^n\right \}=Re\left ( e^{\frac{nix}{2}}2^{n}\cos^{n}\frac{x}{2} \right ).$$
So
$$\sum_{k=0}^{n}{n \choose k}\cos(kx)=2^n\cos^n\left(\frac{x}{2}\right)\cos\left(\frac{nx}{2}\right)$$
 A: Use of generating functions can help. Suppose that
$$ f(x)=\sum_{n=0}^\infty a_n\,\frac{x^n}{n!} \tag{1} $$
is the exponential generating function (e.g.f.) of $\,\{a_n\}.\,$ Then we have
$$ f(x)\,e^x = \sum_{n=0}^\infty b_n\,\frac{x^n}{n!} \tag{2} $$
which is the e.g.f. of the related sequence
$$ b_n=\sum_{k=0}^n{n \choose k}a_k. \tag{3} $$
The e.g.f. for the cosine sequence is
$$ {\textstyle\frac12}(e^{x \exp(i\,t)} + e^{x \exp(-i\,t)}) =
  \sum_{n=0}^\infty \cos(n\,t)\, \frac{x^n}{n!}. \tag{4} $$
Thus, we get the result that if
$$ c_n := \sum_{k=0}^{n}{n \choose k}\cos (k\,t), \tag{5} $$
then the e.g.f of $\,\{c_n\}\,$ is
$$ g(x):={\textstyle\frac12}(e^{x\exp(i\,t)}+e^{x\exp(-i\,t)})e^x. \tag{6} $$
Since $\, 1 + e^{i\,t} = 2\cos(t/2)e^{i\,t/2},\,$ this implies
$$ g(x) = {\textstyle\frac12}(e^{2\cos(t/2)\exp(i\,t/2)\,x} +
e^{2\cos(t/2)\exp(-i\,t/2)\,x}) \tag{7} $$ which simplifies to
$$ c_n = 2^n\cos^n\left(\frac{t}{2}\right)\cos\left(\frac{nt}{2}\right). \tag{8} $$
A: Here's a induction proof. I think the proof using complex numbers is better, but the next proof illustrates a useful method.
We'll prove the following in parallel
$$\sum_k\binom{n}{k}\cos(k\ x) = 2^n\cos^n\left(\frac{x}{2}\right)\cos\left(\frac{n\ x}{2}\right)$$
$$\sum_k\binom{n}{k}\sin(k\ x) = 2^n\cos^n\left(\frac{x}{2}\right)\sin\left(\frac{n\ x}{2}\right)$$
For $n=0$ the equalities are obvious.
For the inductive step we have
\begin{align}
&\sum_k\binom{n+1}{k}\cos(k\ x)  = 
\sum_k\binom{n}{k}\cos(k\ x)+ \sum_k\binom{n}{k-1}\cos(k\ x)\\
&=\sum_k\binom{n}{k}\cos(k\ x)+ \sum_k\binom{n}{k}\cos((k+1)\ x)\\
&=\sum_k\binom{n}{k}\cos(k\ x)+ \cos(x)\sum_k\binom{n}{k}\cos(k\ x)- \sin(x)\sum_k\binom{n}{k}\sin(k\ x)\\
&\stackrel{IH}= 2^n\cos^n\left(\frac{x}{2}\right)\cos\left(\frac{n\ x}{2}\right)
+\cos(x)2^n\cos^n\left(\frac{x}{2}\right)\cos\left(\frac{n\ x}{2}\right)
-\sin(x)2^n\cos^n\left(\frac{x}{2}\right)\sin\left(\frac{n\ x}{2}\right)\\
&= 2^n\cos^n\left(\frac{x}{2}\right)\left(\cos\left(\frac{n\ x}{2}\right)+\cos(x)\cos\left(\frac{n\ x}{2}\right)-\sin(x)\sin\left(\frac{n\ x}{2}\right)\right)\\
&= 2^n\cos^n\left(\frac{x}{2}\right)\left(\cos\left(\frac{n\ x}{2}\right)+\cos\left(\frac{(n+2)\ x}{2}\right)\right)\\
&= 2^n\cos^n\left(\frac{x}{2}\right)2\cos\left(\frac{x}{2}\right)\cos\left(\frac{(n+1)\ x}{2}\right) = 2^{n+1}\cos^{n+1}\left(\frac{x}{2}\right)\cos\left(\frac{(n+1)\ x}{2}\right)
\end{align}
where we used $\cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b)$ and $\cos(a)+\cos(b) = 2 \cos(\frac{a+b}{2})\cos(\frac{a-b}{2})$.
A similar calculation using $\sin(a+b) = \sin(a)\cos(b)+\cos(a)\sin(b)$, $\sin(a)+\sin(b) = 2 \sin(\frac{a+b}{2})\cos(\frac{a-b}{2})$ and the IH proves that $$\sum_k\binom{n+1}{k}\sin(k\ x) = 2^{n+1}\cos^{n+1}\left(\frac{x}{2}\right)\sin\left(\frac{(n+1)\ x}{2}\right)$$
and that completes the induction.
