Is it possible to find period of $f(x)=\sum_{k=1}^{n}(\cos (kx)+\sin(kx))$?

Question

Suppose $n\in \mathbb{N}$ , $n <\infty$ .Is it possible to find period of $f:\mathbb{R}\mapsto \mathbb{R} \\f(x)=\sum_{k=1}^{n}(\cos (kx)+\sin(kx))$ ?
such that :$$f(x)=\cos x+ \sin x +\cos (2x)+\sin (2x)+...+\cos(nx)+\sin(nx)$$ for little $n $ it easy to find, but what can we say for an unknown $n$ ?
I know I should take L.c.m. Of periods . for fixed $n$ we can write $$T_1=\frac{2\pi}{1},T_2=\frac{2\pi}{2},T_3=\frac{2\pi}{3} ,...T_n=\frac{2\pi}{n} \\ T=2\pi.\frac{\frac{[1,2,3,...,n]}{1}+\frac{[1,2,3,...,n]}{2}+\frac{[1,2,3,...,n]}{3}+...\frac{[1,2,3,...,n]}{n}}{[1,2,3,...,n]}$$

Answer 1

Hint: You can sum these by observing

$$\cos(kx)=\operatorname{Re}[\exp(ikx)]$$ $$\sin(kx)=\operatorname{Im}[\exp(ikx)]$$

and by using the geometric series.

In which $\exp(ikx)=\left[\exp(ix)\right]^k=\cos(kx)+i\sin(kx)$ is Euler's Formula. The operators $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ are the real and imaginary part of $z$.

It is easy to sum

$$\sum_{k=1}^{n}\left[\exp(ix)\right]^k=\frac{\exp(ix)-[\exp(ix)]^{n+1}}{1-\exp(ix)}$$

Determine the real and imanginary part of this expression and you will obtain:

$$\sum_{k=1}^n \cos(kx) = \dfrac{\sin \frac{nx}{2} \cos \frac{n+1}{2}x}{\sin \frac x2}$$

$$\sum_{k=1}^{n}\sin(kx)=\frac{\sin \frac{nx}{2}\sin \frac{n+1}{2}x}{\sin \frac x2}$$