Compute the period of this function $f(x)=7+3\cos{(\pi x)}-8\sin{(\pi x)}+4\cos{(2\pi x)}-6\sin{(2\pi x)}$ I'm starting my journey into Fourier Series. I am given this function: $$f(x)=7+3\cos{(\pi x)}-8\sin{(\pi x)}+4\cos{(2\pi x)}-6\sin{(2\pi x)}$$
Following my book, this function has a period of $T=2$ (this is the book I'm reading).
However from what I know:


*

*A function $f(x)$ is periodic with period $T$, if and only if each of its summands is periodic with period $T$.

*A function $g(x)$ is said to be periodic with period $T$ when it satisfies: $g(x+T)=g(x)$


So, for the function above, I can't seem to understand why the author says its period is 2, since: (please correct the following statements if I'm wrong)


*

*$\cos{(\pi x+T)}=\cos{(\pi x)} \Leftrightarrow T=\boxed{2\pi}, 4\pi,...$

*$\sin{(\pi x+T)}=\sin{(\pi x)} \Leftrightarrow T=\boxed{2\pi}, 4\pi,...$

*$\cos{(2\pi x+T)}=\cos{(2\pi x)} \Leftrightarrow T=\boxed{2\pi}, 4\pi,...$

*$\sin{(2\pi x+T)}=\sin{(2\pi x)} \Leftrightarrow T=\boxed{2\pi}, 4\pi,...$


I derived the above periods using the formulas for the sine and cosine of a sum of 2 angles. For example:
$$\begin{align} \cos{(\pi x)} & = \cos{(\pi x+T)} \\ & = \cos{(\pi x)}\underbrace{\cos{(T)}}_{=1}-\sin{(\pi x)}\underbrace{\sin{(T)}}_{=0} \end{align} $$
Which is satisfied only when $T=\boxed{2\pi}, 4\pi,... = 2\pi k$
So... why did the book say $f(x)$ has a period equal to $T=2$? Where am I going wrong?

You need $$7+3\cos{(\pi x)}-8\sin{(\pi x)}+4\cos{(2\pi x)}-6\sin{(2\pi x)}=$$
$$=7+3\cos{(\pi (x+T))}-8\sin{(\pi (x+T))}+4\cos{(2\pi( x+T))}-6\sin{(2\pi (x+T))}$$ for all real $x$, which indeed gives $T=2$.
