$$
f(x) =
\begin{cases}
\frac{\pi}{4}-\frac{x}{2} & [0,\pi] \\
-\frac{3\pi}{4}+\frac{x}{2}, & (\pi,2\pi)
\end{cases}
$$
Is it right to compute only $a_n \text{ and } a_0$ coefficient for fourier series because $f(x)$ is even for fourier? How can I proove it since $f(x)!=f(-x)$
For $2\pi$-periodic function $f(x)$ defined as: $$f(x) = \begin{cases} \frac{\pi}{4}-\frac{x}{2} & x\in [0,\pi] \\ -\frac{3\pi}{4}+\frac{x}{2}, & x\in(\pi,2\pi) \end{cases}$$
We can calculate $$f(2\pi-x) = \begin{cases} \frac{\pi}{4}-\frac{2\pi-x}{2} & x\in [\pi,2\pi] \\ -\frac{3\pi}{4}+\frac{2\pi-x}{2}, & x\in (0,\pi) \end{cases}$$
$$ = \begin{cases} \frac{-3\pi}{4}+\frac{x}{2} & x\in [\pi,2\pi] \\ +\frac{1\pi}{4}-\frac{x}{2}, & x\in (0,\pi) \end{cases}$$
Thus we proved $$f(2\pi-x)=f(x)\text{ (1)}$$.
Since for $2\pi$-periodic function $f(-x)$ we have $$f(-x)= f((-x)+2\pi)=f(2\pi-x)\text{ (2)}$$
Combining (1) and (2), we proved $$f(x)= f(-x)$$.
Thus the coefficients $b_n$ that go with $\sin(2nx)$ are zero.