Let $f$ be a $2 \pi$ Riemann integrable function defined on $\mathbb{R}$. If the Fourier series of the function $f$ can be written as $$f(\theta) \thicksim \hat{f}(0) + \sum_{n\ge1}[\hat{f}(n) + \hat{f}(-n)]\ \cos n\theta + i[\hat{f}(n) - \hat{f}(-n)]\ \sin n\theta. $$
If $f$ is even, then $\hat{f}(n) = \hat{f}(-n)$, and we get a cosine series. If $f$ is odd, then $\hat{f}(n)= -\hat{f}(-n)$, and we get a sine series.
Suppose that $f(\theta + \pi)$ = $f(\theta)$ for all $\theta \in \mathbb{R}$. Show that $\hat{f}(n) = 0$ for all odd $n$.
Could anyone help me please?