Prove that if $f(x)$ is even and $f(\frac{\pi}{2} + x) = - f(\frac{\pi}{2} - x)$ then its Fourier in the interval $(\pi, \pi)$ represents an expansion (Fourier expansion) in the cosine of odd multiple arcs.
My Attempt
If $f(x)$ is even then $f(x) = f(-x)$. At $x=0$ we have $ f(\frac{\pi}{2}) = -f(\frac{\pi}{2})$ then implies that $f(\frac{\pi}{2}) = 0$. Let $x = \pi$ then $f(\frac{3\pi}{2}) = -f(\frac{\pi}{2}) = 0$. Hence the argument can be extended to conclude $f(\frac{\pi}{2} + n\pi) = 0$. If this is an even function, then we must use a fourier cosine series making $b_n = \frac{2}{\pi}\int_{0}^{\pi} f(x)\cos(nx)dx$... I do not really understand what they want me to show in this proof from here.