From $$1=\sum_{k\geq 1} a_k \sin((k\pi+\frac{\pi}{2})x),$$ I want to find $a_k.$
My unsuccessful approach is first multiplying both side by $\cos((k\pi+\frac{\pi}{2})x)$. That is, $$\cos((k\pi+\frac{\pi}{2})x)=\frac{1}{2}\sum_{k\geq 1}(\sin \pi x)a_k.$$
Can anyone give me a trick to find $a_k$? Thanks in advance.
Hint: The functions $ {\cos(k\pi x)}_{k=0}^{\infty} $ are orthogonal on the interval $[-1,1]$, that means
$$ \int_{-1}^{1}\cos(k\pi x) \cos(m \pi x) dx = \begin{cases} 0 &: k\ne m \\ 2 &: k=m=0\\ 1 &: k=m \ne 0 \end{cases}\, .$$