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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.

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Notice that $\sin(k\pi+\frac{\pi}{2})=\cos(k\pi)$. Then, you're simply finding the Fourier cosine series of the function $f(x)=1$ (which should be fairly straightforward using the definition). – icurays1 Nov 13 '12 at 5:27
up vote 0 down vote accepted

Hint: The functions $ {\cos(k\pi x)}_{k=0}^{\infty} $ are orthogonal on the interval $[-1,1]$, that means

$$ \int_{-1}^{1}\cos(k\pi x) \cos(m \pi x) dx = \begin{cases} 0 &: k\ne m \\ 2 &: k=m=0\\ 1 &: k=m \ne 0 \end{cases}\, .$$

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