If $\sin(\pi x)=a_0+\sum\limits_{n=1}^{\infty}a_n \cos(n\pi x)$ for $0
If $\sin(\pi x)=a_0+\sum\limits_{n=1}^{\infty}a_n \cos(n\pi x)$ for $0<x<1$ then what is the value of $(a_0+a_1)\pi$?
Honestly I have no idea about how to approach this problem. I tried the expansion of $\sin x$ on the left hand side and then expand $\cos x$ on the right but end up with a mess. Then again $\cos x=(\sin x)'$, but by converting each term on the right hand side into derivative of sine does not give any sensible identity to draw out something. Really out of ideas on this one. Is there any way to do this? Please give me some hints. Thanks for your time.
 A: Apply the integration $\int_0^1 \cos(\pi x) f(x) dx$, then 
$$\int_0^1 \cos(\pi x) \sin(\pi x) dx=a_0 \int_0^1 \cos(\pi x) dx +\sum_{n=1}^{\infty}a_n \int_0^1 \cos(\pi x) \cos(n\pi x) dx$$
Since $\int_0^1 \cos(\pi x) \cos(n\pi x) dx = 0$ for $n \ne 1$, you have
$$\int_0^1 \cos(\pi x) \sin(\pi x) dx=a_0 \int_0^1 \cos(\pi x) dx +a_1 \int_0^1 \cos^2 (\pi x)\  dx$$ which is $0 = 0 + 0.5 a_1$, hence $a1 = 0$.
Likewise just integrate:
$$\int_0^1 \sin(\pi x) dx=a_0  +\sum_{n=1}^{\infty}a_n \int_0^1 \cos(n\pi x) dx$$ which gives $2/\pi = a_0+ 0$.
So the result is $(a_0+a_1)\pi  = 2$.
A: The region matters. The region is everything. It may seem odd to approximate the odd function $\sin \pi x$ with sums of even functions $\cos n\pi x$ - but that's not what we're really doing.
Continue that $\sin$ to be periodic of period $1$, and we get $\sin(\pi(x+1))=-\sin x$ on $[-1,0]$; the function we're actually trying to estimate is $|\sin x|$. On that note, extend the interval to $[-1,1]$. We are now looking at a conventional Fourier series for the even function $|\sin \pi x|$ on a full period $[-1,1]$.
By standard Fourier series results, the coefficients $a_n$ in $f(x)=a_0+\sum_{n=1}^{\infty}a_n\cos(n\pi x)$ ($f$ assumed even on $[-1,1]$) are given by
\begin{align*}a_0 &= \frac12\int_{-1}^1 f(x)\,dx\\
a_n &= \int_{-1}^1 f(x)\cos(n\pi x)\,dx\end{align*}
since $\int_{-1}^1 1\,dx=2$ and $\int_{-1}^1 \cos^2(n\pi x)\,dx=1$.
Calculating the terms we're interested in for $f(x)=|\sin\pi x|$,
$$a_0=\frac12\int_{-1}^1 |\sin(\pi x)|\,dx = \int_0^1 \sin(\pi x)\,dx = \frac2{\pi}$$
$$a_1 = \int_{-1}^1 \cos(\pi x)|\sin(\pi x)|\,dx = 2\int_0^1 \cos(\pi x)\sin(\pi x)\,dx = \int_0^1 \sin(2\pi x)\,dx =0$$
Then $\pi(a_0+a_1)=\pi\left(\frac 2{\pi}+0\right)=2$.
We could calculate more terms if we felt like it. For all odd $n$, we have $\cos(\pi(1-x))=-\cos(\pi x)$, leading to $a_n=0$. For even $n$, we have the product-to sum identity $\cos(n\pi x)\sin(\pi x) = \frac12(\sin((n+1)\pi x) - \sin((n-1)\pi x))$ so
\begin{align*} a_n &= 2\int_0^1 \cos(n\pi x)\sin(\pi x)\,dx = \int_0^1 \sin((n+1)\pi x) - \sin((n-1)\pi x)\,dx\\
&= \frac{2}{(n+1)\pi}-\frac{2}{(n-1)\pi} = \frac{-4}{(n^2-1)\pi}\end{align*}
With some experience in Fourier series, we can recognize that $O(n^{-2})$ decay rate as characteristic of a function with a jump discontinuity in the first derivative.
