Q:

Fourier Cosine expansion of $\sin x$

Instructor mentioned something in class that got me thinking...
Why is it possible for $\sin x$ to have a Fourier Cosine expansion in $[0, \pi]$? It seems strange to me that you're able to express $\sin x$ as an infinite sum of cosines. Maybe I'm just missing the (perhaps very obvious) point here.

A:

A picture is worth a thousand words :
Red curve : $y_1=\sin(x)$
Blue curve : $\quad y_2=\frac{2}{\pi}-\frac{2}{\pi}\sum_{n=2}^\infty\frac{\cos{n\pi}+1}{n^2-1}\cos(nx) \quad$ : Fourier cosine expansion of $y_1$ on $[0,\pi]$. 
$y_1=y_2$ on the range of definition $[0,\pi]$. 
Outside this range, $y_2$ is the periodic repetition of the pattern defined on $[0,\pi]$.

Since $\cos(n\pi)+1=(-1)^n+1$ is null for odd $n$ ,$\quad y_2=\frac{2}{\pi}-\frac{4}{\pi}\sum_{k=1}^\infty\frac{1}{4\:k^2-1}\cos(2kx) \quad$ 
OTHER EXAMPLE : 
Fourier cosine series of $\sin(x)$ with range of definition $[0,\frac{2\pi}{3}]$ :

