Fourier function expansion for extension over a $2\pi$ period

Question

So I am currently looking at a fourier expansion for

$$f(x)=\left\{\begin{array}{ccl}\sin x &\text{ if }& x\in[0,\pi]\\0 & \text{ if } & x\in[\pi,2\pi]\end{array}\right.$$

I am working out $$a_n(f)=\frac{1}{\pi}\int_{0}^{2\pi}f(x) \cos (nx) dx = \frac{1}{\pi}\int_{0}^{\pi}\sin x \cos nx dx$$

Now i was unsure where to go from here and I looked at the answer where the next step is given as:

$$\frac{1}{2\pi}\int_{0}^{\pi} \sin ([n+1]x) - \sin ([n-1]x) dx$$

I really dont understand how they got to this, i feel like its probably a simple trick im missing but i dont understand the $n+1$ or $n-1$

many thanks

Answer 1

Probably it is easier to write $f(x)$ as $$ f(x)=\frac{1}{2}\left(\sin(x)+\left|\sin(x)\right|\right) $$ then recall that: $$ \left|\sin(x)\right| = \frac{2}{\pi}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos(2nx)}{4n^2-1}.$$

Answer 2

Integrating we find \begin{align} a_n&= \frac{1}{\pi}\int_{0}^{\pi}\sin x \cos (nx) \,\mathrm dx\\ &=\frac{1}{2\pi}\int_{0}^{\pi}\left[ \sin ((n+1)x) - \sin ((n-1)x)\right]\,\mathrm dx\\ &=\frac{1}{2\pi}\left[\frac{\cos((n+1)x)}{n+1}-\frac{\cos((n-1)x)}{n-1} \right]_0^\pi\\ &=\frac{1}{2\pi}\left[\frac{-2 (n \sin x \sin(n x)+\cos x \cos(n x))}{n^2-1}\right]_0^\pi\\ &=\frac{1}{\pi}\left[\frac{n \sin x \sin(n x)+\cos x \cos(n x)}{1-n^2}\right]_0^\pi\\ &=\frac{1}{\pi}\frac{\cos(n \pi)+1}{1-n^2}=\frac{1}{\pi}\frac{(-1)^n+1}{1-n^2} \end{align} In similar way $$ b_n= \frac{1}{\pi}\int_{0}^{\pi}\sin x \sin(nx) \,\mathrm dx=\frac{1}{\pi}\frac{\sin(n \pi)+1}{1-n^2}=\frac{1}{\pi}\frac{1}{1-n^2} $$