How do I show this fourier series relationship?

Question

I want to show that $$\int^a_0\sin(mx)\sin(nx)\ dx = 0$$

for $m, n$ positive integers and $m\neq n$, and $a = k\pi$, i.e. the integer multiple of $\pi$.

I have tried expressing the equation in a fourier series:

$$f(x) = \frac{a_0}{2} + \sum^\infty_1 a_n \cos(n\pi x)$$ since the equation is an even function. However, I'm unable to show that $f(x)$ is equivalent to zero in the interval mentioned. I can reason that for half wavelengths, the summation should end up in zero because of how the cosine wave is like. But how do I remove $a_0 /2$?

Answer 1

HINT: use that $$\sin(mx)\sin(nx)=\frac{1}{2} (\cos (m x-n x)-\cos (m x+n x))$$

Answer 2

Hint

Use the identity $$\sin(\alpha)\sin(\beta)=\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}$$

Which is derived from subtracting the two identities below:

$$\begin{cases} \cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\\ \cos(\alpha-\beta)=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta) \end{cases}$$