The two solutions that are available as of now involve the complex form of solving for the Fourier Coefficient. I want to use the inner product.
Attempt:
Given that $f(x)$ is even this means $f(-x) = f(x)$.
Therefore: $$<f(x), \sin(\pi nx)>_{L^2} = \int_{-1}^{1}f(-x)sin(\pi n x)dx \Rightarrow \int_{-1}^{1}f(x)sin(\pi n x) $$.
And this is where I'm stuck. I feel that there is most likely a property of integration that I'm not using. But perhaps there is more.
Make the change of variable $y=-x$. Then you get:
$$\begin{aligned} \int_{-1}^1f(x)\sin(n\pi x)\,dx&=\int_{1}^{-1}f(-y)\sin(-n\pi y)(-dy)=\int_{-1}^1f(y)\sin(-n\pi y)\,dy\\ &=-\int_{-1}^1f(x)\sin(n\pi x)\,dx \end{aligned}$$
hence the integral equals zero.