Find Fourier series for $f(x) = \sin(x)\cos(3x)$

Question

Earlier I found the Fourier series of $\sin(x)\cos(x)$ by using the trig identity $2\sin(x)\cos(x)=\sin(2x)$ Since $a_0=a_n=0$ and $b_n=1$ when $n=2$, then I found the Fourier series to be: $\sum_{n=1}^\infty \frac{1}{2}\sin(nx)$ where $n=2$ thus $F(x)=\frac{1}{2}\sin(2x)$ which was the original function, which makes sense. However, I can't figure out a way to use a trig identity to make $f(x)=\sin(x)\cos(3x)$ without having to manually solve for $b_n=\frac{1}{\pi} \int_{-\pi}^{\pi} \sin(x)\cos(3x)\sin(nx)\;\mathrm dx$. I understand that $a_0=a_n=0$, but I don't know what to without solving the previously mentioned integral.

Answer 1

Hint $$\sin\alpha\cos\beta=\frac12(\sin(\alpha+\beta)+\sin(\alpha-\beta))$$