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

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.

• Okay so I now used that new trig identity and I obtained 2 answers. When $n=4, F(x)=\frac{1}{2}sin(4x)$ and when $n=2, F(x)=frac{-1}{2}sin(2x)$. Else, $F(x)=0$. – Joseph Aleshaiker Feb 28 '18 at 1:49

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