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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.

Rócherz
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Joseph Aleshaiker
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  • 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 01:49

1 Answers1

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

Teh Rod
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