0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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, $F(x)=0$. $\endgroup$ – Joseph Aleshaiker Feb 28 '18 at 1:49
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.