odd half range extension of
$$f(x) = \sin x \cos(2x)\text{ with limits $0$ to $\pi$}$$
2026-04-26 01:13:13.1777165993
What is the odd Fourier extension of $\sin x \cos(2x)$?
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The question is perfectly readable as is, but see MathJax basic tutorial and quick reference for the future.
If I understand "odd half range extension" correctly, you are supposed to extend $f$ to an odd function on $[-\pi,\pi]$, and then find the Fourier series of that.
Since $\sin (-x)\cos (-2x)=-\sin x \cos 2x$, the function $f$ is naturally an odd function: we use the same formula $\sin x \cos 2x$ on the whole interval $[-\pi,\pi]$.
Now you have two ways to find the Fourier series
Correct but tedious: calculate the coefficients by integrating $\int_{-\pi}^\pi f(x)\sin nx$
Clever: use a product-to-sum identity $$\sin \theta\cos \phi = \frac12 (\sin (\theta+\phi)+\sin(\theta-\phi))$$ where you will put $\theta=x$ and $\phi=2x$. After some cosmetic simplifications, such as $\sin(-x)=-\sin x$, you have your Fourier series.