Find Fourier Series of the function $f(x)= \sin x \cos(2x) $

Question

Find Fourier Series of the function $f(x)= \sin x \cos(2x) $ in the range $ -\pi \leq x \leq \pi $

any help much appreciated

I need find out

$a_0$ and $a_1$ and $b_1$

I can find $a_0$ which is simply integrating something with respect to the limits I can get as far as

$$\frac{1}{2} \int_{-\pi}^\pi \ \frac12 (\sin (3x)-\sin(x)) dx$$

How would I integrate the above expression ?

secondly how would I calculate $a_1$ and $b_1$ but despite knowing the general formula to find the fourier series Im having trouble applying them to this question

Answer 1

If you want to use integration to find $a_0$, $a_1$, and $b_1$, you can use

$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}\sin x\cos{2x}dx=\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{2}(\sin{3x}-\sin x) dx=0$ since the integrand is an odd function.

$a_1=\frac{1}{\pi}\int_{-\pi}^{\pi}\sin x\cos{2x}\cos x dx=\frac{1}{\pi}\int_{-\pi}^{\pi}(\sin x\cos x)\cos{2x}dx=\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{2}\sin{2x}\cos{2x}dx=$ $\:\:\:\:\:\:\:\:\:\:\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{4}\sin{4x}dx=0$ for the same reason.

$b_1=\frac{1}{\pi}\int_{-\pi}^{\pi}\sin x\cos{2x}\sin xdx=\frac{1}{\pi}\int_{-\pi}^{\pi}\sin^{2} x\cos{2x}dx=\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{2}(1-\cos{2x})\cos{2x}dx$

Answer 2

Hint: $$\sin x \cos 2x = \frac{1}{2} \left( \sin 3x - \sin x \right).$$

Answer 3

Instead of integrating, it is easier to just use the trigonometric product formula $$ \sin(a) \cos(b) = \frac{\sin(a+b)+\sin(a-b)}{2} $$ to rewrite the function into a form that it is own Fourier expansion.