<s>[INST] Expand a function in a Fourier Series Expand the function
$$ f(x) = \sin^6{x} \cdot \cos^3{x} $$
in a Fourier series on the segment $[-\pi, \pi]$.
I think we should start from the fact that this function is even, but then the $ a_n $ coefficient turns out to be very big
 [/INST] If you know complex numbers:
Let
$$z:=\cos(x)+i\sin(x)$$
Then
$$\frac{1}{x}=\cos(x)-i\sin(x) \\
\cos(x)=\frac{1}{2}(z+\frac{1}{z})\\
\sin(x)=\frac{1}{2i}(z-\frac{1}{z})
$$
$$f(x) = \sin^6{x} \cdot \cos^3{x} = \frac{-1}{2^9}(z+\frac{1}{z})^3(z-\frac{1}{z})^6\\
= \frac{-1}{2^9}(z^3+3z+3\frac{1}{z}+\frac{1}{z^3})(z^6-6z^4+15z^2 -20+15\frac{1}{z^2} -6 \frac{1}{z^4}+\frac{1}{z^6})\\
$$
Now open the brackets and use 
$$z^n+\frac{1}{z^n}=2 \cos(nx) \\
z^n-\frac{1}{z^n}=2i \sin(nx)$$
Note A faster computation is to observe that 
$$\frac{-1}{2^9}(z+\frac{1}{z})^3(z-\frac{1}{z})^6=\frac{-1}{2^9}(z+\frac{1}{z})^3(z-\frac{1}{z})^3(z-\frac{1}{z})^3\\
=\frac{-1}{2^9}(z^2-\frac{1}{z^2})^3(z-\frac{1}{z})^3\\
=\frac{-1}{2^9}(z^6-3z^2+3\frac{1}{z^2}-\frac{1}{z^6})(z^3-3z+3\frac{1}{z}-\frac{1}{z^3})\\
$$
</s>