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
 A: Repeatedly use the identities $\sin^{2}x=1-\cos^{2}x$, $\cos^{2}x=\frac {1-\cos (2x)} 2$ and $\cos^{3}x=\frac {\cos (3x)+3\cos x} 4$ to write  the function as finite Fourier series. 
A: This is not an answer to your question because I assume you want to calculate your Fourier series manually. Nevertheless it may be useful to know how you can do this using a CAS.
Using Maxima 5.40.0 http://maxima.sourceforge.net I get
(%i1) f1:sin(x)^6*cos(x)^3;
                                   3       6
(%o1)                           cos (x) sin (x)
(%i2) f2:trigreduce(f1);
               (- cos(9 x)) + 3 cos(7 x) - 8 cos(3 x) + 6 cos(x)
(%o2)          -------------------------------------------------
                                      256
(%i3) f3:multthru(f2);
                   cos(9 x)    3 cos(7 x)   cos(3 x)   3 cos(x)
(%o3)           (- --------) + ---------- - -------- + --------
                     256          256          32        128

when I use the basic commands to manipulate expressions.
But simpler si the following
(%i4) poissimp(f1);
                   cos(9 x)    3 cos(7 x)   cos(3 x)   3 cos(x)
(%o4)          (- --------) + ---------- - -------- + --------
                     256          256          32        128

The commands are described in the Maxima Reference Manual.
The following numerical test should increase your confidence in this solution:
(%i5) sum(abs(ev(float(f1)-float(f3),\
    x=-%pi+k/N*2*%pi,numer=true,tr_numer=true)),k,0,N),N=100;
(%o5)                       1.665336423076783E-15

We calculate the absolute value of the difference of $f1$ and $f3$ at 100 equidistant points between $-\pi$ and $\pi$ and sum them up. These sum is smaller than $2\cdot 10^{-15}$. The values of $\sin$ and $\cos$ in these interval vary between $-1$ and $1$.
A: 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 
$$\frac{-1}{2^9}(z+\frac{1}{z})^3(z-\frac{1}{z})^6=\frac{-1}{2^9}(z+\frac{1}{z})^3(z-\frac{1}{z})^3(z-\frac{1}{z})^3\\
=\frac{-1}{2^9}(z^2-\frac{1}{z^2})^3(z-\frac{1}{z})^3\\
=\frac{-1}{2^9}(z^6-3z^2+3\frac{1}{z^2}-\frac{1}{z^6})(z^3-3z+3\frac{1}{z}-\frac{1}{z^3})\\
$$
