Finding the Fourier Series of $\sin(x)^2\cos(x)^3$ I'm currently struggling at calculation the Fourier series of the given function
$$\sin(x)^2 \cos(x)^3$$
Given Euler's identity, I thought that using the exponential approach would be the easiest way to do it.
What I found was: $$\frac{-1}{32}((\exp(2ix)-2\exp(2ix)+\exp(-2ix))(\exp(3ix)+3\exp(ix)+3\exp(-ix)+\exp(-3ix)))$$
Transforming it back, the result is:
$$ -\frac{1}{18}(\cos(5x)+\cos(3x)+2\cos(x))$$
(I've checked my calculations multiple times, I'm pretty sure it's correct.)
Considering the point $x = 0$ however, one can see that the series I found doesn't match the original function.
Could someone help me find my mistake?
 A: 1) Trigonometric identities:
$$
\sin^2 x\cos^3x=(\sin x\cos x)^2\cos x=\left(\frac{\sin 2x}{2}\right)^2\cos x=\frac{1}{4}\sin^22x\cos x
$$
$$
=\frac{1}{4}\left(\frac{1-\cos 4x}{2}\right)\cos x=\frac{\cos x}{8}-\frac{\cos 4x\cos x}{8}
$$
$$
=\frac{\cos x}{8}-\frac{\cos 5x+\cos 3x}{16}
$$
$$
=\frac{\cos x}{8}-\frac{\cos 3x}{16}-\frac{\cos 5x}{16}
$$
2) Complex exponential:
$$
\sin^2x\cos^3x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^2\left(\frac{e^{ix}+e^{-ix}}{2}\right)^3
$$
$$
=-\frac{1}{32}(e^{2ix}-2+e^{-2ix})(e^{3ix}+3e^{ix}+3e^{-ix}+e^{-3ix})
$$
$$
=-\frac{1}{32}(e^{5ix}+e^{3ix}-2e^{ix}-2e^{-ix}+e^{-3ix}+e^{-5ix})
$$
$$
=-\frac{1}{32}(2\cos 5x+2\cos 3x-4\cos x)
$$
$$
=\frac{1}{16}(2\cos x-\cos 3x-\cos 5x)
$$
Note: you made a mistake when you expanded $(e^{ix}-e^{-ix})^2$. I have no idea how you ended up with this $18$. You probably meant $16$.
A: I used trig idendities and came to 
\[ \sin(x)^2 \cdot \cos(x)^3 = \frac{1}{16} \cdot \left( 2\cos(x) - \cos(3x)-\cos(5x)\right)\]
This surely works for $x=0$, and Mathematica gives the same.
A: You don't show enough for me to diagnose anything.  I can tell you by using the double-angle and add-multiply formulae, I get
$$\sin^2{x} \cos^3{x} = \frac{1}{16} (2 \cos{x}-\cos{3 x} - \cos{5 x})$$
The way I did this is to rewrite the LHS as
$$\begin{align} \cos^3{x}-\cos^5{x} &= \cos{x} (\cos^2{x}-\cos^4{x})\\&= \frac12 \cos{x} (1+\cos{2 x}) - \frac14 \cos{x} (1+\cos{2 x})^2 \\ &= \frac12 \cos{x}(1+\cos{2 x}) - \frac14 \cos{x}(1+2 \cos{2 x} + \cos^2{2 x}) \\ &=  \frac14 \cos{x} - \frac18 \cos{x} (1+\cos{4 x})\\ &= \frac18 \cos{x} - \frac{1}{16} \cos{3 x} - \frac{1}{16} \cos{5 x}\end{align}$$
The result follows.
