How can $a_n$ for the trigonometric Fourier series be $0$ for a non-odd square signal? Can $a_n$ for the trigonometric Fourier series be $0$ for a non-odd wave? Since at the moment the square wave is not even nor odd. Square wave is given by:
$$
s_2(t)=\begin{cases}
 -4, &amp; 0\leq t\leq 2.5\\
6,&amp; 2.5\leq t&lt;5
\end{cases}
$$

The signal looks like this:



When I find $a_n$ it gives me $0$ but when you get a value of zero, doesn't this mean that the function is odd? 
OR DOES THE ZERO ALSO IMPLY THAT IT IS NOT AN EVEN and clearly the wave isnt even?


EDIT:

If this wave is odd, how is it odd because I thought it had to be $s(t)=-s(-t)$ so for e.g. $6$ doesn't = $-4$?
Indeed, if the cosine coefficients ($a_n$) of the Fourier series of a function $s(t)$ are all zero, then the function is the sum of sine terms, which are all odd, implying that the function $s(t)$ is odd.

In your case, however, not all $a_n$ are zero; in particular, $a_0=2$. This means that the wave has a "DC" or mean value of $a_0/2=1$. Since $a_n=0$ for $n&gt;0$, then the function $$u(t)=s(t)-a_0/2$$ is indeed odd, and now $a_n=0$ for all $n$.