Phase spectrum of a Fourier transform I am trying to compute the phase spectrum of the signal 
$$
s(t)=\frac{A}{\pi}\left[H\left(t+\frac{\tau}{2}\right)-H\left(t-\frac{\tau}{2}\right)\right]
$$

The Fourier transform is 
$$
S(j\omega)=\frac{A\tau}{\pi}\cdot\frac{\sin\left(\omega\displaystyle\frac{\tau}{2}\right)}{\omega\displaystyle\frac{\tau}{2}}$$

My answer for this would be $\phi_s(\omega)=0$, since the imaginary part of $S(j\omega)$ is zero, but according to my textbook, the answer is 

$$\phi_s(\omega)=\begin{cases}0&amp;S(j\omega)&gt;0\\\pm\pi&amp;S(j\omega)&lt;0\end{cases}$$

How can I get to this answer and interpret it correctly?
A real-valued negative number has a phase of $\pm\pi$, because $e^{\pm j\pi}=-1$. So for all frequencies for which the real-valued spectrum is negative, the corresponding phase angle equals $\pm \pi$:

$$S(j\omega)=\begin{cases}|S(j\omega)|,&amp;&amp;S(j\omega)\ge 0\\
|S(j\omega)|e^{\pm j\pi},&amp;&amp;S(j\omega)&lt;0\end{cases}$$