The trajectory of the ISS only appears to be sinusoidal when it is mapped onto the flat plane. In reality the orbit around Earth looks like that:

The answer to why the ISS' orbit appears to be looking like a sine wave is harmonic motion and circular motion. The ISS is moving along the circumference of a circle (the orbit) around the earth:
$\omega$ is the angular speed, $\Phi$ is the initial angle (phase), $r$ is the radius of the orbit and $P$ is the position of the ISS. The angle at time t is given by $ \omega t + \Phi$.
If you observe the position $P$ at time $t$ you can express the $x$- and $y$-coordinate as
$\ P_{x}(t)=rcos(\omega t+\Phi)$ and similarly $\ P_{y}(t)=rsin(\omega t+\Phi)$
In your case the ISS moves in circular motion around Earth and it's position is simply $P$ at time $t$. If you observe this point, standing at the origion $O$ of the above circle and project that onto the $OY$ axis you get
$\ y(t)=rsin(\omega t+\Phi)$.
Then plotting the function you get something like this:

Note that the border between day and night also appears to be curved due to the same reason. The line isn't closed because while the ISS revolves around Earth once, Earth itself has rotated by a bit.
When mapping the surface of a sphere onto the plane you can only keep one line straight. The equator of the earth appears as a straight line in both pictures.
Also keep in mind that this is a simplification of different projection methods used and only explains why the orbit seems to look like a sine-wave. In reality more complex projection methods are used.
This answer is a rewrite using my original answer and the technical details from where_is_tftp's answer.