Quadrature modulation
The idea is that you can transmit two signals, $I(t)$ and $Q(t)$, over the same bandwidth and at the same time, and still recover each independently of the other. The math is pretty simple. If the transmitted signal is $s(t)=I(t)\cos(2\pi f_ct)-Q(t)\sin(2\pi f_ct)$, then (ignorning noise, ignoring a factor of 1/2 and assuming coherent reception) the receiver can calculate 
\begin{align*}
r(t)\cos(2\pi f_ct) &amp;= I(t)\cos^2(2\pi f_ct)-Q(t)\sin(2\pi f_ct)\cos(2\pi f_ct)\\
&amp;=I(t)+I(t)\cos(2\pi 2f_ct)-Q(t)\sin(2\pi 2f_ct).
\end{align*}
After a low-pass filter, the receiver has recovered $I(t)$. Likewise,
\begin{align*}
r(t)\sin(2\pi f_ct) &amp;= I(t)\cos(2\pi f_ct)\sin(2\pi f_ct)-Q(t)\sin^2(2\pi f_ct)\\
&amp;=I(t)\sin(2\pi 2f_ct)-Q(t)+Q(t)\sin(2\pi 2f_ct).
\end{align*}
from which $Q(t)$ can be recovered by an low-pass filtering and multiplication by -1.

The identities used here are: $\cos(A)\cos(B)=0.5(\cos(A+B)+\cos(A-B))$, $\cos(A)\sin(B)=0.5(\sin(A+B)-\sin(A-B))$ and $\sin(A)\sin(B)=0.5(\cos(A-B)-\cos(A+B))$.

Of course, you can also define the transmitted signal as $v(t)=I(t)\cos(2\pi f_ct)+Q(t)\sin(2\pi f_ct)$ and avoid the sign inversion when estimating $Q(t)$. The main reason not to, though, is that the complex envelope of $s(t)$ is $I(t)+jQ(t)$. Since most of the time it is preferrable to work with the complex envelope, $s(t)$ is chosen to produce the most convenient expression for it.