Usually in radio systems in order to move the frequency response of the message signal to a new frequency band centered at f, the message signal is mixed with $\cos(2\pi ft)$.
What would change in frequency if we multiply the message signal to $\sin(2\pi ft)$ instead of $\cos(2\pi ft)$?
It does the same thing (i.e. shift the frequency to another band), though mathematically it differs a bit. Fourier transforms of the two choices are as shown below:
$$ \begin{array}{|c|c|}\hline f(x)\cos(ax) & \displaystyle\frac{\hat{f}\left(\xi-\frac{a}{2\pi}\right)+\hat{f}\left(\xi+\frac{a}{2\pi}\right)}{2}\\ \hline f(x)\sin(ax) & \displaystyle\frac{\hat{f}\left(\xi-\frac{a}{2\pi}\right)-\hat{f}\left(\xi+\frac{a}{2\pi}\right)}{2i}\\\hline \end{array} $$
The negative frequencies not only get shifted, but also flipped in polarity
Now for the derivation of these formulas: Using the Great Mr. Euler's identity: $e^{iax} = \cos(ax) + i \sin(ax)$
$$\cos(ax) = \frac{e^{iax} + e^{-iax}}{2}$$ and $$\sin(ax) = \frac{e^{iax} - e^{-iax}}{2i}$$
\begin{align} \textrm{F.T.}\left\{f(x) \cos(ax)\right\} &= \int\limits_{-\infty}^{\infty} f(x) \cos(ax) e^{-i2\pi\xi x} dx\\ &= \int\limits_{-\infty}^{\infty} f(x) \frac{e^{iax} + e^{-iax}}{2} e^{-i2\pi\xi x} dx\\ &= \frac{\int\limits_{-\infty}^{\infty} f(x) e^{-i2\pi(\xi + \frac{a}{2\pi} )x} dx + \int\limits_{-\infty}^{\infty} f(x) e^{-i2\pi(\xi - \frac{a}{2\pi})x} dx}{2}\\ &= \frac{\hat{f}(\xi + \frac{a}{2\pi} ) + \hat{f}(\xi - \frac{a}{2\pi} )}{2} \end{align}
You can derive the formula for $\sin$.
Short answer: a phase shift of 90°, since $\cos(t) = \sin(t+\frac\pi2)$.
