Frequency shifting property
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:

$$
\begin{array}{|c|c|}\hline
f(x)\cos(ax) &amp; \displaystyle\frac{\hat{f}\left(\xi-\frac{a}{2\pi}\right)+\hat{f}\left(\xi+\frac{a}{2\pi}\right)}{2}\\
\hline
f(x)\sin(ax) &amp; \displaystyle\frac{\hat{f}\left(\xi-\frac{a}{2\pi}\right)-\hat{f}\left(\xi+\frac{a}{2\pi}\right)}{2i}\\\hline
\end{array}
$$

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\} &amp;= \int\limits_{-\infty}^{\infty} f(x) \cos(ax) e^{-i2\pi\xi x} dx\\
&amp;= \int\limits_{-\infty}^{\infty} f(x) \frac{e^{iax} + e^{-iax}}{2} e^{-i2\pi\xi x} dx\\
&amp;= \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}\\
&amp;= \frac{\hat{f}(\xi + \frac{a}{2\pi} ) + \hat{f}(\xi - \frac{a}{2\pi} )}{2}
\end{align}

You can derive the formula for $\sin$.