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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 $\cos(2\pi ft)$?

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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 $\sin$.

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    $\begingroup$ Nice; got my upvote! I think this question would benefit even more if you'd add the relation between $e^{jx}$ and $\cos$ and $\sin$ explicitly, and hence explained how these transform table entries happen. $\endgroup$ – Marcus Müller Sep 1 '16 at 21:06
  • $\begingroup$ Ping: Srini, I think you can get a lot more upvotes if you actually derive these relationships in your question! $\endgroup$ – Marcus Müller Sep 2 '16 at 7:23
  • $\begingroup$ @Marcus, I am a bit latex-lazy, so was postponing the edit till the weekend :) Your ping made me cut short my laziness. $\endgroup$ – Srini Sep 2 '16 at 14:46
  • $\begingroup$ I wish I could upvote twice :) $\endgroup$ – Marcus Müller Sep 2 '16 at 18:41
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Short answer: a phase shift of 90°, since $\cos(t) = \sin(t+\frac\pi2)$.

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