Discussion
If \(\tilde{F}(s)\) is the Fourier transform of \(f(t)\), then the Fourier transform of :
\[ f(t)\cos at \]
is :
Answer: C
The transform is \(\tfrac{1}{2}[\tilde{F}(s + a) + \tilde{F}(s - a)]\).
Write \(\cos at = \tfrac{1}{2}(e^{iat} + e^{-iat})\). Multiplying \(f(t)\) by \(e^{\pm iat}\) shifts its Fourier transform by \(a\) in one direction or the other (the modulation theorem). The sum of the two shifts, halved, is the transform of \(f(t)\cos at\). With either sign convention for the transform, the two shifts are \(s + a\) and \(s - a\), so the formula is the same.
Options A and B are the transforms of \(f(t)e^{\pm iat}\), one exponential only. The difference in option D belongs to \(f(t)\sin at\), up to a factor \(\pm i\).
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