Take the 2-minute tour ×  How can I find the Fourier transform
of  $$ f(x) = ( \cos(x) )^3$$  I know that for $ g(x) = \cos(x)
$  $$ F \Big\{ g(x) \Big\} = F \Big\{ \cos(x) \Big\} = \pi \Big
[ \delta(w-\pi / 2) + \delta(w+\pi / 2) \Big ]$$  But using this
pair of Fourier transform how to obtain the $ F \Big\{ f(x)
\Big\} $ ?? Is there a direct/simple way to do that?
share|improve this question  2 Answers 2  up vote 10 down vote
accepted  One way would be to use the power-reduction
trigonometric identity:  $$ \cos^3(x) = \frac{3 \cos(x) +
\cos(3x)}{4} $$  Due to the linearity property of the Fourier
transform, you can transform each term separately and take their
weighted sum to get the transform of the entire expression. The
relationship we will use (from line 304 here) is:  $$
\mathcal{F}\{\cos(ax)\} = \pi\left(\delta(\omega - a) +
\delta(\omega + a)\right) $$  Which assumes that you're using
the non-unitary, angular frequency definition of the Fourier
transform:  $$ \mathcal{F}\{x(t)\} = X(\omega) = \int_{