How about solution?

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1 Answer
Aug 13, 2018

cos (omega_0t) harr pi(delta(omega-omega_0) + delta(omega+omega_0))

Explanation:

Let f(t) be a real function of the proper class. By definition the Dirac distribution delta(t) is such that:

< delta, f > = f(0)

so:

int_(-oo)^(+oo) delta(t) e^(-i omega t) dt = 1

and if we translate the distribution:

int_(-oo)^(+oo) delta(t-t_0) e^(-i omega t) dt = e^(-i omega t_0)

So the Fourier transform of delta(t-t_0) is e^(-i omega t_0).

By the symmetry property of the transform if we let: t_0 = -omega_0 then the Fourier transform of e^(-i omega_0 t) is 2pidelta(omega-omega_0)

But using the expression of the cosine as:

cos (omega_0t) = (e^(i omega_0 t) + e^(-i omega_0 t))/2

and the linearity of the transform then:

cos (omega_0t) harr pi(delta(omega-omega_0) + delta(omega+omega_0))