How about solution?

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1 Answer
Aug 13, 2018

cos(ω0t)π(δ(ωω0)+δ(ω+ω0))

Explanation:

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

<δ,f>=f(0)

so:

+δ(t)eiωtdt=1

and if we translate the distribution:

+δ(tt0)eiωtdt=eiωt0

So the Fourier transform of δ(tt0) is eiωt0.

By the symmetry property of the transform if we let: t0=ω0 then the Fourier transform of eiω0t is 2πδ(ωω0)

But using the expression of the cosine as:

cos(ω0t)=eiω0t+eiω0t2

and the linearity of the transform then:

cos(ω0t)π(δ(ωω0)+δ(ω+ω0))