Problem: (From Oppenheim/Wisllsky, 4.22 b.) Find $ F^{-1} (cos(4w+\frac{\pi}{3})). $ ($ F^{-1} $ meaning the inverse Fourier Transform of said function.)
Solution: First, observe that $ cos(4w+\frac{\pi}{3}) = \frac{1}{2} (e^{j4w} e^{j\frac{\pi}{3}} + e^{-j4w} e^{-j\frac{\pi}{3}}) $. First, let's try to apply the formula for the inverse Fourier transform directly:
$ F^{-1} (cos(4w+\frac{\pi}{3})) = \int_{-\infty}^\infty cos(4w+\frac{\pi}{3}) e^{jwt} dt $
$ = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{1}{2} (e^{j4w} e^{j\frac{\pi}{3}} + e^{-j4w} e^{-j\frac{\pi}{3}}) e^{jwt} dt $
$ = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{1}{2} (e^{j4w} e^{j\frac{\pi}{3}} + e^{-j4w} e^{-j\frac{\pi}{3}}) e^{jwt} dt $
$ = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{1}{2} e^{j(t+4)w} e^{j\frac{\pi}{3}} + e^{j(t-4)w} e^{-j\frac{\pi}{3}} dt $
Noting that integrating this is a very difficult undertaking, let's try another approach. Say, for instance, that we were to find x(t) such that $ F(x(t)) = e^{-jwt_0} $. We'll take the highly-educated guess that $ x(t) = \delta(t-t_0) $:
$ F(\delta(t-t_0)) = \int_{-\infty}^\infty \delta(t-t_0) e^{-jwt} dt = e^{-jwt_0} $
So, we can conclude that since $ F(\delta(t-t_0)) = e^{-jwt_0} $, $ F^{-1}(e^{-jwt_0}) = \delta(t-t_0) $. Applying this to our problem at hand:
$ F^{-1} (cos(4w+\frac{\pi}{3})) = F^{-1} ( \frac{1}{2} (e^{j4w} e^{j\frac{\pi}{3}} + e^{-j4w} e^{-j\frac{\pi}{3}})) $
$ = \frac{1}{2} (F^{-1} (e^{j4w}) e^{j\frac{\pi}{3}} + F^{-1} (e^{-j4w}) e^{-j\frac{\pi}{3}}) $
$ = \frac{1}{2} ( e^{j\frac{\pi}{3}} \delta(t + 4) + e^{-j\frac{\pi}{3}} \delta(t-4) ) $
