Let the signal x(t) be equal to:
$ x(t) = cos(2\pi t) \, $
The Fourier Transform of a signal in Continuous Time is defined by:
$ X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \, $
Using this, we obtain:
$ X(\omega) = \int_{-\infty}^{\infty}cos(2\pi t)e^{-j\omega t}dt \, $
Knowing that cos(t) is equal to: $ \frac{e^{jt}+e^{-jt}}{2} $:
$ X(\omega) = \int_{-\infty}^{\infty}\frac{1}{2}(e^{j2\pi t}+e^{-j2\pi t})e^{-j\omega t}dt \, $
$ X(\omega) = \pi \delta(\omega - 2\pi) + \pi \delta(\omega - 2\pi) \, $
Note: This could also be done knowing that the Fourier transform of $ e^{j\omega_0 t} = 2\pi \delta(\omega - \omega_0) \, $.
