The Continuous-Time Fourier Transform (CTFT) of the function $\text{trig}(t)$ depends on the specific trigonometric function you're referring to. I'll compute the CTFT of some common trigonometric functions:

1. CTFT of $\cos(\omega_0 t)$:

$\cos(\omega_0 t) = \frac{e^{j\omega_0 t} + e^{-j\omega_0 t}}{2}$

Using the CTFT definition:

$\mathcal{F}\{\cos(\omega_0 t)\} = \int_{-\infty}^{\infty} \cos(\omega_0 t) e^{-j\omega t} dt$

This results in two Dirac delta functions:

$\mathcal{F}\{\cos(\omega_0 t)\} = \pi \left[ \delta(\omega - \omega_0) + \delta(\omega + \omega_0) \right]$

2. CTFT of $\sin(\omega_0 t)$:

$\sin(\omega_0 t) = \frac{e^{j\omega_0 t} - e^{-j\omega_0 t}}{2j}$

By applying the CTFT, we get:

$\mathcal{F}\{\sin(\omega_0 t)\} = j\pi \left[ \delta(\omega + \omega_0) - \delta(\omega - \omega_0) \right]$

Summary:

• CTFT of $\cos(\omega_0 t)$: $\pi [\delta(\omega - \omega_0) + \delta(\omega + \omega_0)]$
• CTFT of $\sin(\omega_0 t)$: $j\pi [\delta(\omega + \omega_0) - \delta(\omega - \omega_0)]$

Would you like more details on the Fourier Transform process