$\sin (t \omega)$ is not an Energy Signal, then how come its Fourier transform do exist?
You are right that such integrals are meaningless unless they are interpreted as distributions. And this is what we need to do, because - as you know - the Fourier transform of a sine function involves delta impulses. Let me try to make this a bit more intuitive:

The inverse Fourier transform of the delta function $\delta(\omega)$ (in the frequency domain) is given by

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega)e^{i\omega t}d\omega= \frac{1}{2\pi}e^{i0\cdot t}=\frac{1}{2\pi}$$

So we have the Fourier transform relation (time domain $\Longleftrightarrow$ frequency domain)

$$1\Longleftrightarrow 2\pi\delta(\omega)$$

Using the shifting property we obtain

$$e^{i\omega_0 t}\Longleftrightarrow 2\pi\delta(\omega-\omega_0)$$

And since

$$\sin(\omega_0t)=\frac{1}{2i}[e^{i\omega_0 t}-e^{-i\omega_0 t}]$$

we get for its Fourier transform

$$\sin(\omega_0t)\Longleftrightarrow \frac{\pi}{i}[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)]$$