I found that transform of Sin(t) look like THIS
If its amplitude look like THIS
How does its phase look like?
I have found one question on stack which is similar to my question but does not properly answer my question: fft phase plot of pure sine function, why so messy?
The Fourier transform of $\sin(\omega_0 t)$ is $$ \mathcal{F}\left[\sin(\omega_0 t)\right] = -\frac{i}{2}\delta(\omega -\omega_0) + \frac{i}{2}\delta(\omega + \omega_0) $$ where $\delta$ is the Dirac delta function. This is seen in your first image. The magnitude of this function is $$ |\mathcal{F}\left[\sin(\omega_0 t)\right]| = \frac{1}{2}\delta(\omega -\omega_0) + \frac{1}{2}\delta(\omega + \omega_0) $$ as seen in your second image. So then what about the phase? Well, the delta function is zero everywhere except where its argument is zero. That means that for all $\omega \ne \pm \omega_0$, $\mathcal{F}\left[\sin(\omega_0 t)\right](\omega) = 0$. What's the phase of zero? It's undefined. So your phase plot won't be much of a plot.
The reason the plot in the linked question is so noisy is that numerical error is causing the values of $\mathcal{F}\left[\sin(\omega_0 t)\right]$ to not compute to exactly zero. Instead it computes to an extremely small magnitude complex number whose phase is essentially random.
