Polar form of the Fourier transform of $\sin(t)$
I read two question here. One is about the graphical representation of the Fourier transform of a sinusoidal function, and the other is about the magnitude (and phase) of such a Fourier transform. For ordinary functions, these two would be equivalent, but for Fourier transforms containing Dirac delta impulses they are not. The reason is that mathematically you can't compute the magnitude of a Dirac delta impulse. It is not a function, so there are no function values to take the magnitude of.

A standard graphical representation of the Fourier transform of the function $\sin(\omega_0t)$ would usually just be two arrows at $\omega=\omega_0$ and $\omega=-\omega_0$, one pointing up, the other one pointing down, where the $y$-axis is labeled by $j$ (which makes a separate phase plot unnecessary). An example of such a graph is shown here.

Even though you can't compute the magnitude of a Dirac impulse, you could write down a representation of the Fourier transform of a sinusoid consisting of a sum of Dirac impulses with positive areas and a phase term. For the function $f(t)=\sin(\omega_0t)$ such a representation would be

$$F(\omega)=\pi[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]e^{-j\frac{\pi}{2}\text{sign}(\omega)}\tag{1}$$

(Note that I use the non-unitary definition of the Fourier transform, which is different from the unitary version used by Wolfram Alpha. This is why there's a factor of $\pi$ in $(1)$ instead of $\sqrt{\frac{\pi}{2}}$.)

A similar representation of your example $A\cdot\mathcal{F}\{\cos(\omega_0t)\}$ with complex $A=|A|e^{j\phi_A}$, where $\phi_A$ is the argument of the complex number $A$, is given by

$$A\cdot\mathcal{F}\{\cos(\omega_0t)\}=|A|\pi [\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]e^{j\phi_A}\tag{2}$$