$12\frac{\sin 45^\circ}{\sin 60^\circ}$ Need help breaking this down. Otherwise known as $12\dfrac{\left(\frac{1}{\sqrt2}\right)}{\left(\frac{\sqrt3}{2}\right)}$
How do you simplify this multi level fractional radical expression into $4\sqrt{6}$.

Let's write things down and go through it step by step:
$$12 \frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}} = 12\frac{1}{\sqrt{2}}\frac{2}{\sqrt{3}}.$$
This is because when we divide by a fraction, we are actually just multiplying by the reciprocal. We can now change things up a little. We have
$$12\frac{1}{\sqrt{2}}\frac{2}{\sqrt{3}} = \frac{24}{\sqrt{6}}.$$
This is just by multiplying $12$ and $2$ and also multiplying $\sqrt{2}$ and $\sqrt{3}$ (since they have the same power).
Notice that we want a $\sqrt{6}$ in the numerator, but we have it in the denominator. This suggests that we probably should multiply by a "clever form of 1":
$$\frac{24}{\sqrt{6}} = \frac{24}{\sqrt{6}}\frac{\sqrt{6}}{\sqrt{6}}$$
If we multiply the denominator we get $\sqrt{6}^2 = 6$ since the square of a square root gives us what we started with. Can you take it from here?
