<s>[INST] $12\frac{\sin 45^\circ}{\sin 60^\circ}$ 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 $4\sqrt{6}$.
 [/INST] Here's yet another approach: $\frac1{\sqrt{2}}$ is often (equivalently) written as $\frac{\sqrt{2}}2$ (an equivalency that can be easily seen by multiplying both numerator and denominator by $\sqrt{2}$).  Using this form turns the original formula into $12\cdot\dfrac{\frac{\sqrt{2}}2}{\frac{\sqrt{3}}2}$ $= 12\cdot\dfrac{\sqrt{2}}{\sqrt{3}}$ (by multiplying numerator and denominator by an easy factor of two).  Now, again 'clear the radical' in the denominator by multiplying by $1=\dfrac{\sqrt{3}}{\sqrt{3}}$: $12\cdot\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}$ $= 12\cdot\dfrac{\sqrt{6}}{3}$ $=4\sqrt{6}$.
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