$12\frac{\sin 45^\circ}{\sin 60^\circ}$ Need help breaking this down.

Question

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 $4\sqrt{6}$.

Answer 1

Keep at it step by step:

$\frac{12}{\sqrt2}\frac{2}{\sqrt3}$ … dividing by something is the same as multiplying by its inverse.

$\frac{24}{\sqrt6}=\frac{24\sqrt6}{6}$ … multiply top and bottom by $\sqrt6$.

And I'm pretty sure that you can work out $24/6$ by yourself.

Answer 2

$$ 12\cdot \frac{1}{\sqrt{2}}\cdot\frac{2}{\sqrt{3}} = 12\cdot \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}\sqrt{2}}{\sqrt{3}} = 12\cdot\frac{\sqrt{2}}{\sqrt{3}} = 12\cdot\frac{\sqrt{2}\sqrt{3}}{\sqrt{3}\sqrt{3}}= 12\cdot\frac{\sqrt{6}}{3} =\cdots\cdots $$

Answer 3

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?

Answer 4

$$12\dfrac{\left(\frac{1}{\sqrt2}\right)}{\left(\frac{\sqrt3}{2}\right)}=12\cdot\frac{1}{\sqrt2}\div\frac{\sqrt3}{2}=12\cdot\frac{2}{\sqrt{6}}=\frac{24}{\sqrt{6}}=\frac{24}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}}=\frac{24\sqrt{6}}{6}=4\sqrt{6}$$

Answer 5

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 $1=\dfrac{\sqrt{3}}{\sqrt{3}}$: $12\cdot\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}$ $= 12\cdot\dfrac{\sqrt{6}}{3}$ $=4\sqrt{6}$.