Subtracting square roots I'm studying trigonometry but I don't understand one example I have here:
$\cos(\frac{5}{12}\pi) = \cos(\frac{1}{6}\pi + \frac{1}{4}\pi)$
$= \cos(\frac{1}{6}\pi)\cos(\frac{1}{4}\pi) - \sin(\frac{1}{6}\pi)\sin(\frac{1}{4}\pi)$
$=(\frac{1}{2}\sqrt{3})(\frac{1}{2}\sqrt{2})-(\frac{1}{2})(\frac{1}{2}\sqrt{2})$
$=\frac{1}{4}\sqrt{2}(\sqrt{3}-1)$
I got the result $\frac{1}{4}\sqrt{6}-\frac{1}{4}\sqrt{2}$, which is the same as the result above. However, I don't know how this calculation became $=\frac{1}{4}\sqrt{2}(\sqrt{3}-1)$.
Please, anyone can explain to me?

As fast as light:
$$\frac{1}{4}\sqrt{6} - \frac{1}{4}\sqrt{2} = \frac{1}{4}(\sqrt{6} - \sqrt{2})$$
Now you will use the property $\sqrt{6} = \sqrt{3\cdot 2} = \sqrt{3}\cdot \sqrt{2}$ to get the result, i.e.
$$\frac{1}{4}\sqrt{2}(\sqrt{3} - 1)$$
