Expressing in the form $A \sin(x + c)$ 

Express in the form $A\sin(x+c)$
  a) $\sin x+\sqrt3\cos x$; b) $\sin x-\cos x$
sol: a) $A=\sqrt{1+3}=2$, $\tan c=\frac{\sqrt 3}1$, $c=\frac\pi3$. So $\sin x+\sqrt3\cos x=2\sin(x+\frac\pi3)$
  b) $\sqrt 2\sin(x-\frac\pi4)$

Can someone please explain the method used in the provided solution above? (I'm not familiar with this way of solving whatsoever.)
Thanks in advance =]

Using the appropriate formula for $\sin$ you have $A \sin(x+c) = A \sin x \cos c + A \cos x \sin c$. You need to determine $A,c$ so the formula holds true for a), b).
Equating $A \sin x \cos c + A \cos x \sin c = \sin x + \sqrt{3} \cos x$ gives $A \cos c = 1$, $A \sin c = \sqrt{3}$. This gives $\tan c = \frac{A \sin c}{A \cos c} = \sqrt{3}$. If $\tan c = \sqrt{3}$, then $\sin c = \frac{\sqrt{3}}{2}$ and $\cos c = \frac{1}{2}$. This gives $A \frac{1}{2} = 1$, so $A = 2$. You can check that $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}, \cos \frac{\pi}{3} = \frac{1}{2}$, from which it follows that $\sin x + \sqrt{3} \cos x = 2 \sin ( x + \frac{\pi}{3})$.
Similarly, $A \sin x \cos c + A \cos x \sin c = \sin x - \cos x$ gives $A \cos c = 1$, $A \sin c = -1$. This gives $\tan c = \frac{A \sin c}{A \cos c} = -1$, which in turn gives $\sin c = -\frac{1}{\sqrt{2}}$, $\cos c = \frac{1}{\sqrt{2}}$. Then $A \cos c = A \frac{1}{\sqrt{2}} = 1$ gives $A = \sqrt{2}$. You can check that $\sin (-\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}, \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, from which it follows that $\sin x - \cos x = \sqrt{2} \sin ( x - \frac{\pi}{4})$.
