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 =]

Solution for item b of the question (this is the item on Problem Set 1 on MITOCW 18.01SC:
$$A\sin(x+c) = A\sin x \cos c + A \sin c \cos x$$
We can see that if $A \cos c = 1$ in the first term and $A \sin c = -1$ in the second term then we will end up with $\sin x - \cos x$, so we will have shown that $f(x)$ can be written in the form $A \sin(x+c)$.
We have two equations in two unknowns (A and c) and so we can solve for these variables. Square both sides of each equation:
$$A^2 \cos^2 c = 1$$
$$A^2 \sin^2 c = 1$$
Sum the two equations
$$A^2(\cos^2 x + \sin^2 x) =  2$$
Use $\cos^2 x + \sin^2 x = 1$
$$A^2 = 2 \Rightarrow A = \pm \sqrt{2}$$
Solution 1: $A = \sqrt{2}$, $c = -\frac{\pi}{4}$
$$\cos c = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} \Rightarrow c = \frac{\pi}{4} \text{ or } c = \frac{7 \pi}{4}$$
$$\sin c = -\frac{1}{\sqrt{2}}= -\frac{\sqrt{2}}{2} \Rightarrow c = \frac{5\pi}{4} \text{ or } c = \frac{7 \pi}{4}$$
Therefore the value of $c$ that satisfies both equations is $\frac{7\pi}{4}$, which is the same as $-\frac{\pi}{4}$
Solution 2: $A = -\sqrt{2}$, $c = \frac{3\pi}{4}$
$$\cos c = -\frac{1}{\sqrt{2}}= -\frac{\sqrt{2}}{2} \Rightarrow c = \frac{3\pi}{4} \text{ or } c = \frac{5 \pi}{4}$$
$$\sin c = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} \Rightarrow c = \frac{\pi}{4} \text{ or } c = \frac{3 \pi}{4}$$
Therefore the value of $c$ that satisfies both equations is $\frac{3\pi}{4}$.
