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 =]