$-\sin x -\cos x = -\sqrt{2}\sin (x+{\pi\over4})$ 
$-\sin x -\cos x = -\sqrt{2}\sin (x+{\pi\over4})$

How does the cosine disappear and how did sin x turn into $\sin (x+{\pi\over4})$?

A very useful formula:
$$a\sin x+b\cos x=\sqrt{a^2+b^2}\cos(x-\arctan (a/b))$$
$$\sqrt{a^2+b^2}\cos(x)\cos(\arctan a/b)+\sin (x)\sin(\arctan a/b)\space\space (1)$$
$$=\frac{b\sqrt{a^2+b^2}\cos(x)}{\sqrt {a^2+b^2}}+ \frac{a\sqrt{a^2+b^2}\sin x}{\sqrt{a^2+b^2}}\space\space (2) \space\space (3)$$
$$=b\cos x+a\sin x$$
$\cos(A-B)=\cos A\cos B+\sin A\sin B $ $(1)$ [A and B are arbitrary]
$\cos(\arctan a/b)=\frac{b}{\sqrt{a^2+b^2}} \space (2);\sin(\arctan a/b)=\frac{a}{\sqrt{a^2+b^2}}\space (3)$
