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

How does the cosine disappear and how did sin x turn into $\sin (x+{\pi\over4})$?
 A: Hint:
$$\cos(x)=\sin\left(\frac{\pi}{2}-x\right)$$
A: HINT: use that $\sin(x+\pi/4)=1/2\,\sin \left( x \right) \sqrt {2}+1/2\,\cos \left( x \right) \sqrt{2} $
and 
$$\sin \left( x \right) +\cos \left( x \right) -\sqrt {2}\sin \left( x+
\pi /4 \right) 
=0$$
A: The sine and cosine are two facets of the same function, and morph into each other when you apply a "phase shift": by the addition formula
$$\sin(x+\phi)=\sin(x)\cos(\phi)+\cos(x)\sin(\phi),$$
a shifted sine is a linear combination of a sine and a cosine. For specific values of the shift, one of the terms vanishes. For example,
$$\sin(x+\frac\pi2)=\cos(x).$$
For $\pi/4$, the terms get the same amplitude,
$$\sin(x+\frac\pi4)=\frac1{\sqrt2}(\sin(x)+\cos(x)).$$
More generally, a sum of sinusoids and cosinusoids with arbitrary amplitudes can always be reduced to a single sinusoid/cosinusoid with a certain amplitude and phase, and conversely.

A: The equation you've written is not true for all $x$. For instance, when $x=0$, the left-hand side is $-1$, while the right-hand side is $-\sqrt{2}/2$.
A: $$\sin x\cos a + \cos x \ sin a = \sin(x+a)$$
$$-\sin x -\cos x = - (\sin x +\cos x ) = -(\sin x\cos a /cos a + \cos x \ sin a / \ sin a)$$
choosing $\cos a = \ sin a$ , which means $a = \pi/4$, we have
$$-(\sin x\cos a /cos a + \cos x \ sin a / \ sin a) = -\sin(x+\pi/4)\times 2/\sqrt{2}  $$
A: 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]
$\cos(\arctan a/b)=\frac{b}{\sqrt{a^2+b^2}} \space (2);\sin(\arctan a/b)=\frac{a}{\sqrt{a^2+b^2}}\space (3)$
