System of equations involving sin and cos I'm trying to solve the following system:
$$
\sin(x) + \cos(y) = 0.6\\
\cos(x) - \sin(y) = 0.2\\
$$
Solving for y in terms of x:
$$
y=\arccos(0.6-\sin(x))=\arcsin(\cos(x) -0.2)
$$
Therefore:
$$
\frac{\pi}{2} - \arcsin(0.6-\sin(x)) = \arcsin(\cos(x) - 0.2)
$$
Forgot to add:
$$
\frac{-\pi}{2}\le x,y \le \frac{\pi}{2}
$$
Not too sure how to proceed from here, any hints?

here is one way to do this. 
$$\begin{align}\sin x + \cos y &= 0.6\\
  \sin (x) + \sin(\pi/2- y) &= 0.6 \\
 \sin(\pi/4+x/2-y/2)\cos(x/2+y/2-\pi/4) &= 0.3\tag 1\end{align}$$
in the same way
$$\begin{align}\cos x - \sin y &= 0.2\\
  \cos(x) - \cos(\pi/2- y) &= 0.2 \\
 \sin(\pi/4+x/2-y/2)\sin(x/2+y/2-\pi/4) &= -0.1\tag 2\end{align}$$
dividing $(1)$ by $2$, we have  $$\tan(x/2+y/2-\pi/4) = -\frac13\tag 3 $$ which gives you $$x+y=\pi/2 -2\tan^{-1}(1/3)+ 2k\pi \tag 4 $$
using $(3)$ in $(2)$ we find that $$\sin(\pi/4+x/2-y/2)\sin(x/2+y/2-\pi/4) = \sin(\pi/4+x/2-y/2)\left(\mp\frac1{\sqrt{10}}\right)=-0.1 $$
 so that $$x-y = -\pi/2 \pm 2\sin^{-1}(1/\sqrt{10}), -\pi/2 \mp 2\sin^{-1}(1/\sqrt{10}) \tag 5$$
solving $(4)$ and $5$ you get four pairs of solutions. i will do one pair.
$$x = 0, y = \pi/2-2\tan^{-1}(1/3) $$  you can verify that $$\sin x + \cos y = 0.6, \cos x - \sin y = 0.2 $$ 
