Can all points in the plane be represented like this? Solving a task regaring affine geometry, I've come across a problem:
Is it  true that, for every point $(x,y)\in \mathbb{R}^2$, there  exist $t\in \mathbb{R}, \alpha\in[0,2\pi]$, such that 
$$x = t\cos\alpha + \sin \alpha \\ y = t\sin\alpha + \cos \alpha?$$
I don't have any idea how to prove this or find a counterexample. All hints appreciated! 

Let's play around and see what happens.
I write $a$ for $\alpha$,
$c$ for $\cos \alpha$,
and $s$ for $\sin \alpha$
cause I'm lazy.
We have
$x = t c + s$
and 
$y = t s + c$.
$x^2+y^2 = t^2 c^2+2tcs + s^2 + t^2 s^2 + 2tcs + c^2
= t^2+1+4tsc$.
$xy = t^2 cs + tc^2+t s^2 + sc
= cs(t^2+1)+t$.
We can solve for $cs$ in each of these.
$cs = (x^2+y^2-t^2-1)/(4t)$
and
$cs = (xy-t)/(t^2+1)$
so
$(x^2+y^2-t^2-1)/(4t) = (xy-t)/(t^2+1)$.
Now we can solve for $t$.
$4t(xy-t) = (t^2+1)(x^2+y^2-t^2-1)$
or
$4txy-4t^2 = (t^2+1)(x^2+y^2) - (t^2+1)^2
= t^2(x^2+y^2)+(x^2+y^2)-t^4-2t^2-1
$
so
$t^4-t^2(x^2+y^2+2)+4txy+1=0$.
This is a quartic, unfortunately,
and I don't see an obvious root.
We can also do this:
$xs-yc = (tcs + s^2)-(tsc+c^2)
= s^2-c^2
= 2s^2 - 1
$,
so
$yc = xs-2s^2+1$.
Squaring this,
and using $c^2 = 1-s^2$,
we get another quartic for $s$.
In other words,
we can get equations for
$t$
and $\sin \alpha$,
but they are quartics.
