We start with the expression below, and wonder how it could possibly be equivalent to the far more simple expression $xcos(\theta) - ysin(\theta)$.
1) $x^{'} = \sqrt{x^2+y^2} \cdot cos(\theta + acos(x/\sqrt{x^2+y^2}))$
Then we substitute $\sqrt{x^2+y^2}$ with $r$ for convenience.
2) $x^{'} = rcos\Big(\theta + acos\big(\frac{x}{r}\big)\Big)$
Next we will using the following identity.
$cos(\alpha+\beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$
3) $x^{'} = r\Big(cos(\theta)cos\big(acos\frac{x}{r}\big) - sin(\theta)sin\big(acos\frac{x}{r}\big)\Big)$
Notice that $cos\Big(acos\big(\frac{x}{r}\big)\Big) = \frac{x}{r}$
4) $x^{'} = r\Big(cos(\theta)\frac{x}{r} - sin(\theta)sin\big(acos\frac{x}{r}\big)\Big)$
Now we will use the identity $sin(acos(\gamma)) = \sqrt{1-\gamma^2}$.
5) $x^{'} = r\Big(cos(\theta)\frac{x}{r} - sin(\theta)\sqrt{1-\big(\frac{x}{r}\big)^2}\Big)$
Distribute $r$.
6) $x^{'} = cos(\theta)x - rsin(\theta)\sqrt{1-\big(\frac{x}{r}\big)^2}$
Recall that $r$ was just a placeholder for $\sqrt{x^2+y^2}$. We will now substitute this value back into the second $r$.
7) $x^{'} = cos(\theta)x - rsin(\theta)\sqrt{1-\big(\frac{x}{\sqrt{x^2+y^2}}\big)^2}$
Now it is time to exponentiate both parts of the fraction.
8) $x^{'} = cos(\theta)x - rsin(\theta)\sqrt{1-\frac{x^2}{x^2+y^2}}$
Find a common denominator.
9) $x^{'} = cos(\theta)x - rsin(\theta)\sqrt{\frac{x^2+y^2}{x^2+y^2}-\frac{x^2}{x^2+y^2}}$
Subtract.
10) $x^{'} = cos(\theta)x - rsin(\theta)\sqrt{\frac{y^2}{x^2+y^2}}$
Now we will split the radical over the fraction. (Hang in there, we're almost done.)
11) $x^{'} = cos(\theta)x - rsin(\theta)\frac{y}{\sqrt{x^2+y^2}}$
Now lets substitute $\sqrt{x^2+y^2}$ for the last remaining $r$.
12) $x^{'} = cos(\theta)x - \sqrt{x^2+y^2}sin(\theta)\frac{y}{\sqrt{x^2+y^2}}$
It cancels!
13) $x^{'} = cos(\theta)x - sin(\theta)y$
Rearrange variables.
14) $x^{'} = xcos(\theta) - ysin(\theta)$
Congratulations on making it this far! Now you know how to find this reduced equation without using the geometric proof.