We introduced the complex numbers as elements of $ \mathrm{Mat}(2\times 2, \mathbb{R})$ with
$ \mathbb{C} \ni x = \left(\begin{array}{cc} a & -b \\ b & a \\ \end{array}\right) = \frac{1}{\sqrt{a^2+b^2}} \left(\begin{array}{cc} \frac{a}{\sqrt{a^2+b^2}} & \frac{-b}{\sqrt{a^2+b^2}} \\ \frac{b}{\sqrt{a^2+b^2}} & \frac{a}{\sqrt{a^2+b^2}} \\ \end{array}\right) $
Then we concluded that $0 \leq \frac{a}{\sqrt{a^2+b^2}} \leq 1$ and $0 \leq \frac{b}{\sqrt{a^2+b^2}} \leq 1$ and therefore we could find an $\alpha$ so that $\cos \alpha = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin \alpha = \frac{b}{\sqrt{a^2+b^2}}$. Then we can write the matrix with $\cos$ and $\sin$ and can write it as the Euler form as well. So far so good.
My question ist about the following:
Why is it that we cand find an $\alpha$ so that $\cos \alpha = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin \alpha = \frac{b}{\sqrt{a^2+b^2}}$ for every possible value combination of $a$ and $b$? Can't be there a combination of $a$ and $b$ where we can't find one and the same angle $\alpha$ so that the identites are true?