Given the following equations:
\begin{array} ((\sin \beta \cos \alpha - i\sin \beta \sin \alpha)b + a\cos \beta = a \\ (\sin \beta \cos \alpha + i \sin \beta \sin \alpha)a - b\cos \beta = b \end{array} with the condition that $|a|^2 + |b|^2 = 1$. How does it follow that $$a = \cos(\frac{\beta}{2})e^{i \theta_a} ~~~\text{and}~~~b = \sin(\frac{\beta}{2})e^{i \theta_b}?$$
Thanks for any assistance.
Your equations \begin{eqnarray} \sin\beta\cdot e^{-i\alpha}b+(\cos\beta-1)a&=&0,\\ \sin\beta\cdot e^{i\alpha}a-(\cos\beta+1)b&=&0 \end{eqnarray} after using the double angle formulas becomes \begin{eqnarray} 2\sin\frac{\beta}{2}\left(\cos\frac{\beta}{2}\cdot e^{-i\alpha}b-\sin\frac{\beta}{2}\cdot a\right)&=&0,\\ 2\cos\frac{\beta}{2}\left(\sin\frac{\beta}{2}\cdot e^{i\alpha}a-\cos\frac{\beta}{2}\cdot b\right)&=&0 \end{eqnarray} which is equivalent to the single equation $$ \sin\frac{\beta}{2}\cdot e^{i\alpha}a-\cos\frac{\beta}{2}\cdot b=0= \begin{bmatrix}e^{i\alpha}\sin\frac{\beta}{2} & -\cos\frac{\beta}{2}\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix} $$ that has a solution $$ \begin{bmatrix}a\\b\end{bmatrix}=\lambda\begin{bmatrix}\cos\frac{\beta}{2}\\e^{i\alpha}\sin\frac{\beta}{2}\end{bmatrix}. $$ It is straightforward to show that $|a|^2+|b|^2=|\lambda|^2$ and the result follows.