[Math] Trigonometric eigenvalue equation

Question

In solving an eigenvalue problem, I've come to following equation ($\lambda=1$):
$$\begin{pmatrix}
\cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta)
\end{pmatrix}\begin{pmatrix}
a \\ b
\end{pmatrix}=\begin{pmatrix}
a \\ b
\end{pmatrix}$$
Now, the solution says, "This matrix equation can be reduced to a single equation":
$$a \sin(\frac{1}{2}\theta)=b\cos(\frac{1}{2}\theta)$$
I've been rotating trigonometric formulas to get to this, but I simply can't find the way. Could you help me with this, or at least give me a hint?

Answer 1

$$\begin{pmatrix} \cos(\theta) & \sin(\theta) \\ \sin\theta & -\cos\theta \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix}\\=\begin{pmatrix} a\cos\theta+b\sin\theta \\ a\sin\theta-b\cos\theta \end{pmatrix}=\begin{pmatrix} a \\ b \end{pmatrix}$$ Which gives us the following system of equations: $$a\cos\theta+b\sin\theta=a\tag{1}$$ $$a\sin\theta-b\cos\theta=b\tag{2}$$

So from $(1)$ you have:

$$b\sin\theta=a(1-\cos\theta)$$

$$2b\sin(\frac \theta{2})\cos(\frac \theta{2})=a(2\sin^2(\frac \theta{2}))$$ $$b\cos(\frac \theta{2})=a\sin(\frac \theta{2})$$