Trigonometric eigenvalue equation 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?
 A: Since the matrix has the eigenvalue 1, it's enough to consider only the first row. Rearrange to get $(1-\cos(\theta))a = \sin(\theta)b$ which if you expand by the right formulae, give you $2(\sin(\theta/2)^2)a = 2\sin(\theta/2)\cos(\theta/2)b$. Now, if your matrix has to be non-identity, then this will give you the equation want.
A: $$\begin{pmatrix}
  \cos(\theta) & \sin(\theta) \\ \sin\theta & -\cos\theta
 \end{pmatrix}\begin{pmatrix}
  a \\ b
 \end{pmatrix}\\=\begin{pmatrix}
  a\cos\theta+b\sin\theta \\ a\sin\theta-b\cos\theta
 \end{pmatrix}=\begin{pmatrix}
  a \\ b
 \end{pmatrix}$$
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:
$$b\sin\theta=a(1-\cos\theta)$$
$$2b\sin(\frac \theta{2})\cos(\frac \theta{2})=a(2\sin^2(\frac \theta{2}))$$
$$b\cos(\frac \theta{2})=a\sin(\frac \theta{2})$$
A: The matrix system corresponds to the following system of equations
\begin{equation}
\cos(\theta)\,a + \sin(\theta) b = a \,,
\\
\sin(\theta)\,a - \cos(\theta) b = b \,
\end{equation}
Multiplying the first equation by $\sin(\theta)$ and the second by -$\cos(\theta)$ gives
\begin{equation}
\sin(\theta)\,\cos(\theta)\,a + \sin(\theta)^2 b = a\,,
\\
-\sin(\theta) \cos(\theta) \,a + \cos(\theta)^2 b = -b \,
\end{equation}
Add the two equations, and the simplify the resulting equation, you get
$$ b( 1+\cos(\theta) ) = a\,\sin(\theta) $$
Using the identity $ \cos(\theta) = 2\cos(\theta /2)^2 -1 $ and $\sin(\theta) = 2 \sin(\theta /2) \cos(\theta /2) $ in the above equation yields
$$ 2 b \cos(\theta /2)^2 = 2 a \sin( \theta /2 ) \cos( \theta /2 )$$
Dividing both sides of the equation by $\cos(\theta /2)$ gives the result
$$ b \cos(\theta /2) = a \sin( \theta /2 ) $$
