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?

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3 Answers

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.

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$$\begin{pmatrix} \cos(\theta) & \sin(\theta) \\ \sin\theta & -\cos\theta \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix}\\=\begin{pmatrix} a\cos\theta+b\sin\theta \\ a\sin\theta-b\cos\theta \end{pmatrix}=\begin{pmatrix} a \\ b \end{pmatrix}$$ 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:

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

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

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The matrix system corresponds to the following system of equations $$\cos(\theta)\,a + \sin(\theta) b = a \,, \\ \sin(\theta)\,a - \cos(\theta) b = b \,$$ Multiplying the first equation by $\sin(\theta)$ and the second by -$\cos(\theta)$ gives $$\sin(\theta)\,\cos(\theta)\,a + \sin(\theta)^2 b = a\,, \\ -\sin(\theta) \cos(\theta) \,a + \cos(\theta)^2 b = -b \,$$

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

$$b \cos(\theta /2) = a \sin( \theta /2 )$$

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