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Actually, there are only two kinds of 2-order orthogonal matrices: $$\left( \begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{array} \right)$$ or $$\left( \begin{array}{cc} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \\ \end{array} \right).$$

Why?

If we wrote equation fulfilling $A^{2}=I$, where $$A=\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right),$$ the result would be like: $$a^{2}+b^{2}=1$$$$c^{2}+d^{2}=1$$$$ac+bd=0.$$ But how to deal with those equations?

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Perhaps Ted's hint didn't ring any bells.

We get $a=\sin x$, $b=\cos x$ for some $x$, and $c=\sin y$, $d=\cos y$ for some $y$. Then $$0=ac+bd=\sin x\sin y+\cos x\cos y=\cos(x-y)$$ which pretty much tells you what $x-y$ has to be.

It should be easy from there.

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HINT: So $(a,b)$ and $(c,d)$ are both points on the unit circle. How can you represent every such point?

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