Just wanted to make sure my thinking is correct. The exercise key gives me a different answer than the one I'm getting and I'm not sure where I'm going wrong. Please don't give me the answer right away since this is homework.
The exercise gives us an orthogonal matrix A
$$A = \begin{bmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2}\\
\frac{\sqrt{3}}{2} & -\frac{1}{2}\\
\end{bmatrix}$$
Which is a reflection given that det(A) = -1
That corresponds to the reflection matrix R
$$
R = \begin{bmatrix}
cos(\theta) & sin(\theta)\\[0.3em]
sin(\theta) & -cos(\theta)\\[0.3em]
\end{bmatrix}
$$
To find the angle I take whichever length is preserved, which is $sin(\frac{\sqrt{3}}{2})$ and calculate
$$arcsin(\frac{\sqrt{3}}{2})$$
This gives me an angle of $\pi/3$
However the key states the answer is $\pi/6$ which is confusing me.
Thanks for your time.
Best Answer
If reflection is with respect to a line making angle $\theta$ with the $x$-axis its matrix for the standard basis is $$\pmatrix{\cos2\theta & \sin2\theta\cr \sin2\theta & -\cos2\theta\cr}$$. For example $y$-axis makes angle $\pi/2$ with $x$-axis, it sends $(x,y)$ to $(-x,y)$, please check with te definition if it agrees now.