$F(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$. What is $F\left(\frac{\pi}{6}\right) F\left(\frac{\pi}{3}\right)$?

Explanation

1. 01

   Multiplying out, $F(\alpha)F(\beta)$ has $\cos\alpha\cos\beta - \sin\alpha\sin\beta = \cos(\alpha + \beta)$ on the diagonal, and similarly $\sin(\alpha + \beta)$ off it: $F(\alpha)F(\beta) = F(\alpha + \beta)$.

2. 02

   So the product is $F\left(\frac{\pi}{2}\right) = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.