I'm given that $T$ is a transformation in $\mathbb{R}^2$ that rotates vectors 30 degrees counterclockwise about the origin, i.e. $T = \begin{pmatrix} \cos{30} & -\sin{30} \\ \sin{30} & \cos{30} \end{pmatrix}$, and the bases $B=\{\begin{pmatrix} \cos{20} \\ \sin{20} \end{pmatrix}, \begin{pmatrix} -\sin{20} \\ \cos{20} \end{pmatrix}\}$ and $C=\{\begin{pmatrix} \cos{50} \\ \sin{50} \end{pmatrix}, \begin{pmatrix} -\sin{50} \\ \cos{50} \end{pmatrix}\}$.
I already found that $[T]_B^C=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ but when I attempt $[T]_C^B$, I find that rotating the vectors of the C basis yields $\begin{pmatrix} \cos{80} \\ \sin{80} \end{pmatrix}$ and $\begin{pmatrix} -\sin{80} \\ \cos{80} \end{pmatrix}$, where it's suddenly difficult to express these two vectors in terms of the vectors from the B basis.
Is there a simpler way to approach this? Am I missing some more intuitive geometric interpretation or am I missing something else?
Use the angle addition formulas:
$$\begin{bmatrix} \cos80 \\ \sin80 \end{bmatrix} = \begin{bmatrix} \cos60\cos20-\sin60\sin20 \\ \sin60\cos20+\cos60\sin20 \end{bmatrix} = \cos60\begin{bmatrix} \cos20 \\ \sin20 \end{bmatrix} + \sin60\begin{bmatrix} -\sin20 \\ \cos20 \end{bmatrix}$$
and
$$\begin{bmatrix} -\sin80 \\ \cos80 \end{bmatrix} = \begin{bmatrix} -\sin60\cos20-\cos60\sin20 \\ \cos60\cos20-\sin60\sin20 \end{bmatrix} = -\sin60\begin{bmatrix} \cos20 \\ \sin20 \end{bmatrix} + \cos60\begin{bmatrix} -\sin20 \\ \cos20 \end{bmatrix}$$
so the matrix is
$$[T]_C^B = \begin{bmatrix} \cos60 & -\sin60 \\ \sin60 & \cos60 \end{bmatrix}$$
This is computational, but it suggests a geometric interpretation: $T$ rotates $30$ degrees, and the change of basis rotates $50-20=30$ degrees, so the composition rotates $30+30=60$ degrees. It also explains why the opposite change of basis produces the identity matrix: $30-30=0$ degrees.