How to prove that $ \bar e_2 = e_1 \sin (w) + e_2 \cos(w)$ where $w$ is the angle between $e_1$ and $e_2$

Question

In the book of Linear Algebra by Werner Greub, at page 201, it is asked that

Consider an oriented inner product space of dimension 2. Given two positive orthonormal bases $e_1, e_2$ and $\bar e_1, \bar e_2$, prove that
$$\begin{align*}\bar e_1 &= e_1 \cos(w) - e_2 \sin(w) \\ \bar e_2 &= e_1 \sin (w) + e_2 \cos(w)\end{align*}$$ where $w$ is the oriented angle between $e_1$ and $\bar e_1.$

Note: $$\cos(w) = \frac{(e_1, \bar e_1)}{|e_1|\cdot|\bar e_1|}, \quad \sin(w) = \frac{\Delta(e_1, \bar e_1)}{|e_1| \cdot |\bar e_1|}$$ where $\Delta$ is the oriented determinant function.

I have proved the first statement easily, but to prove the second, I have argued that the coefficient matrix of this system has to be a orthogonal matrix and proved the second part in that way.

So my question is that how can we prove the second part differently ?

Answer 1

Perform a rotation through the angle $w$ on basis $e_1, e_2$, the result is basis $\bar{e}_1, \bar{e}_2$. The rotation can be represented as a rotation matrix as

$$ R = \begin{bmatrix} \cos(w) & -\sin(w) \\ \sin(w) & \cos(w) \\ \end{bmatrix} $$

and

$$ \begin{bmatrix} \bar{e}_1 \\ \bar{e}_2 \\ \end{bmatrix} = R \begin{bmatrix} e_1 \\ e_2 \\ \end{bmatrix} $$