How to prove that the cosine of the angle of two vectors is preserved after rotating the vectors with the same angle?

How can we with linear algebra prove that if two vectors in 2 dimensions form an angle $\phi$ then by multiplying those two vectors with the same rotation matrix the cosine of the formed angle will be preserved?

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You can maybe use the fact that matrix corresponding to the rotation is orthogonal: $$R^T = R^{-1}$$ and that the norm of a vector after rotation is the same as the original one: $$||R \mathbf{v}|| = ||\mathbf{v}||$$
So, by direct computation, for two vectors $\mathbf{v}$ and $\mathbf{w}$, you have:
$$\cos(\alpha_{R\mathbf{v},R\mathbf{w}}) = \frac{<R\mathbf{v},R\mathbf{w}>}{||R\mathbf{v}|| \ ||R\mathbf{w}||} = \frac{\mathbf{v}^T R^T R \mathbf{w}}{||\mathbf{v}|| \ ||\mathbf{w}||} = \frac{\mathbf{v}^T \mathbf{w}}{||\mathbf{v}|| \ ||\mathbf{w}||} = \cos(\alpha_{\mathbf{v},\mathbf{w}})$$
We have ${\bf a}\cdot{\bf b} = {\bf a}^T{\bf b} = ab\cos\phi$. Let $R$ be the rotation matrix. Then $$\begin{eqnarray*} {\bf a}^T {\bf b} &\rightarrow& (R{\bf a})^T {R\bf b} \\ &=& {\bf a}^T R^T R {\bf b} \\ &=& {\bf a}^T {\bf b}, \end{eqnarray*}$$ since $R$ is orthogonal.