Rotation Matrix and Triple Angle Formulas?

Question

Define $R_{\theta}:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ as the rotation matrix by angle $\theta$, where

$$R_{\theta} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$

Observe that

$$ (R_\theta)^2 = \begin{pmatrix} \cos^2\theta-\sin^2\theta & -2\sin\theta\cos\theta \\ 2\sin\theta\cos\theta & \cos^2\theta-\sin^2\theta \end{pmatrix} = \begin{pmatrix} \cos2\theta & -\sin2\theta \\ \sin2\theta & \cos2\theta \end{pmatrix}=R_{2\theta} $$

This all makes sense of course since if you rotate a vector by $\theta$ twice, the net result should be a rotation by $2\theta$. The algebra of it all can be verified with the double angle formulas.

However, how do you prove that

$$ (R_\theta)^3 = R_{3\theta} $$

or perhaps that

$$ (R_\theta)^n = R_{n\theta} $$

Are there triple angle formulas that can be used to make the algebra work? n-tuple angle formulas?

Answer 1

If $R_{\theta}:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is the rotation matrix by angle $\theta$, where

$$R_{\theta} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$

Then we can write

$$R_{\theta} = \begin{pmatrix} \cos\theta & 0 \\ 0 & \cos\theta \end{pmatrix} + \begin{pmatrix} 0 & -\sin\theta \\ \sin\theta & 0 \end{pmatrix}$$

$$ = \cos\theta\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \sin\theta\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$

Noting that $$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ is the identity matrix and $$J=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$ satisfies $$J^2=-\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = -I$$

Thus we have $$R_{\theta} =\cos{\theta}I+\sin{\theta}J$$ with $J^2=-I$

We can then make the analogy with De Moivre's equation $$e^{i \theta}=\cos{\theta}+i\sin{\theta}$$

This suggests that $$e^{I \theta}=R_{\theta}=\cos{\theta}I+\sin{\theta}J$$

This can be proved by considering power series as in the case of DeMoivre's theorem. Clearly then we have $$R_{n\theta}=e^{I n\theta}={(e^{I \theta})}^n=R_{\theta}^n$$

Alternatively you can avoid power series and use trig formulas or De Moivre's theorem directly by showing that $$R_{\theta+\alpha}=\cos{(\theta+\alpha)}I+\sin{(\theta+\alpha)}J = R_{\theta}R_{\alpha}$$ on expanding the sin and cosine terms. $R_{n\theta}=R_{\theta}^n$ then follows immediately.

Answer 2

You may use these identities

$\cos x \cos y - \sin x \sin y = \cos(x+y)$

$\sin x \cos y + \cos x \sin y = \sin(x+y)$

and use $\theta $ and $2\theta$ in place of $x$ and $y$. For $R_{n\theta}$, try proving it by induction, assuming $(R_{\theta})^n = R_{n\theta}$ to be true and find $(R_{\theta})^{n+1}$