# Rotation Matrix and Triple Angle Formulas?

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?

• Isn't it really about the sums-of-angle formulas? This makes it clearer since $2\theta+\theta=3\theta$. May 23, 2019 at 2:42
• en.m.wikipedia.org/wiki/De_Moivre's_formula
– amd
May 23, 2019 at 3:24

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 $$(R_{\theta})^{n+1}$$

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.