If $A=\begin{bmatrix}\cos heta & \sin heta\ -\sin heta & \cos heta\end{bmatrix}$, prove that $A^3=\begin{bmatrix}\cos 3 heta & \sin 3 heta\ -\sin 3 heta & \cos 3 heta\end{bmatrix}$.

Question

If $A=\begin{bmatrix}\cos	heta & \sin	heta\ -\sin	heta & \cos	heta\end{bmatrix}$, prove that $A^3=\begin{bmatrix}\cos 3	heta & \sin 3	heta\ -\sin 3	heta & \cos 3	heta\end{bmatrix}$.

Accepted Answer

>

> [!TLDR]
> $A$ rotates by angle $	heta$; multiplying gives $A^2$ (angle $2	heta$) and then $A^3$ (angle $3	heta$), using the sum formulas for sine and cosine.
>
> **Concept.** The matrix $A=\begin{bmatrix}\cos	heta&\sin	heta\-\sin	heta&\cos	heta\end{bmatrix}$ compounds angles under multiplication: $A^n$ has angle $n	heta$.
>
> **Step 1 — compute $A^2=A\cdot A$.**
> $$A^2=\begin{bmatrix}\cos	heta&\sin	heta\-\sin	heta&\cos	heta\end{bmatrix}\begin{bmatrix}\cos	heta&\sin	heta\-\sin	heta&\cos	heta\end{bmatrix}.$$
> Top-left: $\cos^2	heta-\sin^2	heta=\cos2	heta$. Top-right: $\cos	heta\sin	heta+\sin	heta\cos	heta=\sin2	heta$. Bottom-left: $-\sin	heta\cos	heta-\cos	heta\sin	heta=-\sin2	heta$. Bottom-right: $-\sin^2	heta+\cos^2	heta=\cos2	heta$.
> $$A^2=\begin{bmatrix}\cos2	heta&\sin2	heta\-\sin2	heta&\cos2	heta\end{bmatrix}.$$
>
> **Step 2 — compute $A^3=A^2\cdot A$.**
> $$A^3=\begin{bmatrix}\cos2	heta&\sin2	heta\-\sin2	heta&\cos2	heta\end{bmatrix}\begin{bmatrix}\cos	heta&\sin	heta\-\sin	heta&\cos	heta\end{bmatrix}.$$ …

Q.If $A=\begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}$ , prove that $A^3=\begin{bmatrix}\cos 3\theta & \sin 3\theta\\ -\sin 3\theta & \cos 3\theta\end{bmatrix}$ .

Concept understanding — Matrix Rotation Power

Matrix Rotation Power

The intuition

The rotation case

Proving it by induction

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