# The Powers of the Matrix with Cosine and Sine Functions

## Problem 567

Prove the following identity for any positive integer $n$.
$\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix}^n=\begin{bmatrix} \cos n\theta & -\sin n\theta\\ \sin n\theta& \cos n\theta \end{bmatrix}.$

Contents

## Hint.

Recall the addition theorem of trigonometric functions (sum formula)
\begin{align*}
\sin(x+y)&=\sin x \cos y +\cos x \sin y\\
\cos(x+y)&=\cos x \cos y -\sin x \sin y.
\end{align*}

Use the induction on $n$.

## Proof.

We prove the identity by induction on $n$.
The base case $n=1$ is clear.

Suppose that the identity is true for $n=k$.
Then we have
\begin{align*}
&\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta & \cos \theta
\end{bmatrix}^{k+1}\6pt] =& \begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta \end{bmatrix}^{k}\\[6pt] =&\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos k\theta & -\sin k\theta\\ \sin k\theta & \cos k\theta \end{bmatrix} \\ &\text{(by the induction hypothesis)}\\[6pt] =&\begin{bmatrix} \cos \theta \cos k\theta – \sin \theta \sin k \theta & -\cos \theta \sin k\theta-\sin \theta \cos k \theta\\ \sin \theta \cos k \theta + \cos \theta \sin k \theta & -\sin \theta \sin k\theta +\cos \theta \cos k \theta \end{bmatrix} \\ &\text{(by matrix multiplication)}\\[6pt] =&\begin{bmatrix} \cos (\theta+k\theta) & -\sin (\theta+ k\theta)\\ \sin (\theta+k\theta) & \cos (\theta+k\theta) \end{bmatrix}\\ &\text{(by the addition theorem of trigonometric functions)}\\[6pt] =&\begin{bmatrix} \cos (k+1)\theta & -\sin (k+1)\theta\\ \sin (k+1)\theta & \cos (k+1)\theta \end{bmatrix}. \end{align*} This proves that the identity holds for n=k+1. Thus by induction, the identity is true for all positive integers n. Sponsored Links ### More from my site • Powers of a Diagonal Matrix Let A=\begin{bmatrix} a & 0\\ 0& b \end{bmatrix}. Show that (1) A^n=\begin{bmatrix} a^n & 0\\ 0& b^n \end{bmatrix} for any n \in \N. (2) Let B=S^{-1}AS, where S be an invertible 2 \times 2 matrix. Show that B^n=S^{-1}A^n S for any n \in […] • Compute the Product A^{2017}\mathbf{u} of a Matrix Power and a Vector Let \[A=\begin{bmatrix} -1 & 2 \\ 0 & -1 \end{bmatrix} \text{ and } \mathbf{u}=\begin{bmatrix} 1\\ 0 \end{bmatrix}. Compute $A^{2017}\mathbf{u}$.   (The Ohio State University, Linear Algebra Exam) Solution. We first compute $A\mathbf{u}$. We […]
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