The Powers of the Matrix with Cosine and Sine Functions

Linear algebra problems and solutions

Problem 567

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

 
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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 $n$.


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