Induction Proof and Trigonometric Identities

Question

Use trigonometric identities and induction to prove that

$\left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right)^{n} = \left(\begin{array}{cc} \cos (n \theta) & -\sin (n\theta)\\ \sin (n \theta) & \cos (n \theta) \end{array} \right)$

for all $n \in \mathbb{N}.$

Answer 1

I would use mathematical induction.

Of course it is true for $n=1$. So I now assume that it is true for $n=k$, and prove for $n=k+1$.

This means: $\pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta}^k\pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta} = \pmatrix{\cos k\theta & -\sin k\theta \\ \sin k\theta & \cos k\theta}\pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta} $. With some linear algebra, the top row of the RHS will read as:

$\pmatrix{\cos k\theta \cos\theta -\sin k\theta \sin \theta & -\cos k\theta\sin \theta -\sin k\theta \cos\theta \\ X & Y}$

Recall that: $\cos(A+B) = \cos A \cos B - \sin A \sin B$, and you are done

Answer 2

I'll not use induction.

Let $\sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}$, and $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$,

then $\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}=\cos\theta I-i\sin\theta\sigma_y=\exp(-i\theta\sigma_y)$.

The last equity has used the fact that $\sigma_y^2=I$.

Finally you have your $n$-th power of the matrix equals to $\exp(-in\theta\sigma_y)$, expand it you are done with the proof.

Answer 3

For proof by induction; these are the $\color{red}{\mathrm{three}}$ steps to carry out:

Step 1: Basis Case: For $n=1 \implies$ LHS $=\left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right)^{n}$

$=\left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right)^{1}$

$=\left(\begin{array}{cc} \cos (1 \theta) & -\sin (1\theta)\\ \sin (1 \theta) & \cos (1 \theta) \end{array} \right)$

$=\left(\begin{array}{cc} \cos (\theta) & -\sin (\theta)\\ \sin ( \theta) & \cos ( \theta) \end{array} \right)=$ RHS. So statement holds for $n=1$.

Step 2: Inductive Assumption: Assume statement is true for $n=k$ such that

$\left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right)^{n}$

$=\left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right)^{k}$

$=\left(\begin{array}{cc} \cos (k \theta) & -\sin (k\theta)\\ \sin (k \theta) & \cos (k \theta) \end{array} \right)\tag{1}$

Step 3: Prove Statement holds for $n=k+1$ such that

$\left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right)^{n}$

$=\left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right)^{k+1}$

$=\left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right)^{k}$ $\times \left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right)^{1}$

$=\left(\begin{array}{cc} \cos (k \theta) & -\sin (k\theta)\\ \sin (k \theta) & \cos (k \theta) \end{array} \right) \times \left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right)$ [using our inductive assumption $(1)$]

$=\left(\begin{array}{cc} \cos\theta\cos (k \theta)-\sin\theta \sin(k\theta) & -\left(\sin \theta\cos (k \theta)+\sin (k \theta)\cos \theta\right)\\ \cos\theta\sin (k \theta)+\sin\theta \cos(k\theta) & \cos\theta\cos (k \theta)-\sin\theta \sin(k\theta) \end{array} \right)$

$=\color{blue}{\left(\begin{array}{cc} \cos (\theta(k+1)) & -\sin (\theta(k+1))\\ \sin (\theta(k+1)) & \cos(\theta(k+1)) \end{array} \right)}$

Where in the last step I used the fact that $$\cos(A \mp B)=\cos A \cos B \pm \sin A \sin B$$ and $$\sin(A \pm B)=\sin A \cos B \pm \cos A \sin B$$ and $\color{blue}{\mathrm{this}}$ is the same result if $n=k+1$ is substituted into the RHS of your original disposition. Hence statement is true for all $n \in \mathbb{N}.$

QED