Matrix Group induction proof and order of elements question $H$ is the set of the matrices $A$ of the form
$$A= \begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}$$ 
where $0\leq \theta < 2\pi$  is a group with respect to matrix multiplication.
Question part (1) Prove by induction that  for any $\theta$ and any positive integer $n$,
$$\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{n} = \begin{pmatrix} \cos(n\theta) & \sin (n\theta) \\ -\sin (n\theta) & \cos (n\theta)\end{pmatrix}$$ 
When $n =1$ we have $$\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{1} = \begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}$$ 
and clearly this holds for the base case. 
Now I'm going to assume true when $n=k$ and try and prove for $k+1$. So we have 
$$\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{k}\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{1} = \begin{pmatrix} \cos(k\theta + \theta) & \sin (k\theta + \theta) \\ -\sin (k\theta + \theta) & \cos (k\theta + \theta)\end{pmatrix}$$ 
Now I'm sure I can cheese this, but is there some clever way to pull this apart? I want to just be able to say something like the determinant of the left side is $1^{k+1}$ and its the same on the right ...
Question part (2) Find an element of $H$ of order 5.
Here I'm totally lost, I seem to think that every element is of order 1 cause the determinant is 1.
Question part (3) Find an element of $H$ of infinite order.
Again I'm lost; I want to somehow make the determinant 0 for this but it can never be 0 for any $\theta$.
 A: Hints:
For part 1, you didn't actually use the inductive hypothesis yet. Remember that you are assuming that the claim is true when $n=k$, i.e. that
$$\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{k} = \begin{pmatrix} \cos(k\theta) & \sin (k\theta) \\ -\sin (k\theta) & \cos (k\theta)\end{pmatrix}.$$
Thus, what you want to prove can be rewritten as
$$\begin{pmatrix} \cos(k\theta) & \sin (k\theta) \\ -\sin (k\theta) & \cos (k\theta)\end{pmatrix}\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{1} = \begin{pmatrix} \cos(k\theta + \theta) & \sin (k\theta + \theta) \\ -\sin (k\theta + \theta) & \cos (k\theta + \theta)\end{pmatrix}.$$
You'll want to just multiply this out and use some trig identities.

For parts 2 and 3, you should note that multiplying a vector $\begin{pmatrix} a \\ b\end{pmatrix}$ in the plane $\mathbb{R}^2$ by the matrix $\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}$ produces the vector that is $\begin{pmatrix} a \\ b\end{pmatrix}$, except now rotated counterclockwise by an angle of $\theta$.
Can you think of an angle $\theta$ such that rotating the plane around the origin by $\theta$ 5 times in a row is equivalent to doing nothing?
Can you think of an angle $\theta$ such that no matter how many times you rotate the plane by it, it never comes back to its original orientation?
