Validity of a formula for the $n$-th power of a general $2 \times 2$ matrix I am taking an optics course and at one point$^1$ we need to find the $n$-th power of the $2 \times 2$ matrix
$$M := \begin{bmatrix} a & b \\ c & d\end{bmatrix}$$
where $a, b, c, d$ are real numbers. 
My professor (who is not very mathematically rigorous) gives the following formula which he calls "Sylvester's law":
$$\begin{bmatrix} a & b \\ c & d\end{bmatrix}^n = \frac{1}{\sin{\theta}} \begin{bmatrix} a \sin{(n \theta)} - \sin{((n-1) \theta)} & b \sin{(n \theta)} \\ c \sin{(n \theta)} & d \sin{(n \theta)} - \sin{((n-1)\theta)}\end{bmatrix} $$
where he defines $\cos{\theta} := \frac{1}{2}(a + d)$ (so presumably, $\theta = \arccos{(\frac{a+d}{2}}$). 
He then claims that $M^n$ is finite if $-1 < \frac{1}{2} (a + d) < 1$, for then $\theta$ is well-defined and $\sin(\theta) \neq 0$.$^2$ 
I was pretty skeptical since I have never seen this formula before and a google search didn't return anything similar. I've tried this formula for a couple of simple cases, and I do not get the correct answer. For example, for $a = 0.1, b = 0.2, c = 0.3, d = 0.4$ and $n = 2$, the formula gives
$$ \begin{bmatrix} -0.95 & 0.1 \\ 0.15 & -0.8 \end{bmatrix}$$
(note: not even positive) while the right value is
$$M^2 = \begin{bmatrix} 0.07 & 0.1 \\ 0.15 & 0.22\end{bmatrix}.$$
Thus the formula cannot be correct as presented. However, I find it hard to believe that he just made the formula up, so I suspect that there is a similar, correct formula (or at least an approximation).
My question: Is this formula correct? If not, is there a similar formula that is either correct or an approximation of the $n$-th power? If so, what is the range of validity?
I realise that if the answer is no then it will be pretty hard to be certain of that. I'm hoping that someone here will have seen this before.
$^1$For those interested: While discussing optical cavities using ray transfer matrices.
$^2$ A weird statement, since the $n$-th power of a finite matrix can never be infinite. I suspect this just has to do with the range of validity of the formula.
 A: @amd is right: we have to work in $SL(2,\mathbb{R})$, i.e., assume that matrix $M := \begin{bmatrix} a & b \\ c & d\end{bmatrix}$ is such that $det(M)=1$. 
Therefore, its characteristic polynomial is 
$$\chi_M(\lambda)=\lambda^2-trace(M)\lambda+det(M)=\lambda^2-trace(M)\lambda+1  $$
Setting $k=\frac12trace(M)=\frac12(a+d)=\begin{cases}\cos(\theta)&if&|k|\leq1\\
\cosh(t)&&otherwise\end{cases}  \ \ \ (1)$
The formula given in your text deals with the first case. It is this one that I will develop. I will indicate at the end how to work in the second case ($\cosh$ instead of $\cos$).
In the first case, 
$$\chi_M(\lambda)=\lambda^2-2 \cos(\theta)\lambda+1=(\lambda-e^{i\theta})(\lambda-e^{-i\theta})\ \ \ \ (0)$$
establishing that the eigenvalues of $M$ are $e^{\pm i\theta}$.
The formula we have to establish is
$$\begin{bmatrix} a & b \\ c & d\end{bmatrix}^n = \frac{1}{\sin{\theta}} \begin{bmatrix} a \sin{(n \theta)} - \sin{((n-1) \theta)} & b \sin{(n \theta)} \\ c \sin{(n \theta)} & d \sin{(n \theta)} - \sin{((n-1)\theta)}\end{bmatrix}$$
It can be simplified into :
$$M^n = \dfrac{\sin(n \theta)}{\sin(\theta)}M-\dfrac{\sin((n-1) \theta)}{\sin(\theta)}I \ \ \ \ (2)$$
You may or may not know the Chebyshev polynomials of the second kind $U_n$ that are defined by :
$$U_{n-1}(\cos\theta):=\dfrac{\sin(n \theta)}{\sin(\theta)}$$
They are the key tools here.
Remark: these expressions are in fact always polynomials in $\cos \theta$; for example, $U_{3}(\cos\theta):=\dfrac{\sin(4 \theta)}{\sin \theta} = 8 \cos^3 \theta- 4\cos \theta$.
In many cases, one replaces $\cos \theta$ by $x$ : in the latter case, we would write $U_3(x)=8 x^3-4x$.
This definition permits a still simpler expression for (2):
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  M^n = U_{n-1}M-U_{n-2}I \ \ \ \ $ (property $P_{n-1}$) $ \ \ \ (3)$
The $U_n$ constitute one of the important families of orthogonal polynomials with easily established recurrence formula:
$$U_n(\cos\theta)=\cos(\theta)U_{n-1}(\cos(\theta))-U_{n-2}(cos(\theta)) \ \ \ (4)$$
Let us prove (3) by recurrence. It is easily established for $n=2$.
Let us assume that (3) is true. Let us establish property $P_{n}$.
Let us multiply LHS and RHS of (3) by $M$:
$$M^{n+1}=U_{n-1}M^2-U_{n-2}M=U_{n-1}(2 \cos(\theta)M-I)-U_{n-2}M$$
(the second equality results from the application of Cayley-Hamilton's theorem $M^2-2 \cos \theta M+I=0$ : refer to the characteristic polynomial in formula (0)).
Then
$$M^{n+1}=(2\cos(\theta)U_{n-1}-U_{n-2})M-U_{n-1}I \ \ \ (5)$$
We can now use the recurrence relationship (4) to put (5) under the desired form:
$$M^{n+1}=U_{n}M-U_{n-1}I$$
which is formula (3) at the next index. The proof is thus completed.
In the second case ($\cosh$ instead of $\cos$), we would do a very similar computation with "relative" polynomials $T_n$ (Chebyshev polynomials of the first kind), but maybe the physical meaning would be lost... (as you say, $M^n$ may not remain bounded when $n \rightarrow \infty$).
