How can we calculate the exponential form of a rotation matrix Considering the rotation matrix: 
$$
A(\theta) = \left( \begin{array}{cc}
\cos\space\theta & -\sin \space\theta \\
\sin \space\theta & \cos\space\theta \\
\end{array} \right)
$$
How can I calculate $(A(\theta))^n$ where n ≥ 1 ? I'm not sure what to do nor how to write it. 
 A: Notice that $[A(\theta)]^n = A(n\theta)$, based on geometric interpretation of rotations. If you apply $A(\theta)$ twice, it is the same as rotating by $\theta$ twice, or $2\theta$. The same is true for an arbitrary number of rotations. That shows you have to write $A(\theta)^n$, namely as the same matrix with $\theta$ replaced by $n\theta$.  
A: let us write $z = x+iy.$ then the differential equation can be written as 
$\frac{dz}{dt} = e^{i\theta}z, z(0) = z_0$ which has a solution 
$$z = e^{t\cos \theta}\left(\cos(t\sin\theta) +  i\sin(t\sin\theta)\right)(x_0+iy_0)$$ splitting into real and imaginary we get 
$$x =  e^{t\cos \theta}\left(x_0\cos(t\sin \theta) - y_0\sin (t\sin \theta)\right),
\ 
y =  e^{t\cos \theta}\left(x_0\sin(t\sin \theta) + y_0\cos (t\sin \theta)\right)$$
writing in matrix form 
$$e^{At} = e^{t\cos \theta}
\pmatrix{ \cos(t\sin \theta) &- \sin (t\sin \theta)\\
 \sin(t\sin \theta) & \cos (t\sin \theta) }$$
A: $A(\theta)$ is called the rotation matrix simply because it rotates a point in the plane by an angle $\theta$. What happens when I apply this matrix $n$ times? I simply rotate the point $n$ times. How much has it rotated in the end? By an angle $n\cdot\theta$. If you can get this sort of intuitive grasp, you can say the following:
Let $P_n$ be the property
$$[A(\theta)]^n=A(n\theta),\ \ n\ge 1$$
Let's now prove by induction on $n$ that $P_n$ is true.
Base case
$P_1$ is trivial, so let's look at $P_2$:
\begin{align}[A(\theta)]^2&=A(\theta)\cdot A(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\cdot\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\\
&=\begin{pmatrix}\cos^2\theta-\sin^2\theta&-2\sin\cos\theta\\2\sin\cos\theta&\cos^2\theta-\sin^2\theta\end{pmatrix}\\
&=\begin{pmatrix}\cos(2\theta)&-\sin(2\theta)\\\sin(2\theta)&\cos(2\theta)\end{pmatrix}\\
&=A(2\theta)
\end{align}
Therefore $P_1$ and $P_2$ are true.
Inductive Step
Assume $P_n$ is true, and consider the case of $P_{n+1}$:
\begin{align}
[A(\theta)]^{n+1}&=[A(\theta)]^n\cdot A(\theta)=A(n\theta)\cdot A(n\theta)\\
&=\begin{pmatrix}\cos(n\theta)&-\sin(n\theta)\\\sin(n\theta)&\cos(n\theta)\end{pmatrix}\cdot\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\\
&=\begin{pmatrix}\cos(n\theta)\cos\theta-\sin(n\theta)\sin(\theta)&-\cos(n\theta)\sin\theta-\sin(n\theta)\cos\theta\\\sin(n\theta)\cos\theta+\cos(n\theta)\sin\theta&-\sin(n\theta)\sin\theta+\cos(n\theta)\cos\theta\end{pmatrix}
\end{align}
Now we use the following four trigonometric identities:
\begin{cases}\cos\theta\cos(n\theta)&=&\frac{1}{2}(\cos(\theta-n\theta)+\cos(\theta+n\theta))\\\sin\theta\sin(n\theta)&=&\frac{1}{2}(\cos(\theta-n\theta)-\cos(\theta+n\theta))\\
\cos\theta\sin(n\theta)&=&\frac{1}{2}(\sin(\theta+n\theta)-\sin(\theta-n\theta))\\
\sin\theta\cos(n\theta)&=&\frac{1}{2}(\sin(\theta+n\theta)+\sin(\theta-n\theta))\end{cases}
Which allow us to write:
$$[A(\theta)]^{n+1}=\begin{pmatrix}\cos(\theta+n\theta)&-\sin(\theta+n\theta)\\\sin(\theta+n\theta)&\cos(\theta+n\theta)\end{pmatrix}=A\left(\left(n+1\right)\theta\right)$$
Therefore $P_{n+1}$ is true.
Conclusion
We have shown that the property $P$ is true for $n=1$ (trivially; but also for $n=2$), and for $n+1$, if $P_n$ holds. Therefore, by the axiom of induction, $P_n$ is true for all $n\ge 1$.
$$[A(\theta)]^n=A(n\theta),\ \ n\ge 1\quad\Box$$
