How is the entire $SO(2)$ group the standard rotation matrix? In a book I am using, the following is presented,

$$\mathcal{R}(\phi) = \begin{pmatrix} \cos (\phi ) &\sin (\phi ) \\ -\sin (\phi ) &\cos (\phi )\end{pmatrix}$$
The group's name is $\textrm{SO}(2)$, if the angle $\phi$ varies continuously from $0$ to $2\pi$; $\textrm{SO}(2)$ has infinitely many elements and is compact.

From my understanding, $\textrm{SO}(2)$ is the group of orthogonal matrices each of whose determinants are $+1$. I can definitely see how the rotation matrix forms a subgroup, but I don't see how equality (rotation matrix = $\textrm{SO}(2)$) can be shown (especially so easily as to sweep it under the rug so quickly in this book).
Could you tell me how this can be shown?
 A: In order for a matrix to be in SO(2), it has to be (a) orthogonal, and (b) have determinant $1$.
For $(a)$ we need in particular that each row is a unit vector. Every unit vector has the form $(\cos \theta, \sin\theta)$ for some $\theta$, so our matrix necessarily has the form
$$\begin{pmatrix} \cos\theta & \sin\theta \\ \cos \phi & \sin \phi \end{pmatrix} $$
In addition the rows must be perpendicular to each other. Our two unit vectors are perpendicular if and only if $\phi = \theta\pm\frac\pi 2$, up to irrelevant multiples of $2\pi$. By some basic trigonometric identities, this means that the possibilites are now
$$\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \quad\text{ and }\quad \begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix} $$
The first of these always has determinant $1$, so it is in SO(2). The second has determinant $-1$, so it is not in SO(2) (but it is in O(2)).

Things become more complex in higher dimensions. A general form for a member of SO(3) would be
$$ \begin{align} &P^{-1}
\begin{pmatrix} \cos\theta & \sin\theta &0\\-\sin\theta&\cos\theta&0\\0&0&1 \end{pmatrix} P\\[1ex] \text{where }& P = 
\begin{pmatrix} 1&0&0 \\ 0 & \cos\phi & \sin\phi \\ 0 & -\sin\phi & \cos\phi \end{pmatrix}
\begin{pmatrix} \cos\psi & \sin\psi &0\\-\sin\psi&\cos\psi&0\\0&0&1 \end{pmatrix} \end{align} $$
(which you can write out if you want, but it will be horrible), and an element of SO(4) will in general be composed of two independent rotations about different axes:
$$ P^{-1} \begin{pmatrix}\cos \theta&\sin\theta&0&0\\-\sin\theta&\cos\theta&0&0\\
0&0&\cos\phi & \sin\phi \\ 0&0&-\sin\phi&\cos\phi \end{pmatrix} P $$
where $P$ is some orthogonal basis change that I don't even want to try to separate out into components.
The proofs for these higher-dimensional cases are correspondingly more involved.
