Prove this group is Abelian Let $O_2$ consist of all matrices of the form $T_\theta$, where 
$T_\theta =\begin{bmatrix}\cos(\theta)&\sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{bmatrix} $, $\theta\in \Bbb R$.
Prove that $O_2$ under matrix multiplication is an Abelian group.
 A: One pretty easy way:  
note that any matrix of the given form form may be written
$\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} = \cos \theta I + \sin \theta J, \tag{1}$
where
$J = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \tag{2}$
and $I$ is the $2 \times 2$ identity matrix.  Now note that any two matrices of the form
$aI + bJ, \tag{3}$
commute:
$(aI + bJ)(cI + dJ) = (cI + dJ)(aI + bJ), \tag{4}$
which easily follows from the fact that $I$ and $J$ themselves commute, since $I$ is the identity:  $IJ = JI$.  Or simply write the products in (4) out in full:
$(aI + bJ)(cI + dJ) = (ac - bd)I + (bc + ad)J, \tag {5}$
where we have used the fact that
$J^2 = -I. \tag{6}$
Nota Bene: at this point one might notice the curious similarity between (4) and the ordinary multiplication of the complex numbers $a + bi$ and $c + di$!!! ;)
Anywaaayyy . . . .
It is easy to see $I$ is of the form (1), and since it can easily be checked that
$(\cos \theta I + \sin \theta J)^{-1} = \cos \theta I - \sin \theta J = \cos (-\theta) + J\sin (-\theta), \tag{7}$
the set $O_2$ contains the identity and inverses.  Closure under multiplication is had by applying standard trigonometric identities to (5), with $a = \cos \theta_1$, $b = \sin \theta_1$, $c = \cos \theta_2$, $d = \sin \theta_2$.  I'll leave that part of the algebra to you.
If we apply (4) two matrices of the form (1), the next result pops right out:
$\begin{bmatrix} \cos \theta_1 & \sin \theta_1 \\ -\sin \theta_1 & \cos \theta_1 \end{bmatrix} \begin{bmatrix} \cos \theta_2 & \sin \theta_2 \\ -\sin \theta_2 & \cos \theta_2 \end{bmatrix} = \begin{bmatrix} \cos \theta_2 & \sin \theta_2 \\ -\sin \theta_2 & \cos \theta_2 \end{bmatrix} \begin{bmatrix} \cos \theta_1 & \sin \theta_1 \\ -\sin \theta_1 & \cos \theta_1 \end{bmatrix}; \tag{8}$
your "set" $O_2$ is an Abelian group.
Hope this helps.  Cheers, and as always
Fiat Lux!
A: Hint: multiply two such matrices using an angle of $\theta$ and an angle of $\alpha$, say, and apply trig identities to determine if they're equal.
A: The set of matrices of the form
$$
\begin{bmatrix}a&b\\-b&a\end{bmatrix}
$$
is isomorphic to the field of complex numbers. In this interpretation, $O_2$ is the unit circle, the set of complex numbers of absolute value $1$. Since $|zw|=|z|\ |w|$ and $z\bar z=|z|$, this set is a group.
