Is this set of matrices closed under matrix multiplication? 2x2 matrices of the form: 
\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}
x is a real number
multiplying the matrix by itself produces
\begin{bmatrix}\cos^2x-\sin^2x&-2\cos x\sin x\\2\sin x\cos x&\cos^2x-\sin^2x\end{bmatrix}
Is this product matrix still in the set? It seems not to be in my mind, but I'm not sure.
 A: To check if $S:=\left\{\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}:x\in\Bbb R\right\}$ is closed under multiplication you would need to multiply $$\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}\begin{bmatrix}\cos(y)&-\sin(y)\\\sin(y)&\cos(y)\end{bmatrix}$$
not
$$\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}$$
because the two matrices need not be the same. So we need to show
$$\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}\begin{bmatrix}\cos(y)&-\sin(y)\\\sin(y)&\cos(y)\end{bmatrix}=\begin{bmatrix}\cos(x)\cos(y)-\sin(x)\sin(y)&-\cos(x)\sin(y)-\sin(x)\cos(y)\\\sin(x)\cos(y)+\cos(x)\sin(y)&-\sin(x)\sin(y)+\cos(x)\cos(y)\end{bmatrix}$$
is in $S$. Recall the identities:
$$(1)\qquad\sin(x\pm y)=\sin(x)\cos(y)\pm\sin(y)\cos(x)\\(2)\qquad \cos(x\pm y)=\cos(x)\cos(y)\mp \sin(x)\sin(y)$$
A: There is a geometrical interpretation that sheds light on all this :
$$R_x=\begin{bmatrix}\cos x&-\sin x\\ \sin x&  \ \ \cos x \end{bmatrix}$$
is a rotation matrix (https://en.wikipedia.org/wiki/Rotation_matrix).
Thus, what has been done in the answer given by @Dave is plainly 
$$R_x \times R_y = R_{x+y}$$
Using sentences : Rotation with angle $y$ followed by rotation by angle $x$ is rotation by angle $x+y$. 
This is why you have a closed set.
One can say more : the set of rotation matrices is a (commutative) group for multiplication.
