Where did this matrix come from? In my lecture my professor spoke of this function $R$ that takes a vector $\vec u=\left(\begin{matrix}a\\b\end{matrix}\right)$ and rotates by $\frac{\pi}{6}$ radians counter clockwise. Then he talked about a matrix $M=\left(\begin{matrix}\frac{\sqrt{3}}{2} & -\frac12\\\frac12 & \frac{\sqrt{3}}{2}\end{matrix}\right)$. He showed finding the lengths on a triangle with a hypotenuse length $1$, and I have a feeling that the $x$ and $y
$ coordinates is in the first column, but I'm not sure where this second column comes from. If I need to provide any more information let me know. I really don't know what else to say without making this look like a jumbled mess.
 A: A terse answer goes like this:
Rigid rotations of the plane look like this:
$$
A(\theta)=\begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{bmatrix}
$$
Matrices with orthonormal columns, like $A(\theta)$ has, produce all the rotations of the plane around the origin. This matrix produces a rotation of $\theta$ radians when placed to the right of a 1-by-2 row vector. In your case, $\theta=\pi/6$. 
You can explore this by examining where the point $(1,0)$ goes under this rotation, and then argue that everything else rigidly follows. (After rotation, this could be viewed as the length-1 hypoteneuse of a right triangle, which is probably how the proof you saw went.)
A: You know that the effect of the matrix $M$ on a vector $\vec{u}$ is to rotate it counterclockwise by $\frac{\pi}{6}$ radians.  To determine the matrix $M$, let $\vec{e_1} = (1,0)$ and $\vec{e_2} = (0,1)$ be the standard basis vectors.  Then the first column of $M$ is $M \vec{e_1}$ and the second column of $M$ is $M \vec{e_2}$.
To figure out $M \vec{e_1}$, you simply have to rotate $\vec{e_1}$ and write down the resulting vector in coordinates.  For that, you wil need to recall some basic triangle goemetry.  Then you need to perform a similar calculation to determine $M \vec{e_2}$.
