The expression of $f(x,y)$ explicitly Consider the function $f : \mathbb{R^2} \rightarrow \mathbb{R^2}$ that associates to each vector  $(x,y)$ the vector $f(x,y)$ obtained from a counterclockwise rotation of $\theta=30^{\circ}$ of the vector $(x,y)$. 

Write the expressions of $f(x,y)$ explicitly.

How can I prove ? I don't know how to write this expression. 
thanks for your help :-)
 A: Here are two ways, with the second a bit more advanced than the first.


*

*Notice that rotating the vector doesn't change its length, so we may assume (at least at first) that it has length $1$. It joins $0$ to a point on the unit circle, so you can write it as $(x,y)=(\cos\alpha,\sin\alpha)$ for some angle alpha. The image is $f(x,y)=(\cos(\alpha+\theta),\sin(\alpha+\theta))$. By using some identities for the sine and cosine, you find that
$$ \cos(\alpha+\theta)= \cos\alpha\cos\theta-\sin\alpha\sin\theta= x\cos\theta-y\sin\theta\\
\sin(\alpha+\theta)= \cos\alpha\sin\theta+\sin\alpha\cos\theta= x\sin\theta+y\cos\theta,$$
so that gives you
$$ f(x,y) = (x\cos\theta-y\sin\theta, x\sin\theta+y\cos\theta) $$
for a vector $(x,y)$ of length $1$. If it has another length $\ell$, then you can apply the above formula to the length $1$ vector $(x',y')=(x/\ell,y/\ell)$, and then multiply the result by $\ell$. That gives you the exact same formula for $f(x,y)$.

*If you know a bit of linear algebra, you can see that $f$ is a linear map, so you only need to know where it sends the basis vectors $e_1=(1,0)$ and $e_2=(0,1)$. Again, if you draw those two vectors on the unit circle, you'll that $f(e_1)=(\cos\theta,\sin\theta)$ and $f(e_2)=(-\sin\theta,\cos\theta)$, so that
\begin{align} f(x,y)&=f(xe_1+ye_2)\\
&= xf(e_1)+yf(e_2)\\
&= x(\cos\theta,\sin\theta) + y(-\sin\theta,\cos\theta)\\
&= (x\cos\theta-y\sin\theta, x\sin\theta+y\cos\theta).
\end{align}
In any case, for $\theta=30°$ you have $\cos\theta=\sqrt3/2$ and $\sin\theta=1/2$, so that
$$ f(x,y) = \frac12 (\sqrt3 x - y, x + \sqrt3 y). $$
A: You can find $f$ by determining the rotation matrix corresponding to $\theta = 30^{\circ}$. If we call this matrix $R$, we have $f(x, y) = R\begin{bmatrix}x\\y\end{bmatrix}$.
A: Use the Rotation matrix $$R_{\pi/6}=\left(\begin{array}{cc}
\cos(\pi/6) & -\sin(\pi/6) \\
\sin(\pi/6) & \cos(\pi/6)  \end{array}\right)$$ and then follow @Micheal's approach.
