Vector transformation, cylindrical to Cartesian

[tomwilliam]

Homework Statement

I have a result which is in the form (cylindrical coordinates):

$$ A\boldsymbol{e_{\theta }}=kr\boldsymbol{e_{\theta }} $$

And I have to provide the answer in cartesian coordinates.

Homework Equations

I know that the unit vectors:

$$ \boldsymbol{\hat{\theta} }=\begin{bmatrix}-sin\ \theta
\\
cos\ \theta
\end{bmatrix} $$

and that
$$ r=\sqrt{x^{2}+y^{2}} $$

The Attempt at a Solution

$$ kr\boldsymbol{e_{\theta }} =k\left (\sqrt{x^{2}+y^{2}}\right ) $$

$$ \left (-sin\left(tan^{-1}\left(\frac{y}{x}\right ) $$
$$ \right )\boldsymbol{e_{x}} $$
$$ +cos\left (tan^{-1}\left (\frac{y}{x}\right )\boldsymbol{e_{y}} \right )\\ $$

I can't seem to get further than this. I don't know if I've made a mistake, or whether there is some trig identity that can help me simplify further, but I know the final answer and it is much simpler.
Any help much appreciated.
Thanks in advance
P.S. Why does the latex break down when the equation is too long?

 

[jbsteele]

Is there a way to fix this? A:$$kr\boldsymbol{e_{\theta }}=k\left (\sqrt{x^{2}+y^{2}}\right )\begin{bmatrix}-sin\ \theta \\ cos\ \theta \end{bmatrix}$$$$=k\left (\sqrt{x^{2}+y^{2}}\right )\begin{bmatrix}-\sin\left (\tan ^{-1}\left (\frac{y}{x} \right ) \right ) \\ \cos\left (\tan ^{-1}\left (\frac{y}{x} \right ) \right ) \end{bmatrix}$$$$=k\begin{bmatrix}-x\sin\left (\tan ^{-1}\left (\frac{y}{x} \right ) \right ) \\ y\cos\left (\tan ^{-1}\left (\frac{y}{x} \right ) \right ) \end{bmatrix}$$$$=k\begin{bmatrix}-x\frac{y}{\sqrt{x^2+y^2}}\\ y\frac{x}{\sqrt{x^2+y^2}}\end{bmatrix}$$$$=k\begin{bmatrix}-y\\ x\end{bmatrix}$$$$=k\boldsymbol{e_{x}}\times \boldsymbol{e_{y}}$$