Skew coordinate system transformation matrix I am currently working through problems in Susan Lea's Mathematics for Physicists. However I am stuck on a problem and the book is not known to errata. The problem is:


*A Skew (non orthogonal) coordinate system in a plane has x'-axis along the x-axis and y'-axis at an angle $\theta$ to the x axis, where $\theta$ < $\frac{\pi}{2}$. 
(a) Write the transformation matrix that transforms vector components from the Cartesian x-y system to the skew system. 
The solution manual says the new coordinates are:
$y' = \frac{y}{sin\theta}$ 
which I also obtain, but the result:
$x' = x-\frac{y}{tan\theta}$
I simply cannot understand why this is the result. There is no explanation for this and embarrassingly enough I have spent more than 2 hours on this. Can someone please explain this to me, or let me know if it is a mistake. Thank you!
 A: Let the unit vectors in the skew system be $s_1, s_2$. Then the coordinates of $s_1,s_2$
in terms of the standard basis $e_1,e_2$ are $(1,0)^T, (\cos \theta, \sin \theta)^T$.
Let $S=\begin{bmatrix} 1 & \cos \theta \\ 0 & \sin \theta\end{bmatrix}$, then if $n=(n_1,n_2)^T$ are the coordinates of a point in the skew basis, this corresponds to a point $S n$ in the standard basis.
Hence to map back from the standard basis to the skew system we take the inverse.
We have $S^{-1} = {1 \over \sin \theta } \begin{bmatrix} \sin \theta & -\cos \theta\\ 0 &1 \end{bmatrix} = \begin{bmatrix} 1 & -{1 \over \tan \theta}\\ 0 & {1 \over \sin \theta} \end{bmatrix}$.
Hence a point whose coordinates are $(x,y)^T$ in the standard basis will correspond to
coordinates $(x-{y \over \tan \theta}, { y \over \sin \theta})^T$ in the skew system. 
A: Start by drawing a point at some arbitrary point in the plane. For simplicity, I would draw it between $x'$ and $y'$ axes in the first quadrant. Now draw some parallel lines to $x'$, $y'$ and $y$ axes from that point. The perpendicular from your point to the $x$ axis has length $y$, the parallel to the $y'$ axis has length $y'$. The angle between the parallel to $y'$ and $x$ is $\theta$, so in the triangle formed by the parralels to $y'$ and $y$ with the $x$ axis you get $$\sin\theta=\frac{y}{y'}$$ In the same triangle, $$\cos\theta=\frac{x-x'}{y'}$$
From the first equation then $$y'=\frac{y}{\sin\theta}$$
If you plug into the second equation you get $$x'=x-\frac{y\cos\theta}{\sin\theta}$$
