Two coordinate systems of the same orientation are given…

Two coordinate systems of the same orientation are given: $xOy$ (rectangular) and $x'O'y'$ (oblique, coordinate angle $\omega=45^\circ$). The axis $O'x'$ makes an angle of $30^\circ$ with $Ox$, and $O'$ has coordinates $(1,-1)$ in the old system. Find the relations between the old and new coordinates.

Step 1: Direct formulas

The unit vectors of the new axes are $\vec e_1=(\cos30^\circ,\sin30^\circ)$ and $\vec e_2=(\cos75^\circ,\sin75^\circ)$, because $30^\circ+\omega=75^\circ$. From $\overrightarrow{OM}=\overrightarrow{OO'}+x'\vec e_1+y'\vec e_2$:

\[ \cos75^\circ=\frac{\sqrt6-\sqrt2}{4},\qquad \sin75^\circ=\frac{\sqrt6+\sqrt2}{4} \]

\[x=1+\frac{\sqrt3}{2}x'+\frac{\sqrt6-\sqrt2}{4}y',\qquad y=-1+\frac12x'+\frac{\sqrt6+\sqrt2}{4}y'\]

Step 2: Determinant

\[ \det M=\cos30^\circ\sin75^\circ-\sin30^\circ\cos75^\circ=\sin(75^\circ-30^\circ)=\sin\omega=\frac{\sqrt2}{2}. \]

Step 3: Inverse formulas

With $X=x-1$, $Y=y+1$, Cramer's rule gives $x'=\dfrac{X\sin75^\circ-Y\cos75^\circ}{\sin45^\circ}$ and $y'=\dfrac{Y\cos30^\circ-X\sin30^\circ}{\sin45^\circ}$. Using $\tfrac{\sqrt6\pm\sqrt2}{2\sqrt2}=\tfrac{\sqrt3\pm1}{2}$:

\[x'=\frac{(\sqrt3+1)(x-1)-(\sqrt3-1)(y+1)}{2},\qquad y'=\frac{-x+\sqrt3\,y+\sqrt3+1}{\sqrt2}\]

The source's form $x'=\dfrac{(\sqrt6+\sqrt2)(x-1)-(\sqrt6-\sqrt2)(y+1)}{2\sqrt2}$ is the same expression, before cancelling $\sqrt2$.

Step 4: Check

For $M(2,\,0.8)$: $X=1$, $Y=1.8$.

\[ x'=\tfrac{2.732-1.318}{2}\approx0.707,\qquad y'=\tfrac{3.118-1}{1.414}\approx1.497 \]

Back-substitution gives $x=1+0.612+0.387=1.999$ and $y=-1+0.354+1.446=0.800$. $\checkmark$