The Rotation of Axes

Let the $xy$-coordinate system be rotated through an angle $\theta$, such that the range of the angle is $0 < \theta < \frac{\pi }{2}$ about the same origin $O$. The new coordinate system is $XY$-coordinate system as shown in the given diagram. Since triangle $OBP$ is a right triangle with $m\angle BOP = \alpha – \theta$ so,

$\begin{gathered} \frac{{OB}}{{OP}} = \cos \left( {\alpha – \theta } \right) \\ \Rightarrow OB = r\cos \left( {\alpha – \theta } \right)\,\,\,\,\,\,\,\because OP = r\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered}$

$\begin{gathered} \frac{{BP}}{{OP}} = \sin \left( {\alpha – \theta } \right) \\ \Rightarrow BP = r\sin \left( {\alpha – \theta } \right)\,\,\,\,\,\,\,\because OP = r\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right) \\ \end{gathered}$

Since $\left( {OB,BP} \right)$ are the coordinates of the point $P$ with respect to the new coordinate system, $XY$-system, so $OB = X,\,\,BP = Y$.

Putting these values in (i) and (ii), we get the following:
$\begin{gathered} X = r\cos \left( {\alpha – \theta } \right) \\ \Rightarrow X = r\cos \alpha \cos \theta + r\sin \alpha \sin \theta \,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right) \\ \end{gathered}$

$\begin{gathered} Y = r\sin \left( {\alpha – \theta } \right) \\ \Rightarrow Y = r\sin \alpha \cos \theta = r\cos \alpha \sin \theta \,\,\,{\text{ – – – }}\left( {{\text{iv}}} \right) \\ \end{gathered}$

Triangle $OAP$ is also a right triangle with
$x = OA = r\cos \alpha$ and $y = AP = r\sin \alpha$

Putting these values in (iii) and (iv), we have the following equations as
$\boxed{X = x\cos \theta + y\sin \theta ,\,\,\,\,\,\,\,Y = – x\sin \theta + y\cos \theta }$

These equations are used to find the coordinates of a point with respect to the new rotated coordinate system, $XY$-system. Thus, the point $P\left( {x,y} \right)$ with respect to $XY$-plane is $P\left( {x\cos \theta + y\sin \theta ,\, – x\sin \theta + y\cos \theta } \right)$.

Conversely, if the coordinates of a point with respect $XY$-system are given, then its coordinates with respect to the original system can be determined by the equations
$\boxed{x = X\cos \theta – Y\sin \theta ,\,\,\,\,\,\,\,y = X\sin \theta + Y\cos \theta }$

Example: The $xy$-coordinate axes are rotated about the origin through the angle of measure ${30^ \circ }$. The new axes are $OX$ and $OY$. Find the $XY$-coordinates of the point $P\left( {1,6} \right)$.

Solution: Here $x = 1,\,\,y = 6$ and $\theta = {30^ \circ }$. The coordinates of $P$ with respect to the new $XY$-coordinates system are

$\begin{gathered} \,\,\,\,\left( {x\cos \theta + y\sin \theta ,\, – x\sin \theta + y\cos \theta } \right) \\ = \left( {1\cos {{30}^ \circ } + 6\sin {{30}^ \circ },\, – 1\sin {{30}^ \circ } + 6\cos {{30}^ \circ }} \right) \\ = \left( {\left( {\frac{{\sqrt 3 }}{2}} \right) + 6\left( {\frac{1}{2}} \right),\, – \left( {\frac{1}{2}} \right) + 6\left( {\frac{{\sqrt 3 }}{2}} \right)} \right) \\ = \left( {\frac{{\sqrt 3 }}{2} + \frac{6}{2},\, – \frac{1}{2} + \frac{{6\sqrt 3 }}{2}} \right) = \left( {\frac{{\sqrt 3 + 6}}{2},\,\frac{{ – 1 + 6\sqrt 3 }}{2}} \right) \\ \end{gathered}$