# 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,

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:

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

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

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