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,

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

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


rotation-axes

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

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