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 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 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 referred to 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}

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