Show that rotation about an arbitrary pivot point (a, b) through an angle  is equivalent to the rotation about the origin through  followed by a translation through (a, b), where a= a(1-cos )  bsin  and b=b(1-cos) - a sin .

To show that rotation about an arbitrary pivot point (a, b) through an angle \(\Theta\) is equivalent to the rotation about the origin through \(\Theta\) followed by a translation through (a, b), let's use homogeneous coordinates and transformation matrices.

1. Rotation about the Origin:

The rotation matrix \(R\) for rotating a point (x, y) about the origin through an angle \(\Theta\) is given by:

\[ R = \begin{bmatrix} \cos(\Theta) & -\sin(\Theta) \\ \sin(\Theta) & \cos(\Theta) \end{bmatrix} \]

2. Translation:

The translation matrix \(T\) for translating a point (x, y) through (a, b) is given by:

\[ T = \begin{bmatrix} 1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{bmatrix} \]

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