Let $\Delta=\Omega-\phi_d$. From part (iii), $\phi-\phi_d=\Delta+(\phi-\Omega)$ and $\tan(\phi-\Omega)=\cos I\tan f$. The <tangent addition formula> therefore gives
$$
\boxed{
\tan(\phi-\phi_d)
=\frac{\tan\Delta+\cos I\tan f}
{1-\cos I\tan f\tan\Delta}}
$$
or, without singular coordinate tangents,
$$
\boxed{
\tan(\phi-\phi_d)=
\frac{\sin\Delta\cos f+\cos\Delta\cos I\sin f}
{\cos\Delta\cos f-\sin\Delta\cos I\sin f}}.
$$
For $I=\pi/2-I'$ with $I'\ll1$, smooth choice of the angular branch gives
$$
\boxed{\phi-\phi_d\simeq\Omega-\phi_d+I'\tan f\pmod\pi}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 316 / 1 / vi / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-316/1/vi/solution