From a physics textbook: how did they get the eccentric anomaly ψ?

Question

The initial formula was: $$ \begin{align} \cos\psi &= \cos(θ-θ₀) + e \sin^2(θ-θ₀) - e^2 \cos(θ-θ₀)\sin^2(θ-θ₀)\\ &+ e^3 \cos^2(θ-θ₀)\sin^2(θ-θ₀) - ...\\ \end{align} $$ Where e is the eccentricity, $ψ$ is the eccentric anomaly, and $θ-θ₀$ is the true anomaly. They got out of this: $$ \psi = (θ-θ₀) - e \sin(θ-θ₀) + {e^2\over4} \sin2(θ-θ₀)+ \dots $$ I've tried various things with taylor series, but got nowhere, does anyone here know how this was derived?

Answer 1

If you want an expansion up to $e^2$ you can set $$ \psi=\psi_0+e \alpha+e^2\beta+O(e^3), $$ where $\psi_0=\theta-\theta_0$, while $\alpha$ and $\beta$ are unknown coefficients to be found. Using trig addition formulas and Taylor series for sine and cosine we can then write: $$ \begin{align} \cos\psi &=\cos(\psi_0+e \alpha+e^2\beta+O(e^3))\\ &=\cos\psi_0\cos(e \alpha+e^2\beta+O(e^3))-\sin\psi_0\sin(e \alpha+e^2\beta+O(e^3))\\ &=\cos\psi_0\left(1-{1\over2}e^2\alpha^2\right)-\sin\psi_0(e \alpha+e^2\beta)+O(e^3)\\ &=\cos\psi_0-e\alpha\sin\psi_0-e^2\left({1\over2}\alpha^2\cos\psi_0 +\beta\sin\psi_0\right)+O(e^3)\\ \end{align} $$ Comparing this with the given expansion we thus obtain: $$ \begin{align} -\alpha\sin\psi_0 &=\sin^2\psi_0\\ {1\over2}\alpha^2\cos\psi_0+\beta\sin\psi_0 &=\cos\psi_0\sin^2\psi_0 \end{align} $$ which can be solved to get $$ \alpha=-\sin\psi_0,\quad \beta={1\over2}\cos\psi_0\sin\psi_0={1\over4}\sin2\psi_0 $$ If desired, the same method can be used to find the expansion up to $e^3$ terms.