Problem in Derivation of Goldstein 
In Goldstein, chapter three, third derivation, given as, Kepler's equation can be written as
${\rho} = e\sin({\omega}t + {\rho})$,
Now I have to prove that the first approximation to ${\rho}$ is ${\rho_1}$ given by
$$\tan({\rho_1}) = \frac{e\sin({\omega}t)}{1-e\cos({\omega}t)} $$
and also
$$\sin({\rho_2} - {\rho_1}) = -e^{3} \sin({\omega}t + {\rho_1})[1+e\cos({\omega}t)].$$

Now, my approach to this question as below:
$${\rho} = e\sin({\omega}t)\cos({\rho}) + e\cos({\omega}t)\sin({\rho})$$
$${\rho} = e\sin({\omega}t)\cos({\rho}) + e\cos({\omega}t)({\rho} - {\rho^3}/6 + {\rho^5}/120 + \ldots)$$
$${\rho} = e\sin({\omega}t)\cos({\rho}) + e\cos({\omega}t)({\rho}) - e\cos({\omega}t)({\rho^3/6}) + e\cos({\omega}t)({\rho^5/120})$$
$${\rho}[1-e\cos({\omega}t)] = e\sin({\omega}t)\cos({\rho})- e\cos({\omega}t)({\rho^3/6}) + e\cos({\omega}t)({\rho^5/120})$$
$${\rho} = \frac{e\sin({\omega}t)\cos({\rho})}{1-e\cos({\omega}t)}- \frac{e\cos({\omega}t)({\rho^3/6})}{1-e\cos({\omega}t)} + \frac{e\cos({\omega}t)({\rho^5/120})}{1-e\cos({\omega}t)}$$
I don't know how to proceed further. I am not getting where $\tan({\rho_1})$ should be coming from?  Any hint would be helpful.
 A: In general there are many ways to numerically approximate a transcendental solution. In Goldstein, some recommended approximation methods seem strange, but the first time you solve a problem, it's best if you first stick to Goldstein's recommendations.

Successive approximations to $\rho$ can be found by expanding $\sin \rho$ in its Taylor series and replacing $\rho$ its expression (the Taylor series) given by Kepler's equation

This method of approximation does indeed seem odd, but you should probably go through with it. Write:
$$
\begin{align*}
\sin \rho_1 &\approx \rho_1 \\
&= e\sin(\omega t + \rho_1) \\
&= e\sin(\omega t) \cos\rho_1 + e \cos(\omega t) \sin\rho_1 \\
\Longrightarrow \sin\rho_1 (1 - e \cos(\omega t)) &= e \sin(\omega t) \cos\rho_1 \\
\Longrightarrow \tan\rho_1 &= \frac{e \sin (\omega t)}{1 - e \cos(\omega t)}
\end{align*}
$$
Once you solve it this way, we can try your approximation scheme as well, which is an excellent one. When you expand $\sin \rho_1$, via a polynomial expansion, you should also expand $\cos \rho_1$. In general, when you approximation a function applied at a small input value, do it for all occurrences of this small variable. Therefore, you should modify it to:
$$
{\rho} = e\sin({\omega}t)(1 - \frac{\rho^2}{2} + \frac{\rho^4}{24} - \cdots) + e\cos({\omega}t)({\rho} - \frac{\rho^3}{6} + \frac{\rho^5}{120} + \cdots)
$$
When you take the first order approximation here for $ \rho_1 \ll 0.1$, you should drop all terms of order $\mathcal{O}(\rho^2)$ and higher, and find:
$$
\begin{align*}
\rho_1 &\approx e \sin(\omega t) + e \cos(\omega t) \rho_1 \\
\Longrightarrow \rho_1 &= \frac{e \sin(\omega t)}{1 - e \cos (\omega t)}
\end{align*}
$$
For very small values of $\rho_1$, we have that $\tan \rho_1 \approx \rho_1$, so we get something that looks along the lines of the desired result.
For the second part of the question, there is a major typo in the book. Please consult Goldstein Errata. You should get:
$$
\begin{align*}
( \sin(\rho_2 - \rho_1)) &\approx ( \sin\rho_2 - \sin\rho_1) \\
&= -\frac{1}{6}e^3 \sin^3 (\omega t + \rho_1)
\end{align*}
$$
There are two ways of arriving at a decent second-order approximation. As before, you can expand using Goldstein's approximation method. Observe that because $\sin \rho $ is a concave down in the positive reals near zero, our previous approximation is an overestimate. When $\rho < 0$, the function $\sin \rho$ is concave up near zero, and our approximation is now an underestimate. We can try and compensate for this by subtracting the cubic term:
$$
\begin{align*}
\sin \rho_2 &\approx \rho_1 - \frac{\rho_1^3}{6} \\ 
&= e \sin ( \omega t + \rho_1) - \frac{e^3}{6} \sin^3 ( \omega t + \rho_1) \\
\Longrightarrow \sin \rho_2 - e \sin(\omega t + \rho_1) &= - \frac{e^3}{6} \sin^3 ( \omega t + \rho_1) \\
\Longrightarrow \sin \rho_2 - e \sin(\omega t) \cos\rho_1 - e \cos(\omega t) \sin \rho_1 &=- \frac{e^3}{6} \sin^3 ( \omega t + \rho_1) 
\end{align*}
$$
Then, from the earlier equation:
$$
\tan\rho_1 = \frac{e \sin (\omega t)}{1 - e \cos(\omega t)} \Longrightarrow
e \sin(\omega t) \cos \rho_1 = \sin \rho_1 - e  \cos (\omega t) \sin \rho_1 \\
$$
Substitute this result back into the previous equation to find:
$$
( \sin\rho_2 - \sin\rho_1) = -\frac{1}{6}e^3 \sin^3 (\omega t + \rho_1)
$$
Finally, we can once again try your method as well, dropping all terms $\mathcal{O}(\rho^3)$ and higher:
$$
\begin{align*}
{\rho} &= e\sin({\omega}t)(1 - \frac{\rho^2}{2} + \frac{\rho^4}{24} - \cdots) + e\cos({\omega}t)({\rho} - \frac{\rho^3}{6} + \frac{\rho^5}{120} + \cdots) \\
&\approx e \sin(\omega t)(1 - \frac{\rho^2}{2}) + e \cos(\omega t )\rho
\end{align*}
$$
We move all terms onto one side of the equation and normalize $\rho^2$ to find:
$$
0 = \rho^2 + (\frac{2}{e} \csc(\omega t ) - 2 \cot (\omega t)) \rho - 2
$$
which can be solved for $\rho$ using the quadratic formula.
$$
\rho = { \cot (\omega t) - \frac{\csc (\omega t)}{e} \pm \sqrt{\frac{ \csc^2(\omega t)}{e^2} - \frac{2 \cot(\omega t) \csc(\omega t)}{e} +  \cot^2(\omega t) + 2}}
$$
Yikes!
A: I'm writing this in connection with najkim's answer above, just because I noticed something when some terms were simplified using the first identity
For the 2nd part, when najkim gets to the part
$\sin{\rho_2} - e \sin{\omega t}\cos{\rho_1} - e\cos{wt}\sin{\rho_1}=-\frac{1}{6}e^3 \sin^3{(\omega t + \rho_1)}$
The left hand side of this equation can be further simplified to
$\sin{\rho_2} - e \sin{(\omega t + \rho_1)}$
And from the first identity, $\tan{\rho_1} = \frac{e \sin{\omega t}}{1-e \cos{\omega t}}$, we get
$e \sin{\omega t}\cos{\rho_1} = \sin{\rho_1} -e \cos{\omega t}\sin{\rho_1} $
which leads to
$ \sin{\rho_1} = e\sin{(\omega t + \rho_1)}$
So shouldn't the final expression be
$\sin{\rho_2}-\sin{\rho_1}=-\frac{1}{6}e^3 \sin^3{(\omega t + \rho_1)}?$
