I'm failing to understand the derivation of the equation of the centre given on page 11 of Determining planetary positions in the sky for ±50 years to an accuracy of $\lesssim1$ degree with a calculator by Singal and Singal. In what follows $e$ is orbital eccentricity, which for our Solar System planets is small. Starting with$$\Delta\theta=2e\sin\left(\theta_{0}+\Delta\theta\right)-\frac{3}{4}e^{2}\sin\left(2\left(\theta_{0}+\Delta\theta\right)\right)+\ldots$$they go on to say:
Expanding in powers of $\Delta\theta$ and neglecting terms of order $e\left(\Delta\theta\right)^{2}$ and $e^{2}\Delta\theta$ and higher we get$$\Delta\theta\left(1-2e\cos\theta_{0}\right)=\left(2e\sin\theta_{0}-\frac{3}{4}e^{2}\sin2\theta_{0}\right).$$
How do they do that? I've tried taking a Taylor series expansion of the first equation, spent ages fiddling with the results but end up with nothing like the second equation.
If I accept the second equation I am able to go on to find the correct equation of the centre, ie$$\Delta\theta=2e\sin\theta_{0}+\frac{5}{4}e^{2}\sin2\theta_{0}.$$
Got it! I've only just realised $\sin x\cos x=\frac{1}{2}\sin2x$. A Taylor series expansion at $\Delta\theta=0$ gives $$\Delta\theta=-\frac{1}{2}e\sin\theta_{0}\left(3e\cos\theta_{0}-4\right)+2e\Delta\theta\cos\theta_{0}$$ $$=-\frac{3}{2}e^{2}\sin\theta_{0}\cos\theta_{0}+2e\sin\theta_{0}+2e\Delta\theta\cos\theta_{0}$$ $$\Delta\theta-2e\Delta\theta\cos\theta_{0}=-\frac{3}{4}e^{2}\sin2x+2e\sin\theta_{0}$$ $$\Delta\theta\left(1-2e\cos\theta_{0}\right)=2e\sin\theta_{0}-\frac{3}{4}e^{2}\sin2\theta_{0}.$$