Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) - \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $ 
Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) -  \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $

If I would start from the right expression I would be able to reduce it. But I'm not used to do the reverse.
I can see that Wolframalpha expands the left expression in many ways, one of which is the one I want, but are there any trick to do this expansion by hand?
This question is linked to this one.

We'll use the following identities:
$\cos^2(\theta) = \dfrac{1+\cos(2\theta)}{2}$
$\sin^2{\theta} = \dfrac{1-\cos(2\theta)}{2}$
$\sin \theta = \dfrac{e^{i\theta} - e^{-i\theta}}{2i}$
$\cos \theta = \dfrac{e^{i\theta} + e^{-i\theta}}{2}$
$$
(\sin(\theta) - \cos(2\theta))^2 = \sin^2(\theta) + \cos^2(2\theta) - 2 \sin(\theta) \cdot \cos(2\theta)
$$
$$
\sin(\theta) \cdot \cos(2\theta) = \dfrac{e^{i\theta} - e^{-i\theta}}{2i} \cdot \dfrac{e^{2i\theta} + e^{-2i\theta}}{2} \\
= \dfrac{e^{3i\theta} + e^{-i\theta} - e^{i\theta} - e^{-3i\theta}}{4i} \\
= \dfrac{e^{3i\theta} - e^{-3i\theta}}{4i} - \dfrac{e^{i\theta} - e^{-i\theta}}{4i} \\
= \frac1{2} \sin(3\theta) - \frac1{2} \sin(\theta)
$$
As such,
$$
(\sin(\theta) - \cos(2\theta))^2 =  \dfrac{1-\cos(2\theta)}{2} +  \dfrac{1+\cos(4\theta)}{2} + \frac1{2} \sin(3\theta) - \frac1{2} \sin(\theta)
$$
