<s>[INST] 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.
 [/INST] As we know the RHS, let's  try to express $\cos^2(2\theta)$ in terms of $\cos(4\theta)$
$\begin{align}(\sin\theta-\cos2\theta)^2 & = \sin^2\theta - 2\sin\theta\cos2\theta + \cos^22\theta\\& = \sin^2\theta-2\sin\theta\cos2\theta + \frac{1 + \cos4\theta}2 \end{align}$
Now $\cos2\theta = 1 - 2\sin^2\theta$ we use this to get $4\sin^3\theta$
$\begin{align}(\sin\theta-\cos2\theta)^2 & = \sin^2\theta - 2\sin\theta + 4\sin^3\theta +\frac12+\frac{\cos4\theta}2 \\
&=\sin^2\theta +\sin\theta -( 3\sin\theta - 4\sin^3\theta) +\frac12+\frac{\cos4\theta}2 \\&=\sin\theta - \sin3\theta+\frac{\cos4\theta}2 + 1 - \frac12+\sin^2\theta \\&=\sin\theta-\sin3\theta+\frac{\cos4\theta}2 + 1 - \frac{1-2\sin^2\theta}{2} \\&=\boxed{1+\sin\theta-\sin3\theta+\frac{\cos4\theta}2  - \frac{\cos2\theta}{2} }\end{align}$
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