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.

Apply the identities
$$\cos4\theta = 2\cos^22\theta -1,\>\>\>\>\>\cos2\theta = 1-2\sin^2\theta $$
$$\sin3\theta =\sin\theta (2\cos2\theta+1)$$
in simplifying the RHS
\begin{align}
RHS=&1+\sin\theta -\sin\theta (2\cos2\theta+1)
-\frac12(1-2\sin^2\theta)+\frac12(2\cos^22\theta-1)\\
=&\sin^2\theta -2\sin\theta\cos2\theta+\cos^22\theta\\
=&(\sin\theta -\cos2\theta)^2=LHS
\end{align}
