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.
 A: Method I. Write everything in terms of $\sin(\theta)$ and $\cos(\theta)$, and then compare. Tedious, but doable.
Method II. If you know what a Fourier series is, you may determine the coefficients on the right-hand side by integration.
Method III.
First expand the left by $(a+b)^2=a^2+2ab+b^2$.
\begin{align}
(\sin x-\cos(2x))^2
&=\sin^2 x+\cos^2(2x)-2\sin x\cos(2x)
\end{align}
Then use the double angle formulas:
$$
\sin^2x=\frac{1-\cos 2x}{2},\quad \cos^2(2x)=\frac{1+\cos(4x)}{2}
$$
and the general product to sum formula:
$$
\sin x\cos(2x)=\frac12(-\sin(x)+\sin(3x))
$$
A: 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 $4\sin^3\theta$
$\begin{align}(\sin\theta-\cos2\theta)^2 & = \sin^2\theta - 2\sin\theta + 4\sin^3\theta +\frac12+\frac{\cos4\theta}2 \\
&=\sin^2\theta +\sin\theta -( 3\sin\theta - 4\sin^3\theta) +\frac12+\frac{\cos4\theta}2 \\&=\sin\theta - \sin3\theta+\frac{\cos4\theta}2 + 1 - \frac12+\sin^2\theta \\&=\sin\theta-\sin3\theta+\frac{\cos4\theta}2 + 1 - \frac{1-2\sin^2\theta}{2} \\&=\boxed{1+\sin\theta-\sin3\theta+\frac{\cos4\theta}2  - \frac{\cos2\theta}{2} }\end{align}$
A: 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,
$$
(\sin(\theta) - \cos(2\theta))^2 =  \dfrac{1-\cos(2\theta)}{2} +  \dfrac{1+\cos(4\theta)}{2} + \frac1{2} \sin(3\theta) - \frac1{2} \sin(\theta)
$$
A: 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}
