Lowering powers of $\cos^2x \sin^4x$ First, I will be straight up, this is a homework question.
I need to write $\cos^2 x \sin^4 x$ in terms of cosine to the first power. I know that $\sin^4x$ = 
$$ \frac{3-4\cos 2x+\cos 4x}{8}$$
from there I go:
$$ \frac{1+\cos 2x}{2} \cdot \frac{3-4\cos 2x+\cos 4x}{8}$$
$$ \frac{3(1 + \cos 2x) - 4\cos 2x(1 + \cos 2x) + \cos 4x (1 + \cos 2x)} {16} $$
$$ \frac{3 + 3\cos 2x - 4\cos 2x \color{red}{+} 4 \cos^2 2x + \cos 4x + \cos 4x \cos 2x}{16}$$
$$ \frac{3 - \cos 2x + \cos 4x + 4 \frac{1 + \cos 4x}{2} \cos 4x  \cos 2x}{16}$$
Then $$ 4 \frac{1 + \cos 4x}{2}$$ simplifies to $$ 2 + 2\cos 4x$$
adding this into the rest of the fraction gives me
$$ \frac{5 - \cos 2x + 3\cos 4x + \cos 4x \cos 2x}{16}$$
However, the answer sheet that I have says the answer is 
$$ \frac{1 - \cos 2x - \cos 4x + \cos 2x \cos 4x }{16}$$
This is problem $13$ from section $7.3$ of Stewart, Redlin and Watson precalculus $6$th edition
Where is my mistake?

There is a systematic approach to solve such questions, based on complex numbers.
We have $\cos kx=\dfrac{e^{ikx}+e^{-ikx}}2=\dfrac{z^k+z^{-k}}2$ and $\sin kx=\dfrac{e^{ikx}-e^{-ikx}}{2i}=\dfrac{z^k-z^{-k}}{2i}$, where $z=e^{ix}$.
The expression can be rewritten
$$\left(\frac{z+z^{-1}}{2}\right)^2\left(\frac{z-z^{-1}}{2i}\right)^4=\frac{z^6+z^{-6}-2z^4-2z^{-4}-z^2-z^{-2}+4}{64}=\frac{\cos6x-2\cos4x-\cos2x+2}{32}.$$
