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?

I found:
$$cos^2(x)sin^4(x)=\frac{1}{32}(-cos(2x)-2cos(4x)+cos(6x)+x)$$
Verify the indenty:
$$32cos(x)^2sin(x)^4=2-cos(2x)-2cos(4x)+cos(6x)$$
$$6-2cos(2x)-4-4cos(4x)+2cos(4x)+cos(2x)+cos(6x)=2-cos(2x)-2cos(4x)+cos(6x)$$
$$\frac{1}{32}(-cos(2x)-2cos(4x)+cos(6x)+x)=\frac{1}{32}(-cos(2x)-2cos(4x)+cos(6x)+x)$$
The left hand side and right hand side are equal!
