What is the mistake in solving this trigonometry simplification problem?

[fawk3s]

Homework Statement

[2sin3y + (sin3x-siny)(cos4y-cos2y) : cos(x-y)-cos(x+y)] : sin^2x

Homework Equations

sinx-siny=2 * cos x+y/2 * sin x-y/2
cosx-cosy=2 * sin x+y/2 * sin x-y/2
cos(x-y)=cosxcosy+sinxsiny
cos(x+y)=cosxcosy-sinxsiny
cos2x=cos^2x-sin^2x

The Attempt at a Solution

This part

(sin3x-siny)(cos4y-cos2y) : cos(x-y)-cos(x+y)

I get to be 2sin3ycos2x, and therefore

2sin3y + (sin3x-siny)(cos4y-cos2y) : cos(x-y)-cos(x+y)=2sin3y+2sin3ycos2x=2sin3y(1+cos2x)=2sin3y(1+cos^2x-sin^2x)=2sin3y*2cos^2x=4sin3ycos^2x

Though I should get 4sin3ysin^2x, and the final answer to be 4sin3y.
I think this part

2sin3y(1+cos2x)

actually ought to be

2sin3y(1-cos2x)

but why? I don't get the minus from anywhere.
Where do I make the mistake?

Thanks in advance,
fawk3s

 

[eumyang]

I can't make sense at what you wrote. Not enough parentheses. And what does the ":" mean?

However...

Homework Equations

sinx-siny=2 * cos x+y/2 * sin x-y/2
cosx-cosy=2 * sin x+y/2 * sin x-y/2
cos(x-y)=cosxcosy+sinxsiny
cos(x+y)=cosxcosy-sinxsiny
cos2x=cos^2x-sin^2x

The bolded is wrong. It should be
$$\cos x - \cos y = -2 \sin \left( \frac{x + y}{2} \right) \sin \left( \frac{x - y}{2} \right)$$

 

[fawk3s]

Oh, thank you !