How to derive the sum-to-product formula for cos(a-b)-cos(a+b)? I know how to derive the sum-to-product formulas for cos+cos, sin+cos, and cos+sin. But I cannot seem to do it for cos-cos. My problem is the negative out front. I cannot product it from any of my proofs.
Here is my attempt:
I know that cos(a-b) = cosacosb + sinasinb
I know that cos(a+b) = cosacosb - sinasinb.
If I subtract these, I get cos(a-b)-cos(a+b) = 2sinasinb
However, my book tells me that cosa - cosb = -2sin(a+b / 2)sin(a-b /2)...
Where does the negative come from?
additional effort
My book also says I can make a + b = u and a - b = v. By adding and subtracting these from one another, this gets me u+v / 2 and u-v / 2. But none of those come with a negative that I can use to make sin(-x) = -sin(x).

Develop the right side:
$$-2\sin\frac{a+b}2\,\sin\frac{a-b}2=-2\left[\sin\frac a2\cos\frac b2+\sin\frac b2\cos\frac a2\right]\left[\sin\frac a2\cos\frac b2-\sin\frac b2\cos\frac a2\right]=$$
$$\stackrel{\text{squares difference}}=-2\left[\sin^2\frac a2\,\cos^2\frac b2-\sin^2\frac b2\,\cos^2\frac a2\right]=$$
$$=-2\left[\left(1-\cos^2\frac a2\right)\cos^2\frac b2-\left(1-\cos^2\frac b2\right)\cos^2\frac a2\right]=$$
$$=-2\left[\cos^2\frac b2-\cos^2\frac a2\right]=-2\left[\cos^2\frac b2-\sin^2\frac b2+\sin^2\frac b2-\cos^2\frac a2+\sin^2\frac a2-\sin^2\frac a2\right]=$$
$$=-2\left[\cos b-\cos a+\sin^2\frac b2-\sin^2\frac a2\right]=-2\left[\cos b-\cos a-\left(\frac12-\sin^2\frac b2\right)+\frac12-\sin^2\frac a2\right]=$$
$$=-2\left[\cos b-\cos a-\frac12\cos b+\frac12\cos a\right]=\cos a-\cos b$$
We used several times in the above the super-identity
$$\cos2x=\begin{cases}\cos^2x-\sin^2 x\\{}\\2\cos^2 x-1\\{}\\1-2\sin^2x\end{cases}$$
Can you spot the places where we used this?
