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).
 A: Put
$$a=u+v; \;\; b=u-v$$
then
$$u=\frac{a+b}{2};\;\;  v=\frac{a-b}{2}$$
$$\cos(a)-\cos(b)=$$
$$\cos(u+v)-\cos(u-v)=$$
$$\cos(u)\cos(v)-\sin(u)\sin(v)-$$
$$\cos(u)\cos(v)-\sin(u)\sin(v)=$$
$$-2\sin(u)\sin(v)=$$
$$-2\sin(\frac{a+b}{2})\sin(\frac{a-b}{2})$$
and you computed the opposite
$$\cos(u-v)-\cos(u+v)=$$
$$2\sin(u)\sin(v)$$
A: Indeed, we have
$$
\cos(a - b) - \cos(a + b) = 2 \sin(a) \sin(b).
$$
As your book indicates, we can define
$$
u = a - b, \quad v = a + b,
$$
which can be solved for $a$ and $b$ to yield
$$
a = \frac {u + v}2, \quad b = \frac {v - u}{2} = - \frac{u - v}{2}.
$$
The correct solution for $b$ has the negative sign that you were missing.
Now starting with the identity in terms of $a$ and $b$, substitute the first pair of equations into the left hand side, and substitute the second set of equations into the right hand side. We end up with
$$
\cos(u) - \cos(v) = 2 \sin\left(\frac{u + v}{2}\right)\sin\left( -\frac{u - v}{2}\right)
\\ = 
-2 \sin\left(\frac{u + v}{2}\right)\sin\left( \frac{u - v}{2}\right).
$$
A: From complex expressions of $\sin x$ and $\cos x$:
$\sin x =\frac{e^{ix}-e^{-ix}}{2i}$, $\cos x =\frac{e^{ix}+e^{-ix}}{2}$
$2\sin \frac{a+b}{2}\sin \frac{a-b}{2} =2(\frac{e^{^{i\frac{a+b}{2}}}-e^{-{i\frac{a+b}{2}}}}{2i})(\frac{e^{{i\frac{a-b}{2}}}-e^{{-i\frac{a-b}{2}}}}{2i})$
$2\sin \frac{a+b}{2}\sin \frac{a-b}{2} =2\frac{(e^{ia}+e^{-ia})-(e^{ib}+e^{-ib})}{4i^2}=-(\cos a-\cos b)$=$\cos b-\cos a$
A: 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?
