Product-to-sum formulas? My old pre-calculus book says:
$$\sin u\cos v=\frac{1}{2}[\sin (u+v)+\sin(u-v)]$$
and $$\cos u \sin v=\frac{1}{2}[\sin(u+v)-\sin(u-v)]$$ I don't understand why there is a difference, since multiplication is commutative. Can anyone help? Thanks!
 A: Sine is an odd function, be careful with $u$ and $v$:
$$\sin \color{red}{u}\cos \color{blue}{v}=\frac{1}{2}[\sin (\color{red}{u}+ \color{blue}{v})+\sin(\color{red}{u}- \color{blue}{v})] = \frac{1}{2}[\sin ( \color{blue}{v}+\color{red}{u})-\sin( \color{blue}{v}-\color{red}{u})] = \cos  \color{blue}{v} \sin \color{red}{u}.$$
The right side is $  \cos\color{blue}{v} \sin \color{red}{u}$, not  $ \cos\color{red}{u} \sin \color{blue}{v} $.
A: Multiplication is certainly commutative, but that is not what is happening here. You are looking at fixed angles $u, v$; it does NOT follow from commutativity that $\sin u \cos v = \cos u \sin v$. It is true by commutativity that $\sin u \cos v = \cos v \sin u$.
A: Notice the following, in the first one you have $\sin$ of $u$ and $\cos$ of $v$ and in the second you have $\sin$ of $v$ and $\cos $ of $u$. I'll show what happens. Let's use the first formula and derive the second. On the right hand side of the first formula, change every $u$ to $v$ and every $v$ to $u$, we have:
$$\sin v\cos u= \frac{\sin(v+u) + \sin(v - u)}{2}$$
Now, $v-u = -(u-v)$ so that $\sin(v-u)=\sin(-(u-v))$ and we all know that this is the same as writing $\sin(v-u) = -\sin(u-v)$. In that case we get the second formula:
$$\cos u\sin v = \frac{\sin(u+v)-\sin(u-v)}{2}$$
A: Note simply that $\sin (u+v) = \sin (v+u)$ and $\sin (u-v)=-\sin (v-u)$. 
The role of $u$ and $v$ in the two equations you quote has been swapped on the left hand side, but not on the right hand side - hence the change of sign. If the order is swapped on the right hand side of the second equation, so is the sign.
A: In the first one you have $\;\cos\color{red}u\;$ , and in the second one you have $\;\cos\color{blue}v\;$ , and likewise for the sines...
A: $$\sin(u+v)+\sin(u-v)=2\sin u \cos v$$
$$\sin(u+v)-\sin(u-v)=\sin u \cos v + \cos u \sin v - \sin u \cos v +\cos u \sin v=2\cos u \sin v$$
Simply expand and check. 
$u\neq v$. Hence, $\cos u \neq \cos v$. Similarly, $\sin u \neq \sin v$. 
