This question asks me to prove the validity of the trig identity by using the compound angle identities for $\sin(A+B)$ and $\sin(A-B)$
$$\sin P +\sin Q =2\sin\frac{P+Q}{2}\cos\frac{P-Q}{2}$$
I didn't get it at first so looked at the mark scheme and they set $$A+B=P$$ $$A-B=Q$$ and then got $$2\sin A\cos B=\sin(A+B)+\sin(A-B)$$ I get all the algebra, as in when they sub everything in to make it equal but I don't get why they suddenly got $A+B$ to equal $P$ and $Q$?
Also, if they got $\sin(A+B)+\sin(A-B)$ why would they not use the sin addition formula to actually prove it? How can they just say that $2\sin A\cos B=\sin(A+B)+\sin(A-B)$ and then sub $P$ and $Q$ back in to prove the question?
Best Answer
They did it because they could. $A$ and $B$ don't have any significance on their own. If you have any two numbers $x,y$ and can always find two other numbers where $x= A+B$ and $y= A-B$. (just let $A=\frac {x+y}2$ and $B=\frac {x-y}2$). This can be a useful manipultion to make as expression appear simpler.
But we could ignore it.
$P = \frac {P+Q}2 + \frac {P-Q}2$
And $Q = \frac {P+Q}2 -\frac {P-Q}2$.
So $\sin P = \sin (\frac {P+Q}2 + \frac {P-Q}2) = \sin \frac {P+Q}2\cos\frac {P-Q}2 + \cos \frac {P+Q}2\sin \frac {P-Q}2$.
(Interestingly that will true regardless of what $Q$ is. We could take it as a trig identity that $\sin \theta = \sin (\frac {\theta+x}2 + \frac {\theta-x}2) = \sin \frac {\theta+x}2\cos\frac {\theta-x}2 + \cos \frac {\theta+x}2\sin \frac {\theta-x}2$ for all $x$.)
And $\sin Q =\sin (\frac {P+Q}2 - \frac {P-Q}2) = \sin \frac {P+Q}2\cos\frac {P-Q}2 - \cos \frac {P+Q}2\sin \frac {P-Q}2$
And so adding them up you get.... $2 \sin \frac {P+Q}2\cos \frac {P-Q}2$.
....
The people who wrote the proof simply thought it'd be easier to write and to follow if we replace $P$ with $A+B$ and $Q$ with $A-B$ (by letting $A:= \frac {P+Q}2$ and $B:=\frac {P-Q}2$) and prove $\sin(A+B) - \sin(A-B) = 2\sin A\cos B$.
Was it easier to write and to follow? Well, I'll let you be the jury on that.
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tl;dr
I think what the writer of the proof wanted to express but failed to make the point clear was this:
Prove $\sin P + \sin Q = 2\sin \frac {P+Q}2 \cos \sin \frac{P-Q}2$.
Proof: Let's assign labels $A = \frac {P+Q}2$ and $B=\frac {P-Q}2$. If we make this substitution we see that $A+B = \frac {P+Q}2 + \frac {P-Q}2 = \frac {2P}2 = P$ and that $A-B = \frac {P+Q}2 -\frac {P-Q}2 = \frac {2Q}2 = Q$.
Using this substitution our statement to prove becomes simply
$\sin(A+B) + \sin (A-B) = 2\sin A \cos B$.
ANd that follows immediately from the sine addition rule $\sin(A\pm B)=\sin A\cos B \pm \cos A\sin B$
(so $\sin(A+B) + \sin (A-B)= $
$(\sin A\cos B + \cos A\sin B) + (\sin A\cos B -\cos A\sin B) =$
$ 2\sin A\cos B$.)
....
Had the proof been written in that wording would it have been clearer?