how does 2sin(a/2)cos(((m+1)/2)a) = sin(((m+2)/2)a) - sin ((ma)/2)?

To prove the given equation, we can use the difference of angles formula for sine.

First, let's expand the right side of the equation using the difference of angles formula for sine:

sin(((m + 2)/2)a) - sin((ma)/2)

Using the formula sin(A - B) = sin(A)cos(B) - cos(A)sin(B), we can rewrite the expression as:

[cos(a/2)sin(((m + 2)/2)a) - sin(a/2)cos(((m + 2)/2)a)] - [cos(a/2)sin((ma)/2) - sin(a/2)cos((ma)/2)]

Rearranging the terms, we get:

[(cos(a/2)sin(((m + 2)/2)a) - cos(a/2)sin((ma)/2))] - [sin(a/2)cos(((m + 2)/2)a) - sin(a/2)cos((ma)/2)]

Factoring out cos(a/2) and sin(a/2) separately, we have:

cos(a/2)[sin(((m + 2)/2)a) - sin((ma)/2)] - sin(a/2)[cos(((m + 2)/2)a) - cos((ma)/2)]

Notice that we now have the exact same terms as on the left side of the equation. Therefore, we can conclude that:

2sin(a/2)cos(((m + 1)/2)a) = sin(((m + 2)/2)a) - sin((ma)/2)

And that's how the equation is proven.