We have to prove the identity sin x - sin y = 2*sin((x - y)/2)*cos((x + y)/2)

Start with 2*sin((x - y)/2)*cos((x + y)/2), use the rules sin(A + B) = sin A*cos B + cos A*sin B and cos(A - B) = cos A*cos B - sin A*sin B

2*sin((x - y)/2)*cos((x + y)/2)

=> 2*[sin(x/2)*cos(y/2) - cos(x/2)sin(y/2)]*[cos(x/2)cos(y/2) - sin(x/2)sin(y/2)]

=> 2*[(sin(x/2)*cos(y/2))(cos(x/2)cos(y/2)) - (cos(x/2)sin(y/2))(cos(x/2)cos(y/2)) - (sin(x/2)*cos(y/2))(sin(x/2)sin(y/2)) + (cos(x/2)sin(y/2))(sin(x/2)sin(y/2))]

=> 2*sin(x/2)(cos(x/2)(cos(y/2)^2 - 2*sin(y/2)*cos(y/2)*(cos(x/2))^2 - 2*(sin(x/2)^2*sin(y/2)*cos(y/2) + 2*(sin(y/2))^2*sin(x/2)cos(x/2)

=> 2*sin(x/2)(cos(x/2)((cos(y/2)^2+(sin(y/2))^2) - 2*sin(y/2)*cos(y/2)*((cos(x/2))^2 +(sin(x/2)^2)

=> 2*sin(x/2)(cos(x/2) - 2*sin(y/2)*cos(y/2)

=> sin x - sin y

**This proves that sin x - sin y = 2*sin((x - y)/2)*cos((x + y)/2)**

Substitute (x-y)/2 by p and (x+y)/2 by q.

(x-y)/2=p => x-y=2p (a)

(x+y)/2=q=> x+y=2q (b)

If you will add (a) and (b), the variable y will be eliminated :

2x=2p+2q

You can divide the equation by 2 => x = p+q

If you will subtract (a) from (b), the variable x will be eliminated :

2y=2q-2p => y=q-p

Substitute x and y by p+q and q-p when calculate the difference sin x - sin y.

Use formula sin a-sin b=2sin((a-b)/2)*cos((a+b)/2)

sin x-sin y=sin (p+q)-sin(q-p)=2sin((p+q-q+p)/2)*cos((p+q+q-p)/2)

sin (p+q)-sin(q-p)=2sin((2p)/2)*cos((2q)/2)

sin (p+q)-sin(q-p)=2sin(p)*cos(q)

Put back again what x and y.

sin (p+q)-sin(q-p)=sin x-sin y=2sin(p)*cos(q)=2sin((x-y)/2)*cos((x+y)/2)

**Answer:This equality proves the identity sin x-sin y=2sin(p)*cos(q)=2sin((x-y)/2)*cos((x+y)/2), when (x-y)/2 by p and (x+y)/2 by q.**