Question

Prove the identity sin x - sin y = 2 sin( (x - y)/2 ) cos( (x + y)/2 )?

Accepted Answer

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.

Answer

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)

Math

Prove the identity sin x - sin y = 2 sin( (x - y)/2 ) cos( (x + y)/2 )?

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