# Prove that sin x + sin y = 2*sin((x+y)/2)*cos((x-y)/2).

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We have to prove: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 + 2*(sin(y/2))^2*sin(x/2)cos(x/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(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)*1 + 2*sin(y/2)*cos(y/2)*1

=> sin x + sin y

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