sin(2x)=sin(x+x) =sin(x)cos(x)+cos(x)sin(x) =2*sin(x)*cos(x); cos(2x)=cos(x+x) =cos(x)cos(x)-sin(x)sin(x) =[cos(x)]2-[sin(x)]2 Now let x+y=a and x-y=b, then x=(a+b)/2 and y=(a-b)/2. Then sin(a)=sin[(a+b)/2+(a-b)/2] =sin([a+b)/2]*cos[(a-b)/2]+sin[(a-b)/2]*cos[(a+b)/2]; sin(b)=sin[(a+b)/2+(b-a)/2] =sin[(a+b)/2]*cos[(b-a)/2]+sin[(b-a)/2]*cos[(a+b)/2]; sin(x) is an odd function: sin(x)=-sin(-x) and cos(x) is an even function: cos(x)=cos(-x).Hence, sin(b)=sin([a+b)/2]*cos[(a-b)/2]-sin[(a-b)/2]cos[(a+b)/2]. Adding these two formulas gives: sin(a)+sin(b)=2sin[(a+b)/2]*cos[(a-b)/2]