求证sina+sinb=2sin(a+b)/2*cos(a-b)/2
因为:sinx*cosy=(1/2)[sin(x+y)+sin(x-y)] (1)
设a=x+y
b=x-y
则有:x=(a+b)/2
y=(a-b)/2
(1)式就变成:
sin[(a+b)/2]*cos[(a-b)/2]=(1/2)[sina+sinb]
就是:sina+sinb=2sin(a+b)/2*cos(a-b)/2