sinA+sinB=2sin((A+B)/2)cos((A-B)/2的推导过程是怎样的?

由两角和与差的正 弦公式得：sin(a+b)=sinacosb+cosasinb,sin(a-b)=sinacosb-cosasinb,两式相加得：sin(a+b)+sin(a-b)=2sinacosb,设a+b=A,a-b=B,解得：a=(A+B)/2，b=（A-B)/2，代入上式即得：sinA+sinB=2sin[(A+B)/2]cos[(A-B)/2].

把sinA写成sin[(A+B)+(A-B)]/2;把sinB写成sin[(A+B)+(B-A)]/2，再利用sin(x+y)=sinXcosY+cosXsinY展开上面二式，又因为cos(B-A)/2=cos(A-B)/2,sin(B-A)/2=-sin(A-B)/2.相加即得。

令a=x+y b=x-y 则x=(a+b)/2,y=(a-b)/2 sina+sinb=sin(x+y)+sin(x-y) =sinxcosy+sinycosx+sinxcosy-sinycosx =2sinxcosy 将 x=(a+b)/2,y=(a-b)/2代入 得sina+sinb=2[sin(a+b)/2][cos(a-b)/2]