为什么1/2(sin2A+sin2B)=sin(A+B)cos(A-B)
问题描述：
为什么1/2(sin2A+sin2B)=sin(A+B)cos(A-B)
答
1/2(sin2A+sin2B)=1/2（sin(A+B+A-B)+sin((A+B)-(A-B)))=1/2(sin(A+B)cos(A-B)+sin(A-B)cos(A+B)+sin(A+B)cos(A-B)-sin(A-B)cos(A+B)）=1/2*2sin(A+B)cos(A-B)=sin(A+B)cos(A-B)
答
左边展开为1/2(2sinAcosA+2sinBcosB)=sinAcosA+sinBcosB
右边展开为(sinAcosB+cosAsinB)(cosAcosB+sinAsinB)=sinAcosA(cosB)^2+sinAcosA(sinB)^2+sinBcosB(cosA)^2+sinBcosB(sinA)^2=sinAcosA((cosB)^2+(sinB)^2)+sinBcosB((sinA)^2+(cosA)^2)=sinAcosA+sinBcosB
两式相等