求证：1/2(cos2B-cos2A)=sin(A+B)sin(A-B)

题目

求证：1/2(cos2B-cos2A)=sin(A+B)sin(A-B)
(一定要从1/2(cos2B-cos2A）推到sin(A+B)sin(A-B)
不可以像这样：
/2（cos2B-cos2A)=1/2[2(cosB)^2-1-2(cosA)^2+1]=(cosB)^2-(cosA)^2
sin(A+B)sin(A-B)=(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)
=(sinAcosB)^2-(cosAsinB)^2=(cosB)^2[1-(cosA)^2]-(cosA)^2[1-(cosB)^2]
=(cosB)^2-(cosBcosA)^2-(cosA)^2+(cosAcosB)^2
=(cosB)^2-(cosA)^2
综上,化简后 1/2（cos2B-cos2A)=sin(A+B)sin(A-B) ）

答案

左边=1/2{cos[(A+B)-(A-B)]-cos[(A+B)+(A-B)]}=1/2{[cos(A+B)cos(A-B)+sin(A+B)sin(A-B)]-[cos(A+B)cos(A-B)-sin(A+B)sin(A-B)]}=1/2[cos(A+B)cos(A-B)+sin(A+B)sin(A-B)-cos(A+B)cos(A-B)+sin(A+B)sin(A-B)]=1/2[2s...