-m=sin(a+b)sin(a-b)=(sinacosb+cosasinb)(sinacosb-cosasinb)
=(sina)^2(cosb)^2-(cosa)^2(sinb)^2
=(sina)^2(cosb)^2-[1-(sina)^2](sinb)^2
=(sina)^2[(cosb)^2+(sinb)^2]-(sinb)^2
=(sina)^2-(sinb)^2
=1-(cosa)^2-1+(cosb)^2=-[(cosa)^2-(cosb)^2]
所以：(cosa)^2-(cosb)^2=m
-sin(A+B)sin(A-B)=m
1/2 { cos[(A+B)+(A-B)] -cos[(A+B)-(A-B)]}=m
1/2{ cos(2A) -cos(2B)} = m
1/2{ (2cos^2A-1) -（2cos^2B-1)} = m
1/2{(2cos^2A-2cos^2B-2} = m
cos^2A-cos^2B-1 = m
cos^2A-cos^2B = m + 1
=(1/2)[cos(2A)-cos(2B)]
=(1/2)[2(cosA)^2-1-2(cosB)^2+1]
=(cosA)^2-(cosB)^2=m
∴(cosA)^2-(cosB)^2=m追问
化到这步=(1/2)[cos(2A)-cos(2B)]用的是啥公式啊
追答三角函数的积化和差公式
本回答被网友采纳若sin(A+B)sin(B-A)=m,则(cosA)^2-(cosB)^2=?
解答：-m=sin(a+b)sin(a-b)=(sinacosb+cosasinb)(sinacosb-cosasinb)=(sina)^2(cosb)^2-(cosa)^2(sinb)^2 =(sina)^2(cosb)^2-[1-(sina)^2](sinb)^2 =(sina)^2[(cosb)^2+(sinb)^2]-(sinb)^2 =(sina)^2-(sinb)^2 =1-(cosa)^2-1+(cosb)^2=-[(cosa)^2-(co...
已知sin(A+B)sin(B-A)=m,则cosA的平方-cosB的平方
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数学已知sin(A+B)sin(A-B)=M求cos的平方-cos的平方
=sin（A+B）sin（A-B)=(sinAcosB+sinBcosA)(sinAcosB-sinBcosA)=(sinAcosB)^2-(sinBcosA)^2………^2 表示"的平方"=[1-(cosA)^2](cosB)^2-[1-(cosB)^2](cosA)^2 =(cosB)^2-(cosAcosB)^2-(cosB)^2+(cosBcosA)^2 =(cosB)^2-(cosA)^2 =M ...
已知sin(a+b)sin(b-a)=m,则cos平方a-cos平方b等于? A -m B m C -4...
sin(a＋b)sin(b－a)=－(1\/2)[cos2b－cos2a]=m，则：cos2b－cos2a=－2m [2cos²b－1]－[2cos²a－1]=－2m cos²a－cos²b=m 【选B】
已知sin(α+β)sin(α-β)=m,求cos^2α-cos^2β
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sin(α+β)sin(α-β)=2m,(m不等于0),则cos^2α-cos^2β=?(过程)
sin（α+β）sin（α-β）=2m [sinacosb + cosasinb] [sinacosb - cosasinb] =2m sin^2a* cos^2b - sin^2b *cos^2a =2m (1- cos^2a)cos^2b - (1 - cos^b)* cos^2a =2m cos^2a - cos^2b = 2m
证sin(a+b)sin(a-b)=cos^2b-cos^2a
sin(a+b)sin(a-b)=（sinacosb+cosasinb)(sinacosb-cosasinb)=sin^2acos^2b-cos^2asin^2b =(1-cos^2a)cos^2b-cos^2a(1-cos^2b)=cos^2b-cos^2acos^2b-cos^2a+cos^2acos^2b =cos^2b-cos^2a
cos^2a-cos^2b=m,那么sin(a+b)sin(a-b)=?
sin(a+b)sin(a-b)=-[cos(a+b+a-b)-cos(a+b-a+b)]\/2 =(cos2b-cos2a)\/2 ={[2(cosb)^2-1]-[2(cosa)^2-1]}\/2 =(cosb)^2-(cosa)^2 =-m
证明:cos(a+b)cos(a-b)=(cosa)^2-(cosb)^2
cos(a+b)*cos(a-b)=(cosacosb-sinasinb )*(cosacosb+sinasinb)=(cosacosb)^2 -(sinasinb)^2 (cosa)^2 -(cosb)^2 =(cos2a - cos2b)\/2 =-sin(a+b)sin(a-b)原题有误吧
sin(a+b)cos(a-b)=sina*cosa+sinb*cosb
sin(A)*cos(B)= (1\/2)*[sin(A+B)+ sin(A-B)]二倍角公式 ：sin(2A)= 2*sinAcosA 再做题：sin(a+b)cos(a-b)= (1\/2)*{sin[(a+b)+(a-b)]+ sin[(a+b)-(a-b)]} = (1\/2)*[sin(2a)+ sin(2b)]= (1\/2)*[2*sin(a)*cos(a)+ 2*sin(b)*cos(b)]= sin(...
