sin(a+b)sin(a-b)=-1/2[cos(a+b+a-b)-cos(a+b-a-b)]
=-1/2(cos2a-cos2b)
=-1/2[(2cos^2a-1)-(2cos^2b-1)]
=-1/2(2cos^2a-2cos^2b)
=-1/2[2(cos^2a-cos^2b)]
=-(cos^2a-cos^2b)=-1/3
∴(cos^2a-cos^2b)=1/3
=(sinacosb+sinbcosa)(sinacosb-sinbcosa)
=(sinacosb)^2-(sinbcosa)^2
=(cosb)^2(1-(cosa)^2)-(cosa)^2(1-(cosb)^2)
=(cosb)^2-(cosa)^2
=-1/3
(cosa)^2-(cosb)^2=1/3
已知sin(a+b)*sin(a-b)=m,求cos^2a-cos^2b的值
根据积化和差公式：sinαsinβ=-1\/2[cos(α+β)-cos(α-β)]sin(a+b)sin(a-b)=-1\/2[cos(a+b+a-b)-cos(a+b-a-b)]=-1\/2(cos2a-cos2b)=-1\/2[(2cos^2a-1)-(2cos^2b-1)]=-1\/2(2cos^2a-2cos^2b)=-1\/2[2(cos^2a-cos^2b)]=-(cos^2a-cos^2b)所以cos^2a-...
...sin(a+b)sin(a-b)怎么化成 负二分之一(cos2a-cos2b) 给个详细的步...
=-2sin(a+b)sin(a-b)所以 sin(a+b)sin(a-b)=-1\/2(cos2a-cos2b)
sin(a+b)sin(a-b)怎么化成 负二分之一(cos2a-cos2b)
=-2sin(a+b)sin(a-b)所以 sin(a+b)sin(a-b)=-1\/2(cos2a-cos2b)
已知sin(a+b).sin(a-b)=-1\/3,求sin²a-sin²b的值
sin²a-sin²b =(1-cos²a)-(1-cos²b)=cos²b-cos²a 望采纳
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(α+β)sin(α-β)=-1\/3
sin(α+β)sin(α-β)=(sinαcosβ+cosαsinβ)(sinαcosβ-cosαsinβ)=(sinαcosβ)^2-(cosαsinβ)^2 =(sinαcosβ)^2+(sinαsinβ)^2-(sinαsinβ)^2-(cosαsinβ)^2 =sin^2α*(cos^2β+sin^2β)-sin^2β*(sin^2α+cos^2α)=sin^2α-sin^2β=-1\/3 所以 ...
已知sin(α+β)sin(α-β)=1\/3.(1)求cos2β-cos2α的值;(2)求(1\/4...
(1)因为sin(α+β)sin(α-β)=1\/3 而cos2β-cos2α =-2*sin(α+β)sin(β-α)=2sin(α+β)sin(α-β)=2\/3 (2)(1\/4)(sin2α)^2+(sinβ)^2+(cosα)^4 =1\/4*(1-cos^2(2α))+(1-cos2β)\/2+[(1+cos(2α))\/2]^2 =1\/4+(1-cos2β)\/2+1\/4+cos(2α...
sin(a+ b) sin(a- b)=什么公式?
sin(a+b)sin(a-b)=[cos(a+b-a+b)-cos(a+b+a-b)]\/2 =[cos(2b)-cos(2a)]\/2。又由二倍角公式 [cos(2b)-cos(2a)]\/2=[1-2sin^2b-（1-2sin^2a）]\/2=[2sin^2a-2sin^2b）]\/2=sin^2 a-sin^2b。诱导公式 诱导公式是指三角函数中，利用周期性将角度比较大的三角函数，...
已知sin(A+B)·sin(B-A)=m
sin(A+B)·sin(B-A)直接积化和差sin(A+B)·sin(B-A)=-1\/2(cos2B-cos2A)=m cos2A-cos2B=-2m
已知cos(a+B)cos(a-B)=1\/3,那么cos^2a-sin^2B等于
即(cosacosb-sinasinb)(cosacosb+sinasinb=1\/3 cos²acos²b-sin²asin²b=1\/3 cos²a(1-sin²b)-(1-cos²a)sin²b=1\/3 cos²a-cos²asin²b-sin²b+cos²asin²b=1\/3 所以原式=1\/3 选C ...
