已知cos²α-cos²β=1/3,那么sin(α+β)*sin(α-β)=
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已知cos²α-cos²β=1/3,那么sin(α+β)*sin(α-β)=
已知cos²α-cos²β=1/3,那么sin(α+β)*sin(α-β)=
已知cos²α-cos²β=1/3,那么sin(α+β)*sin(α-β)=
sin(α+β)=sinαcosβ+cosαsinβ
sin(α-β)=sinαcosβ-cosαsinβ
sin(α+β)*sin(α-β)=sinα²cosβ²-cosα²sinβ²
=(1-cosα²)cosβ²-cosα²(1-cosβ²)
=cosβ²-cosα²cosβ²-cosα²+cosα²cosβ²
=cosβ²-cosα²
因为 cos²α-cos²β=1/3
所以 cosβ²-cosα²=-1/3
sin(α+β)*sin(α-β)=-1/3
sin(α+β)*sin(α-β)
=(1/2)[cos2β -cos2α) ]
=(1/2)[ 2(cosβ)^2 - 2(cosα)^2 ]
= -1/3
sin(α+β)*sin(α-β)= (sinαcosβ+cosαsinβ)*(sinαcosβ-cosαsinβ)=sin²αcos²β-cos²αsin²β =(1-cos²α )cos²β -cos²α (1-cos²β )=cos²β-cos²α=-1/3