Question

已知sin(a+b)sin(a-b)=-1/3,求cos^2a-cos^2b的值

Answer 1

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)=-1/3
∴(cos^2a-cos^2b)=1/3

Answer 2

sin(a+b)sin(a-b)
=(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