证明cos^2(A+B)—sin^2(A—B)=cos2Acos2B
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证明cos^2(A+B)—sin^2(A—B)=cos2Acos2B
cos^2(A+B)-sin^2(A-B)=[cos(A+B)+sin(A-B)][cos(A+B)-sin(A-B)]=(cosAcosB-sinAsinB+sinAcosB-cosAsinB)(cosAcosB-sinAsinB-sinAcosB+cosAsinB)=(cosA)^2(cosB)^2-sinAsinBcosAcosB-(cosB)^2sinAcosA+(cosA)^2sinBcosB-sinAsinBcosAcosB+(sinA)^2(sinB)^2+(sinA)^2sinBcosB-(sinB)^2sinAcosA+(cosB)^2sinAcosA-(sinA)^2sinBcosB-(sinA)^2(cosB)^2+sinAcosAsinBcosB-(cosA)^2sinBcosB+(sinB)^2sinAcosA+sinAsinBcosAcosB-(cosA)^2(sinB)^2=(cosA)^2(cosB)^2+(sinA)^2(sinB)^2-(sinA)^2(cosB)^2-(cosA)^2(sinB)^2=[(cosA)^2-(sinA)^2][(cosB)^2-(sinB)^2]=cos2Acos2B
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