Question

三角函数!证明1/2(cos2B-cos2A)=sin(A+B)×sin(A-B)

Answer 1

1/2（cos2B-cos2A)

=1/2[2(cosB)^2-1-2(cosA)^2+1]

=(cosB)^2-(cosA)^2.(1)

sin(A+B)sin(A-B)

=(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)

=(sinAcosB)^2-(cosAsinB)^2

=(cosB)^2[1-(cosA)^2]-(cosA)^2[1-(cosB)^2]

=(cosB)^2-(cosBcosA)^2-(cosA)^2+(cosAcosB)^2

=(cosB)^2-(cosA)^2.(2)

(1)=(2)

∴1/2（cos2B-cos2A)=sin(A+B)sin(A-B）