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求证:1/2(cos2B-cos2A)=sin(A+B)sin(A-B)(一定要从1/2(cos2B-cos2A)推到s
题目内容:
求证:1/2(cos2B-cos2A)=sin(A+B)sin(A-B)
(一定要从1/2(cos2B-cos2A)推到sin(A+B)sin(A-B)
不可以像这样:
/2(cos2B-cos2A)=1/2[2(cosB)^2-1-2(cosA)^2+1]=(cosB)^2-(cosA)^2
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
综上,化简后 1/2(cos2B-cos2A)=sin(A+B)sin(A-B) )
求证:1/2(cos2B-cos2A)=sin(A+B)sin(A-B)
(一定要从1/2(cos2B-cos2A)推到sin(A+B)sin(A-B)
不可以像这样:
/2(cos2B-cos2A)=1/2[2(cosB)^2-1-2(cosA)^2+1]=(cosB)^2-(cosA)^2
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
综上,化简后 1/2(cos2B-cos2A)=sin(A+B)sin(A-B) )
(一定要从1/2(cos2B-cos2A)推到sin(A+B)sin(A-B)
不可以像这样:
/2(cos2B-cos2A)=1/2[2(cosB)^2-1-2(cosA)^2+1]=(cosB)^2-(cosA)^2
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
综上,化简后 1/2(cos2B-cos2A)=sin(A+B)sin(A-B) )
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