sinA=1/2,sinB=1/3,sin(A+B)sin(A-B)=?
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sinA=1/2,sinB=1/3,sin(A+B)sin(A-B)=?
sinA=1/2,sinB=1/3,sin(A+B)sin(A-B)=?
sinA=1/2,sinB=1/3,sin(A+B)sin(A-B)=?
=(sinacosb+sinbcosa)(sinacosb-sinbcosa)
=sina^2cosb^2-sinb^2cosa^2
=(1/2)^2[1-(1/3)^2]-(1/3)^2[1-(1/2)^2]
=2/9-1/12
=5/36
由积化和差公式得:
sin(A+B)sin(A-B)=(-1/2)[cos(A+B+A-B)-cos(A+B-A+B)]
=(-1/2)[cos2A-cos2B)
=(-1/2)[1-2sin²A-1+2sin²B)
=sin²A-sin²B
=1/4-1/9
=5/36
(sinAcosB+sinBcosA)(sinAcosB-sinBcosA)
=sinA^2cosB^2-sinB^2cosA^2
=sinA^2cosB^2-sinB^2(1-sinA^2)
=sinA^2cosB^2-sinB^2+sinB^2sinA^2
=sinA^2(cosB^2+sinB^2)-sinB^2
=sinA^2-sinB^2
=1/4-1/9=5/36
5/36
dd
sinA=sin[(A+B)/2 +(A-B)/2]=sin((A+B)/2)*cos((A-B)/2)+cos((A+B)/2)*sin((A-B)/2)=1/2 .......(1)
sinB=sin[(A+B)/2 -(A-B)/2]=sin((A+B)/2)*cos((A-B)/2)-cos((A+B)/2)*sin((A-B)/2)=1/3 .......(2)
(1...
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sinA=sin[(A+B)/2 +(A-B)/2]=sin((A+B)/2)*cos((A-B)/2)+cos((A+B)/2)*sin((A-B)/2)=1/2 .......(1)
sinB=sin[(A+B)/2 -(A-B)/2]=sin((A+B)/2)*cos((A-B)/2)-cos((A+B)/2)*sin((A-B)/2)=1/3 .......(2)
(1)+(2)得2sin((A+B)/2)*cos((A-B)/2)=5/6............(3)
(1)-(2)得2cos((A+B)/2)*sin((A-B)/2)=1/6............(4)
(3)*(4)得2sin((A+B)/2)*cos((A-B)/2)*2cos((A+B)/2)*sin((A-B)/2)=sin(A+B)sin(A-B)=5/36
答案是5/36
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∵sinA=1/2,sinB=1/3
∴sin(A+B)sin(A-B)=(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)
=(sinAcosB)^2-(cosAsinB )^2
=(1/2cosB)^2-(1...
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∵sinA=1/2,sinB=1/3
∴sin(A+B)sin(A-B)=(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)
=(sinAcosB)^2-(cosAsinB )^2
=(1/2cosB)^2-(1/3cosA)^2
=1/4(cosB)^2-1/9(cosA)^2
=1/4〔1-(sinB)^2〕-1/9〔1-(sinA)^2〕
=1/4(1-1/9)-1/9(1-1/4)
=2/9-1/12
=5 /36
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