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已知3sinB=sin(2A+B),求证:tan(A+B)=2tan A.
人气:447 ℃ 时间:2020-03-28 01:03:23
解答
3sinB=sin(2A+B)
3sinB=sinAcos(A+B)+cosAsin(A+B)
3sinB=sinA(cosAcosB-sinAsinB)+cosA(sinAcosB+cosAsinB)
3sinB=sinAcosAcosB-sinAsinAsinB+cosAsinAcosB+cosAcosAsinB
3sinB=2cosAsinAcosB+sinB(cosAcosA-sinAsinA)
3(sinAsinA+cosAcosA)sinB=2cosAsinAcosB+sinB(cosAcosA-sinAsinA)
4sinAsinAsinB+2cosAcosAsinB=2cosAsinAcosB
2sinAsinAsinB+cosAcosAsinB=cosAsinAcosB
2sinAsinAsinB+cosAcosAsinB=2cosAsinAcosB-cosAsinAcosB
2sinAcosAcosB-2sinAsinAsinB=cosAsinAcosB+cosAcosAsinB
2sinA(cosAcosB-sinAsinB)=cosA(sinAcosB+cosAsinB)
2sinAcos(A+B)=cosAsin(A+B)
tan(A+B)=2tan A
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