Question

一个数学三角函数证明题

证明：
1、sin3A=3sinA-（4sinA的三次方）
2、cos3A=(4cosA的三次方）-3cosA

Answer 1

1.sin3A
=sin(2A+A)
=sin2AcosA+cos2AsinA
=2sinAcosAcosA+(1-2sinAsinA)sinA
=2sinA(1-sinAsinA)+(1-2sinAsinA)sinA
=3sinA-(4sinAsinAsinA)
2.cos3A
=cos(2A+A)
=cos2AcosA-sin2AsinA
=(2cosAcosA-1)cosA-2sinAcosAsinA
=(2cosAcosA-1)cosA-2(1-cosA)cosA
=(4cosAcosAcosA)-3cosA

Answer 2

Cos3A=cos(2A+A)=cos2AcosA-sin2AsinA=(2cos2A-1)cosA-2sinAcosAsinA
=2cos3A-cosA-2sin2AcosA
=2cos3A-cosA-2cosA(1-cos2A)
=2cos3A-cosA-2cosA+2cos3A
=4 cos3A-3cosA

sin3a
=sin(2a+a)
=sin2acosa+cos2asina
=2sina(1-sin²a)+(1-2sin²a)sina
=3sina-4sin³a
cos3a
=cos(2a+a)
=cos2acosa-sin2asina
=(2cos²a-1)cosa-2(1-cos²a)cosa
=4cos³a-3cosa

Answer 3

1）sin3A=sin(2A+A)=sin2AcosAa+sinAcos2A=2sinAcosA^2+sinA(1-2sinA^2)=2sinA(1-sinA^2)+sinA(1-2sinA^2)=3sinA-4sinA^3
2)cos3A=cos2AcosA-sinAsin2A=(2cosA^2-1)cosA-2cosAsinA^2=(2cosA^2-1)cosA-2cosA(1-cosA^2)=4cosA^3-3cosA