求证：(sin3A+4cos2A+sinA-4)/(3cosA+cos3A-4sin2A)=tanA应该是用到三倍角公式的,可是我就是做不了

Question

求证：(sin3A+4cos2A+sinA-4)/(3cosA+cos3A-4sin2A)=tanA
应该是用到三倍角公式的,可是我就是做不了

Answer 1

sin3A=3sinA-4sin³A
cos3A=4cos³A-3cosA
左边=[3sinA-4sin³A+4(1-2sin²A)+sinA-4]/(3cosA+4cos³A-3cosA-8sinAcosA)
=(4sinA-4sin³A-8sin²A)/(4cos³A-8sinAcosA)
=(sinA-sin³A-2sin²A)/(cos³A-2sinAcosA)
=[sinA(1-sin²A)-2sin²A]/(cos³A-2sinAcosA)
=(sinAcos²A-2sin²A)/(cos³A-2sinAcosA)
=sinA(cos²A-2sinA)/[cosA(cos²A-2sinA)]
=sinA/cosA
=tanA=右边
命题得证