4(tan角A)=tan(3角A) 求角A
1个回答

4tanA = tan(3A),∴4sinA/cosA = sin(3A)/cos(3A)

∴4sinA·cos(3A) = sin(3A)·cosA

积化和差可得:

原式左边 = 2(sin4A - sin2A)

原式右边 = (1/2)·(sin4A + sin2A)

∴4sin4A - 4sin2A = sin4A + sin2A

∴3sin4A = 5sin2A ,∴6cos2A = 5 ,

∴cos2A = 5/6 = [1 - (tanA)^2]/[1 + (tanA)^2]

令x = tanA ,则:5 + 5x^2 = 6 - 6x^2

∴x = ±√(1/11) = tanA

∴A = ±arctan√(1/11)