[AMATHS 三角函數

1. ( 1 + tan^2 θ)+2tanθ
=sec^2 θ+2tanθ
=1/cos^2θ+2sinθ/cosθ
=(1+2sinθcosθ)/cos^2 θ
=[(cosθ+sinθ)/cosθ]^2
=(1+tanθ)^2
2
( 1+cosθ/ tanθ- secθ) + (1 - cosθ/ tanθ+ secθ)
=[( 1+cosθ)(tanθ+ secθ)+ ( 1-cosθ)(tanθ- secθ)]/(tanθ+ secθ)(tanθ- secθ)
=[(tanθ+ secθ)+(cosθtanθ+cosθsecθ)+ (tanθ- secθ)+(-cosθtanθ+cosθsecθ)]/(tanθ+ secθ)(tanθ- secθ)
=[(2tanθ)+(2cosθsecθ)]/(tanθ+ secθ)(tanθ- secθ)
=2(tanθ+1)/(tan^2θ- sec^2θ)
=- 2 ( 1+ tan θ )
3x^2 + (4sinθ)x - 2cosθ = 0 has two equal roots
(4sinθ)^2+24cosθ=0
16(1-cos^2θ)+24cosθ=0
2(1-cos^2θ)+3cosθ=0
2cos^2θ- 3cosθ- 2= 0
(2cosθ+1)(cosθ-2)=0
cosθ=-1/2=>θ= 120


2009-02-06 01:43:22 補充：
(1+2sinθcosθ)/cos^2 θ
=(cos^2θ+2sinθcosθ+sin^2θ)/cos^2 θ