第1个回答 2018-03-28
θ∈(π/2,π),则sinθ>0
cosθ=-5/13,又sin²θ+cos²θ=1,因此
sinθ=√(1-cos²θ)=√[1-(-5/13)²]=12/13
sin2θ=2sinθcosθ
=2·(12/13)·(-5/13)
=-120/169
cos2θ=2cos²θ-1
=2·(-5/13)²-1
=-119/169
tan2θ=sin2θ/cos2θ
=(-120/169)/(-119/169)
=120/119
cosθ=-5/13,又sin²θ+cos²θ=1,因此
sinθ=√(1-cos²θ)=√[1-(-5/13)²]=12/13
sin2θ=2sinθcosθ
=2·(12/13)·(-5/13)
=-120/169
cos2θ=2cos²θ-1
=2·(-5/13)²-1
=-119/169
tan2θ=sin2θ/cos2θ
=(-120/169)/(-119/169)
=120/119