tanθ=2,2sin^2θ+3sin2θ+1=?怎么两个答案不一样啊
问题描述：
tanθ=2,2sin^2θ+3sin2θ+1=?
怎么两个答案不一样啊
答
tanθ=2
sinθ=2cosθ
(sinθ)^2=4(cosθ)^2
1-(cosθ)^2=4(cosθ)^2
(cosθ)^2=1/5
2(sinθ)^2-3sin2θ+1
=2(2cosθ)^2-6sinθcosθ+1
=8(cosθ)^2-6*2*cosθ*cosθ+1
=-4(cosθ)^2+1
=1/5
答
2sin^2θ+3sin2θ+1=(2sin^2θ+3sin2θ+1)/1=(2sin^2θ+3*2sinθ*cosθ+sin^2θ+cos^2θ)/sin^2θ+cos^2θ=
然后分子,分母,同时除以cos^2θ
=(3tan^2θ+6tanθ+1)/tan^2θ+1=(12+12+1)/5=5