题目

三角函数：化简、求值
1.求证：01.1+tan^α=sec^α
02.(cos α-1)^+sin^α=2-2cos α
03.(cos α-cosβ）^+(sin α-sinβ)^=2-2(cos αcos β+sin αsin β)
04.sin^α+cos^α=1-2sin^αcos^α
05.1+tan^θ=sin θcos θ/tan θ
06.cos^α-sin^α/1-2sin αcos α=1+tan α/1-tan α
2.化简：01.sin^α+cos^α-sin^α-cos^α
02.1+cot θ/1+tan θ
03.1+sin θ/cos θ+cos θ/1+sin θ
04.1+sin θ/cos θ+cot θ

提问时间：2021-01-13

答案

01.sinα^2+cosα^2+sinα^2/cosα^2=
1/cosα^2(sinα^2cosα^2+cosα^4+sinα^2)=
1/cosα^2[cosα^2(sinα^2+cosα^2)+sinα^2]=
1/cosα^2(cosα^2+sinα^2)=
1/cosα^2=secα^2
02.cosα^2-2cosα+1+sinα^2=
1+1-2cosα=2-2cosα
03.cosα^2-2cosαcosβ+cosβ^2+sinα^2-2sinαsinβ+sinβ^2
=cosα^2+sinα^2+cosβ^2+sinβ^2-2cosαcosβ-2sinαsinβ
=2-2(cos αcos β+sin αsin β)
04.1-2sinαcosα=
sinα^2-2sinαcosα+cosα^2
=(sinα-cosα)^2
05.sinθcosθ/tanθ=sinθcosθ*cosθ/sinθ=cosθ^2
左边=sinθ^2+cosθ^2+tanθ^2=1/cosθ^2
2.
01.原式=0
02.原式=tanθ+1/tanθ/tanv+1=tanθ
03.原式=1+2sinθ+sinθ^2+cosθ^2/cosθ*(1+sinθ)=2(1+sinθ)/cosθ*(1+sinθ)
=2/cosθ
04.原式=1+sinθ/cosθsinθ+cosθ/sinθ
=sinθ/cosθ=tanθ