Question

已知sinα+cosα/sinα-cosα=2,求sin^2-2sinαcosα+1的值，请详细些，谢谢

Answer 1

sinα+cosα/sinα-cosα=2 等号左边分子分母都除以cosa 得 tana=3
sina^2-2sinαcosα+1=(sina^2-2sinαcosα+sin²α+cos²α）/（sin²α+cos²α） 【这一步是1的变换注意这个小技巧】分子分母同除以cos²α有
=（tan²a-2tana+tan²a+1）/（tan²a+1）=13/10

Answer 2

sinα+cosα/sinα-cosα=2
sina=3cosa
tana=1/3

sin^2-2sinαcosα+1
=2sin²a+cos²a-2sinacosa
除以 cos²a+sin²a
=[2sin²a+cos²a-2sinacosa]/(cos²a+sin²a)
分子分母同时除以 cos²a 得
=(2(tana)²+1-2tana)/(1+tan²a)
=(2/9+1-2/3)/(1+1/9)
=(5/9)/(10/9)
=5/10
=1/2

Answer 3

(sinα+cosα)/(sinα-cosα)=2分子分母同时除以cosα
(sinα/cosα+cosα/cosα)/(sinα/cosα-cosα/cosα)=2
(tanα+1)/(tanα-1)=2
tanα+1=2(tanα-1)
tanα+1=2tanα-2
tanα=3

sin^2α-2sinαcosα+1
=(1-cos2α)/2-sin2α+1
=-1/2*cos2α-sin2α+3/2
=-1/2*(1-tan^2α)/(1+tan^2α)-2tanα/(1+tan^2α)+3/2
=-1/2*(1-3^2)/(1+3^2)-2*3/(1+3^2)+3/2
=-1/2*(-8)/10-6/10+3/2
=-4/10-6/10+15/10
=5/10
=1/2

Answer 4

把(sinα+cosα)/(sinα-cosα)=2两边乘方得
（1+sin2α）/（1-sin2α）=4
sin2α=3/5
cos2α = 1-2(sinα)^2=4/5
(sinα)^2 = 1/10
∴原式=1/10-3/5+1 = 1/2