Question

sinα+3cosα=2,求(sinα-cosα)/(sinα+cosα)的值

Answer 1

已知sinα+3cosα=2,求(sinα-cosα)/(sinα+cosα)的值
解：设tanφ=3，φ∊(0，π/2)；sinφ=3/√10；cosφ=1/√10；
sinα+3cosα=sinα+tanφcosα=(1/cosφ)[sinαcosφ+cosαsinφ)=(√10)sin(α+φ)=2
故sin(α+φ)=2/√10；cos(α+φ)=√(1-4/10)=√(6/10)；tan(α+φ)=2/√6；
tan(α+φ)=(tanα+tanφ)/(1-tanαtanφ)=(tanα+3)/(1-3tanα)=2/√6
即有(√6)(tanα+3)=2(1-3tanα)；(6+√6)tanα=2-3√6；故tanα=(2-3√6)/(6+√6)；
故(sinα-cosα)/(sinα+cosα)=(tanα-1)/(tanα+1)=[(2-3√6)/(6+√6)-1]/[(2-3√6)/(6+√6)+1]
=(-4-4√6)/(8-2√6)=-(2+2√6)/(4-√6)=-(2+√6)

Answer 2

1-cos^2 α)/(sinα-cosα)-(sinα+cosα)/(tan^2 α-1)
=sin^2a/(sina-cosa)-(sina+cosa)/(sin^2a/cos^2a-1)
=sin^2a/(sina-cosa)-(sina+cosa)/((sin^2a-cos^2a)/cos^2a)
=sin^2a/(sina-cosa)-(sina+cosa)cos^2a/[(sina-cosa)(sina+cosa)]
=sin^2a/(sina-cosa)-cos^2a/(sina-cosa)
=(sin^2a-cos^2a)/(sina-cosa)
=(sina-cosa)(sina+cosa)/(sina-cosa)
=sina+cosa
sinacosa=1/8
(sina+cosa)^2=sin^2a+2sinacosa+cos^2a=1+2*1/8=5/4
原式=sina+cosa=-根号5/2（因为在第三象限，sina<0 cosa<0 sina+cosa<0)