13问答网 > 已知sinαcosα=1⼀8,且α是第三象限角,求(1-cos^2 α)⼀(sinα-cosα)-(sinα+cosα)⼀(tan^2 α-1)的值

已知sinαcosα=1⼀8,且α是第三象限角,求(1-cos^2 α)⼀(sinα-cosα)-(sinα+cosα)⼀(tan^2 α-1)的值

2025-02-27 15:06:53

推荐回答（1个）

回答1：

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)