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

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