已知tanx 2 则sin^2x-sinxcosx
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已知tanx 2 则sin^2x-sinxcosx
![已知tanx 2 则sin^2x-sinxcosx](/uploads/image/z/19112128-16-8.jpg?t=%E5%B7%B2%E7%9F%A5tanx+2+%E5%88%99sin%5E2x-sinxcosx)
sin^2x-sinxcosx
=(sin^2x-sinxcosx)/(sin^2x+cos^2x) (上下同除以cos^2x)
=(tan^2x-tanx)/(tan^2x+1)
=2/5
=(sin^2x-sinxcosx)/(sin^2x+cos^2x) (上下同除以cos^2x)
=(tan^2x-tanx)/(tan^2x+1)
=2/5
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