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已知tanx+1/tanx-1=3,求2sinx-3cosx/4sinx-9cosx;sin^2x-3sinxcosx+1
人气:349 ℃ 时间:2020-02-04 20:55:38
解答
(tanx+1)/(tanx-1)=3,tanx=2.∴(2sinx-3cosx)/(4sinx-9cosx) =(2tanx-3)/(4tanx-9) =(2×2-3)/(4×2-9) =-1.sin^2x-3sinxcosx+1 =(sin^2x-3sinxcosx)/(sin^2x+cos^2x)+1 =(tan^2x-3tanx)/(tan^2x+1)+1 =(4-3×2)/(4+...
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