高一数学在线等

已知tanα/2=2 求（1）tan(α+π/4)的值（2）（6sinα+cosα)/3sinα-2cosα) (3)sin^2α-sinαcosα+5cos^2α

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2009-12-06

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由tanα/2=2 得tanα=(2tanα/2)/[1-(tanα/2)^2] =4/(1-4) =-4/3 (1)tan(α+π/4) =(1+tanα)/(1-tanα) =(1-4/3)/(1+4/3) =-1/7 (2)（6sinα+cosα)/(3sinα-2cosα) =(6tanα+1)/(3tanα-2) =(-8+1)/(-4-2) =7/6 (3)sin^2α-sinαcosα+5cos^2α =(sin^2α-sinαcosα+5cos^2α)/1 =(sin^2α-sinαcosα+5cos^2α)/(sin^2α+cos^2α) =[(tanα)^2-tanα+5]/[(tanα)^2+1] =(16/9+4/3+5)/(16/9+1) =73/25 。
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