tanθ=2,求3*sin^2θ+cos^2θ

cos2θ+sin2θ
=(1-tan^θ)/(1+tan^θ) +2tanθ/(1+tan^θ)
=(1-tan^θ+2tanθ)/(1+tan^θ)
=(1-4+4)/(1+4)
=1/5

万能公式
cos2θ＝(1-tan^θ)/(1+tan^θ)
sin2θ＝ 2tanθ/(1+tan^θ)
一定要记住啊！用起来简单！

3sin^2a+cos^2a
=2sin^2a+sin^2a+cos^2a
=2sin^2a+1

tana=2
sina/cosa=2
sin^2a=4cos^2a
sin^a=4-4sin^a
5sin^2a=4
sin^2a=4/5.

所以：
所求式=2sin^2a+1
=2*4/5+1
=13/5.

[3(sina)^2+(cosa)^2]/[(sina)^2+(cosa)^2]
=[3(tana)^2+1]/[(tana)^2+1]
=[3*4+1]/[4+1]
=13/5

3*sin^2θ+cos^2θ=1+2sin^2θ
tanθ=2,所以sinθ=+2根号5/5或-2根号5/5
所以sin^2θ=4/5
所以原式=8/5+1=13/5

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