Question

若tan(π-α)=1/3,则cos2α/2sinαcosα+cos²α的值为

Answer 1

∵tan(π-α)=1/3
即tanα=-1/3
∴cos2α/(2sinαcosα+cos²α)
=(cos²α-sin²α)/(2sinαcosα+cos²α)
=(1-tan²α)/(2tanα+1)
=[1-(-1/3)²]/[2*(-1/3)+1]
=8/3

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Answer 2

tan(Pai-a)=-tana=1/3, tana=-1/3
原式=cos2α/2sinαcosα+cos²α
=(cos^2a-sin^2a)/(2sinacosa+cos^2a)
=(1-tan^2a)/(2tana+1)
=(1-1/9)/(-2/3+1)
=8/3