Question

已知(π/4+α)=-1/2,试求式子(sin2α-2cos^2α)/(1+tanα)的值。

已知tan(π/4+α)=-1/2,试求式子(sin2α-2cos^2α)/(1+tanα)的值

Answer 1

(sin2α-2cos^2α)/(1+tanα)＝(2sinαcosα-2cosα^2)/(1+tanα)
=(2sinαcosα-2cosα^2)/[(1+tanα)(sinα^2+cosα^2)]
=(2tanα-2)/[(1+tanα)(tanα^2+1)]

因为tan(π/4+α)=-1/2＝(1+tanα)/(1-tanα),解出tanα,代入上式即可。

Answer 2

tan(π/4 +α)
=(tanπ/4+tanα)/(1-tanπ/4*tanα)
=(1+tanα)/(1-tanα)=-1/2
所以tanα=-3

sinα/cosα=tanα=-3
sinα=-3cosα
代入(sinα)^2+(cosα)^2=1
(sinα)^2=9/10
(cosα)^2=1/10

sinα/cosα=-3<0
所以sinαcosα<0
所以sinαcosα=-根号(sinα)^2(cosα)^2=-3/10
sin2α=2sinαcosα=-3/5

所以原式=(-3/5-2*1/10)/(1-3)=2/5