若tan(α+π/4)=3/2 ,则2sin²α-cos²α=

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若tan(α+π/4)=3/2 ,则2sin²α-cos²α=

若tan(α+π/4)=3/2 ,则2sin²α-cos²α=
若tan(α+π/4)=3/2 ,则2sin²α-cos²α=

若tan(α+π/4)=3/2 ,则2sin²α-cos²α=
tan(α+π/4)=3/2
(tanπ/4+tanα)/(1-tanαtanπ/4)=3/2
(1+tanα)/(1-tanα)=3/2
2+2tanα=3-3tanα
5tanα=1
tanα=1/5
2(sinα)^2-(cosα)^2=(2(sinα)^2-(cosα)^2)/((sinα)^2+(cosα)^2)=(2(tanα)^2-1)/((tanα)^2+1)
=(2/25-1)/(1/25+1)=-(23/25)/(26/25)=-23/26

sin(2α+π/2)=3/2X2/(1+3^2/2^2)=12/13
cos(2α+π/2)=-5/13 所以cos2α=12/13
2sin²α-cos²α=1-cos2α-(1+cos2α)/2=1/2-3/2Xcos2α=-23/26

tan(α+π/4)=3/2
(tanα+tanπ/4)/(1-tanπ/4tanα) = 3/2
tanα+1 = 3/2-3/2tanα
tanα = 1/5
(sinα)^2 = 1/26
(cosα)^2 = 25/26
2sin²α-cos²α
=2(1/26) -25/26
=-23/26