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=
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

若tan(α+β)=3,tan(β-π/4)=2,则tan(α+π/4)=? 2tanα／(1-tanαtanα)=-4／3求tanα 2tanα／(1-tanαtanα)=-4／3求tanα要过程已知tanα/2=2,求tanα与tan（α+π/4）若tanα=3,tanβ=2/3,则tan(α+β) = 若tanα=2,tan（β-α）=3,则tan（β-2α）的值为? 若tanα=1/3,tanβ=-1/4,则tan(2α-2β)= 若tanα=1/3,tan（β-α）=-2,则tanβ的值为已知tanα+tanβ=2,tan(α+β)=4 则tanα×tanβ= 已知tanα+tanβ=2,tan(α+β)=4,则tanαtanβ等于设tan(α+β)=2/3,tan(β-π/4)=1/4.则(1+tanα)/(1-tanα)的值为若（1－tanα）/（1+tanα）=4+√5,则tan(π/2－2α)的值若tan（α+β）=2/5.tan(b-π/4)=1/4.则tan（a+π/4）等于证明：tan（α+π/4）+tan（α+3π/4）=2tan2α tan（ x/2+π/4）+tan（x/2-π/4 ）=2tanxtan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-1)^2]/[1-(tan(x 已知tan（α+π/4）=2,则tanα/tan2α= 若tanα=1/2,则tan(α+π/4)=? tanα=2则tan（α-π/6）=5根号3-8,