已知tan(x+π/4）=2,则cos2x=?

已知tan(x+π/4）=2,则cos2x=?
已知tan(x+π/4）=2,则cos2x=?

已知tan(x+π/4）=2,则cos2x=?
tan(x+π/4) = (tanx+tanπ/4)/(1-tanxtanπ/4) = (tanx+1)/(1-tanx) = 2
tanx+1 = 2-2tanx
3tanx = 1
tanx=1/3
cosx = ±1/根号(1+tan^2x) = ±1/根号(1+1/9) = ± 3/根号10
cos2x = 2cos^2x - 1 = 2×(±3/根号10)^2 - 1 = 18/10 - 1 = 4/5

原式=tanx+1/1-tanx=2
tanx=1/3
(tanx)(tanx)=1-cos2x/1+cos2x=1/9
cos2x=4/5