求tan(π/8)

求tan(π/8)
求tan(π/8)

求tan(π/8)
let x = π/8
2x = π/4
tan2x = 2tanx/(1-(tanx)^2)
1 = 2tanπ/8 / (1- (tanπ/8)^2 )
1- (tanπ/8)^2 = 2 tanπ/8
(tanπ/8)^2 + 2tanπ/8 -1 =0
tanπ/8 = -1 +√2
or tanπ/8 = -1+√2 ( rejected)
tanπ/8 = -1 +√2

令tanπ/8=x
则tan(π/4)=1=2x/(1-x²)
x²+2x-1=0
x=-1±√2
显然tanπ/8>0
所以tanπ/8=√2-1