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
设 x = π/8
则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
或tanπ/8 = -1－√2（舍）
因为π/8是第一象限角，tanπ/8是正值
综上所述，tanπ/8 = -1 +√2
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