關於三角函數計算的問題!!!

Let x = 22.5˚

tan (x + x) = 2 tan x / (1-tan^2 x)
tan 45˚ = 2 tan x / (1-tan^2 x)
1 = 2 tan x / (1-tan^2 x)
1-tan^2 x = 2 tan x for 1-tan^2 x does not equal to zero
tan^2 x + 2 tan x -1 = 0
tan^2 x + 2 tan x +1 - 2 = 0
tan^2 x + 2 tan x +1 = 2
(tan x + 1)^2 = 2
tan x + 1 = square root 2
tan x = square root 2 - 1

2008-05-13 14:26:42 補充：
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