a math f.4

若x=22.5度,利用2倍角公式證明 tan^2x+2tanx-1=0,
由此,求tan22.5度 的直(以根式表示)

因為tan2x=2tanx/(1-tan^2x)
1-tan^2x=2tanx/tan2x
tan^2x=1-(2tanx/tan2x)
tan^2(22.5)+2tan(22.5)-1
1-[2tan(22.5)/tan45]+2tan(22.5)-1
=1-2tan(22.5)+2tan(22.5)-1 (tan45=1)
=0
因x=22.5度,tan^2x+2tanx-1=0
考慮y^2+2y-1=0
則y=tan22.5
y=1/2[-2+√(4+4)] 或y=1/2[-2-√(4+4)] (捨去﹐因為tan22.5是正數)

y=1/2[-2+√8]
所以tan22.5=1/2[-2+√8]=√2-1

Part 1

Since x = 22.5度, 2x=45度, 所以 1 = tan(2x)

And 1 = tan(2x)= sin(2x) / cos(2x) = (2 sinx cos x) / (cosx cos x - sin x sin x)

所以 cos x cos x - sin x sin x = 2 sin x cos x
由於x = 22.5度, cos x 不等於 0, 上述方程兩邊每項都給cos x cos x 除
1 - tan x tan x = 2 tan x
therefore, (tanx) ^2 + 2 tan x - 1 = 0.

Part 2

tanx is a root of y^2 + 2y -1 =0

tan x = (-2 + square root (4 - 4(-1)))/2 or (-2 -square root (4 - 4(-1)))/2
= (-2 + 2 square root (2))/2 or (-2 -2square root (2))/2 (rejected since tan x is positive with the given condition that x = 22.5度)
= -1 + square root (2)