## [高中] 正十七邊形尺規 之半角公式應用 麻煩幫我說明最後部分

### [高中] 正十七邊形尺規 之半角公式應用 麻煩幫我說明最後部分

rabbitfate

cos(α)
= [-1 + √17 + sqrt(34 - 2 √17)] / 16
+ sqrt(17 + 3√17 + sqrt(170 - 26√17) - 4sqrt(34 + 2 √17)) / 8

benice

Step1. 先列出維基百科已有的式子：

x1 = cos(α) + cos(4α) ...... (*)
x2 = cos(2α) + cos(8α)
y1 = cos(3α) + cos(5α)
y2 = cos(6α) + cos(7α)
x1 + x2 = (-1 + √17)/4 ...... (1)
y1 + y2 = (-1 - √17)/4 ...... (2)

Step2. 判別 x1 與 y1 的正負號：

cos(α) cos(4α)
= (cos(4α-α) + cos(4α+α))/2 （By 積化和差）
= (cos(3α) + cos(5α))/2
= y1/2 ...... (**)  （由此式可得 y1 ＞ 0

Step3. 計算 x1 x2 與 y1 y2

x1 x2
= [cos(α) + cos(4α)] [cos(2α) + cos(8α)]
= cos(α) cos(2α) + cos(α) cos(8α) + cos(2α) cos(4α) + cos(4α) cos(8α)
= [cos(α) + cos(3α) + cos(7α) + cos(9α) + cos(2α) + cos(6α) + cos(4α) + cos(12α)] / 2 （By 積化和差）
= [cos(α) + cos(3α) + cos(7α) + cos() + cos(2α) + cos(6α) + cos(4α) + cos()] / 2
= [cos(α) + cos(2α) + cos(3α) + cos(4α) + cos(5α) + cos(6α) + cos(7α) + cos(8α)] / 2
= [-1/2] / 2
= -1/4

x1 x2 = -1/4 ...... (3)

y1 y2 = -1/4 ...... (4)

Step4. 解 x1 與 y1

x1 = [-1 + sqrt(17) + sqrt(34 - 2 sqrt(17))] / 8 （負根不合）...... (A)

y1 = [-1 - sqrt(17) + sqrt(34 + 2 sqrt(17))] / 8 （負根不合）...... (B)

Step5. 解 cos(α)：

cos(α) + cos(4α) = x1 ...... (*)
cos(α) cos(4α) = y1/2 ...... (**)

cos(α)
= [2x1 ± sqrt(4x12 - 8y1)] / 4
= [x1 ± sqrt(x12 - 2y1)] / 2

cos(α)
= [-1 + sqrt(17) + sqrt(34 - 2 sqrt(17))] / 16
+ sqrt(17 + 3 sqrt(17) - sqrt(34 - 2 sqrt(17)) - 2 sqrt(34 + 2 sqrt(17))) / 8

benice

rabbitfate