特別角的三角函數值(4)
Author: 藍邦偉  老師

1

1

x

36°

72°

72°

x

1-x

36°

 \frac{ x }{2 } 

特別角的正餘弦三角函數值18°與72°

作一個單位圓

考慮單位圓的內接正十邊形

O: 0.5 * Corner[1.0] + 0.5 * Corner[3.0]
r: 0.4 * (y(Corner[4.0]) - y(Corner[1.0]))
c = Circle(O, r)
n = 2 (slider from 0 to 6, step 1)
A: O + (r * cos(252.0°), r * sin(252.0°))
l = Segment(O, A)
D: O + (r * cos(0.0°), r * sin(0.0°))
C: O + (r * cos((-(36.0))°), r * sin((-(36.0))°))
E: O + (r * cos(36.0°), r * sin(36.0°))
F: O + (r * cos(72.0°), r * sin(72.0°))
G: O + (r * cos(108.0°), r * sin(108.0°))
H: O + (r * cos(144.0°), r * sin(144.0°))
I: O + (r * cos(180.0°), r * sin(180.0°))
J: O + (r * cos(216.0°), r * sin(216.0°))
B: O + (r * cos(288.0°), r * sin(288.0°))
a = Segment(A, B)
b = Segment(B, C)
d = Segment(C, D)
e = Segment(D, E)
f = Segment(E, F)
g = Segment(F, G)
h = Segment(G, H)
i = Segment(H, I)
j = Segment(I, J)
k = Segment(J, A)
m = Segment(O, B)
p, q = AngularBisector(a, l)
K = Intersect(m, q)
s = Segment(A, K)