特別角的三角函數(2)
Author: 藍邦偉  老師

特別角的正餘弦三角函數值(2) 45°

45°

1

1

 \sqrt{ 2 } 

A = (-1.3999999999999995, -1.4599999999999995)
l: 0.6 * (y(Corner[4.0]) - y(Corner[1.0]))
B: A + (l, 0.0)
a = Segment(A, B)
C = Rotate(A, 270.0*°, B)
b = Segment(C, B)
c = Segment(A, C)
t = -1.24 (slider from -1.54 to 3.8839999999999986, step 0.1)
d: x = t
D = Intersect(d, c)
E = Intersect(a, d)
e = Segment(D, E)
多邊形1, e_1, a_1, d_1 = Polygon(A, D, E)