三角函數的性質
Author: 藍邦偉　老師

平方關係，倒數關係，商數關係

三角函數的性質

sin²θ+cos²θ=1

sinθcscθ=1

cosθsecθ=1   

sinθ

cosθ

1

sinθ

cosθ

tanθ

cotθ

secθ

cscθ

平方關係:

1+tan²θ=sec²θ

1+cot²θ=csc²θ

倒數關係:

tanθcotθ=1

商數關係:

tanθ= \frac{ sinθ }{ cosθ} 

cotθ= \frac{ cosθ }{ sinθ} 

n = 1 (slider from 0 to 4, step 1)
α = 0.8726646259971648 rad (slider from 0 to 6.283185307179586, step 0.017453292519943295)
T_1 = Corner(1.0)
T_4 = Corner(4.0)
T_2: (6.0 * Corner[1.0] + 4.0 * Corner[2.0]) / 10.0
T_3: (6.0 * Corner[4.0] + 4.0 * Corner[3.0]) / 10.0
多邊形1, t_1, t_2, t_3, t_4 = Polygon(T_1, T_2, T_3, T_4)
A_0: 1.0 / 2.0 * (T_2 + Corner[3.0])
c = Circle(A_0, 4.0)
a: y = y(A_0)
A, B = Intersect(c, a)
d = Circle(A, 4.0)
e = Circle(B, 4.0)
C, D = Intersect(c, d)
E, F = Intersect(e, c)
b = Segment(D, E)
f = Segment(C, F)
g = Segment(A, B)
h = Segment(F, B)
i = Segment(B, E)
j = Segment(E, C)
k = Segment(C, A)
l = Segment(A, D)
m = Segment(D, F)
p = Circle(A_0, 0.9)
G, H = Intersect(p, b)
I, J = Intersect(p, g)
K, L = Intersect(p, f)
q = Segment(D, G)
r = Segment(I, A)
s = Segment(K, C)
t = Segment(H, E)
a_1 = Segment(J, B)
b_1 = Segment(L, F)
多邊形2, a_0, f_1, d_1 = Polygon(F, D, A_0)
多邊形3, c_1, a_2, a_3 = Polygon(A_0, A, C)
多邊形4, b_2, a_4, e_1 = Polygon(A_0, E, B)
g_1 = Circle(A_0, 1.0)