三角方程式(1)
Author: 藍邦偉  老師

0°≤θ<360°

三角方程式(1)

y= \frac{  \sqrt{ 3 }  }{ 2} 

B

A

B

A

y= \frac{  \sqrt{ 2 }  }{ 2} 

y= \frac{ 1 }{2 } 

B

A

y=- \frac{ 1 }{2 } 

B

A

y=- \frac{  \sqrt{ 2 }  }{2 } 

B

A

y=- \frac{  \sqrt{ 3 }  }{ 2} 

sinθ=  \frac{  \sqrt{ 3 }  }{ 2}  \rightarrow{}θ=?

sinθ= \frac{  \sqrt{ 2 }  }{ 2} \rightarrow{}θ=?

sinθ= \frac{ 1 }{2 } \rightarrow{}θ=?

sinθ=- \frac{ 1 }{ 2} \rightarrow{}θ=?

sinθ=-  \frac{  \sqrt{ 2 }  }{ 2} \rightarrow{}θ=?

sinθ=- \frac{  \sqrt{ 3 }  }{2 }\rightarrow{}θ=? 

O: 0.5 * (Corner[1.0] + Corner[3.0])
u = Vector((x(Corner[1.0]), y(O)), (x(Corner[3.0]), y(O)))
v = Vector((x(O), y(Corner[1.0])), (x(O), y(Corner[3.0])))
L_1: y = y(O) + 0.86603
d = Circle(O, 1.0)
A, B = Intersect(d, L_1)
a = Segment(B, O)
b = Segment(O, A)
P: O + (1.0, 0.0)
n = 3 (slider from 0 to 6, step 1)
β = Angle(P, O, B)
α = Angle(P, O, A)
L_6: y = y(O) - 0.5 * sqrt(3.0)
L_5: y = y(O) - 0.5 * sqrt(2.0)
L_2: y = y(O) + 0.5 * sqrt(2.0)
L_3: y = y(O) + 0.5
L_4: y = y(O) - 0.5
C, D = Intersect(d, L_2)
c = Segment(C, O)
e = Segment(O, D)
E, F = Intersect(d, L_3)
f = Segment(E, O)
g = Segment(O, F)
G, H = Intersect(d, L_4)
h = Segment(O, H)
i = Segment(O, G)
I, J = Intersect(d, L_5)
j = Segment(O, J)
k = Segment(O, I)
K, L = Intersect(d, L_6)
l = Segment(O, L)
m = Segment(O, K)
γ = Angle(P, O, C)
δ = Angle(P, O, D)
ε = Angle(P, O, E)
ζ = Angle(P, O, F)
η = Angle(P, O, H)
θ = Angle(P, O, G)
ι = Angle(P, O, J)
κ = Angle(P, O, I)
λ = Angle(P, O, L)
μ = Angle(P, O, K)