特別角的三角函數(2)
Author: 藍邦偉  老師
Date: 25 12 月 2012

$1$

$1$

$ A $

$ B $

$ C $

$ 45° $

$ \sqrt{2} $

$ C' $

$ B' $

$ \left\lbrace\begin{array}{ll} sin45°=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \ cos45°=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \ tan45°=\frac{1}{1}=1 \end{array}   \right.\  $

$ 特別角的三角函數值\;45° $

t = 4.74 (slider from 1 to 2 * l + x(A), step 0.01)
A = (-2.44, -1.72)
E: A + (t, 0)
l = 5 (slider from -5 to 5, step 1)
B: A + (l, 0)
a = Segment(A, B)
C: A + (l, l)
b = Segment(C, B)
c = Segment(A, C)
d: x = t
D: (x(A) + t, y(A) + t)
e = Segment(D, E)
多邊形1, e_1, a_1, d_1 = Polygon(A, D, E)
α = Angle(C, B, A)
f = Segment(C, D)
g = Segment(B, E)
β = Angle(D, E, B)
γ = Angle(D, E, A)