特別角的三角函數(6)
Author: 藍邦偉  老師
Date: 12 12 月 2012

$2$

$2$

$2$

1

特別角\; 15°與75°

1

$ O $

$ A $

$ B $

$ C $

$ D $

$ 30° $

$ 60° $

$ E $

$ \sqrt{3} $

$ 15° $

$ x $

$ \left\lbrace \begin{array}{ll} tan15°=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}  \ tan75°=\frac{2+\sqrt{3}}{1}=2+\sqrt{3} \ \end{array}  \right.\  $

$ x²=(2+\sqrt{3})²+1²=8+4\sqrt{3} $
$x=\sqrt{8+4\sqrt{3}}=\sqrt{8+2\sqrt{12}}=\sqrt{6}+\sqrt{2}$

$ \left\lbrace  \begin{array}{ll} sin15°=cos75°=\frac{1}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4} \ cos15°=sin75°=\frac{2+\sqrt{3}}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{4} \ \end{array} \right.\  $

$ 4 $

$ \sqrt{6}+\sqrt{2} $

$ 2 $

$ \sqrt{6}+\sqrt{2} $

$ \sqrt{6}-\sqrt{2} $

$75°  $

O = (7.58, 1.64)
c = Circle(O, 3)
a: y = y(O)
A, C_1 = Intersect(c, a)
B = Rotate(A, 240*°, O)
D = Rotate(A, 120*°, O)
多邊形1, d, a_1, b = Polygon(A, B, D)
C = Intersect(a, a_1)
e = Segment(A, B)
f = Segment(B, C)
g = Segment(C, A)
h = Segment(A, D)
i = Segment(D, C)
k = Circle(A, B)
E = Intersect(k, a, 1)
j = Segment(E, B)
n = 6 (slider from 0 to 6, step 1)
l = Segment(E, A)
α = Angle(B, C, A)
p = CircleArc(A, B, E)