2D 座標架
Author: 藍邦偉  老師

$ O $

餘角關係

$θ=%v$

$P(cosθ,sinθ)$

$ sin(90°-θ)=cosθ$
$cos(90°-θ)=sinθ $
$sin(90°+θ)=sin((90°-(-θ))$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=cos(-θ) =cosθ$
$cos(90°+θ)=cos(90°-(-θ))$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=sin(-θ)=-sinθ $ 

$而\;tan(90°-θ)=\frac{sin(90°-θ)}{cos(90°-θ)}$
$\;\;\;\;=\frac{cosθ}{sinθ}=\frac{1}{tanθ}$
$tan(90°+θ)=\frac{sin(90°+θ)}{cos(90°+θ)}$
$\;\;\;\;=\frac{cosθ}{-sinθ}=-\frac{1}{tanθ}$

$Q(cos(90°-θ),sin(90°-θ))$

$R(cos45°,sin45°)$

顯示正切

1

顯示正切

顯示角度

u = Vector((x(Corner[1]), 0), (x(Corner[3]), 0))
v = Vector((0, y(Corner[1])), (0, y(Corner[3])))
O = (0, 0)
Q_2: 0.55 * Corner[1] + 0.45 * Corner[2]
Q_3: (x(Q_2), y(Corner[4]))
poly1, q_1, q_2, q_4, q_3 = Polygon(Corner[1], Q_2, Q_3, Corner[4])
θ = 1.7627825445142729 rad (slider from 0 to 6.283185307179586, step 0.017453292519943295)
r = 3
c: x^(2) + y^(2) = r^(2)
P: (r * cos(θ), r * sin(θ))
T: (r, 0)
a_1 = Segment(O, ((r + 1) * cos(θ), (r + 1) * sin(θ)))
Q: (r * cos(90° - θ), r * sin(90° - θ))
R: (r * cos(45°), r * sin(45°))
a_2 = Segment(O, ((r + 1) * cos(90° - θ), (r + 1) * sin(90° - θ)))
a_3 = Segment((-(r + 1) * cos(45°), -(r + 1) * sin(45°)), ((r + 1) * cos(45°), (r + 1) * sin(45°)))
a = false
A = (-12, 4)
b = false
t_P: "P\;(cos" + (LaTeX[If[b ≟ true, "" + θ, " θ"]]) + ",sin" + (LaTeX[If[b ≟ true, "" + θ, "θ"]]) + ")"
t_Q: "Q(cos" + (LaTeX[If[b ≟ true, "(90°-" + θ + ")", "(90°-θ)"]]) + ",sin" + (LaTeX[If[b ≟ true, "(90°-" + θ + ")", "(90°-θ)"]]) + ")"
text1: "$若標準位置角" + (LaTeX[If[b ≟ true, "" + θ, "θ"]]) + "," + (LaTeX[If[b ≟ true, "(90°-" + θ + ")", "(90°-θ)"]]) + ",與45°的終邊與$
$單位圓的交點分別為P\;(cos" + (LaTeX[If[b ≟ true, "" + θ, "θ"]]) + ",sin" + (LaTeX[If[b ≟ true, "" + θ, "θ"]]) + "),$
$Q(cos" + (LaTeX[If[b ≟ true, "(90°-" + θ + ")", "(90°-θ)"]]) + ",sin" + (LaTeX[If[b ≟ true, "(90°-" + θ + ")", "(90°-θ)"]]) + "$
$與R(cos45°,sin45°),因為\frac{θ+(90°-θ)}{2}=45°,$
$所以∠ROP=∠ROQ,即P,Q兩點對稱於$
$直線y=x.\;$
$\fgcolor{red}{又P(x,y)和Q(y,x)對稱於直線\;y=x}$"