三角函數與柯西不等式
===
## 正弦與餘弦
- $\sin\theta$ 與 $\cos\theta$ 的定義：
- 考慮**單位圓**(半徑為1的圓)上的圓弧$\overset{\LARGE{\frown}}{AB}$，其中B點的座標為$(1,0)$，弧長為 $\theta$，則我們定義 $\cos\theta$ 為 A 點對應的 $x$ 座標，$\sin\theta$ 為 A 點對應的 $y$ 座標。
- 根據畢氏定理，$\sin^2\theta+\cos^2\theta=1$。
- $\cos (-\theta)=\cos \theta$ and $\sin(-\theta)=-\sin \theta$.
- $\sin\theta = \cos (\pi/2-\theta)$.

import matplotlib.pyplot as plt
import numpy as np
from matplotlib.patches import FancyArrowPatch

# 角度 θ
theta = np.pi / 3  # 60 degrees

# 建立圖形與座標軸
fig, ax = plt.subplots(figsize=(6, 6))
ax.set_aspect('equal')
ax.grid(True, which='both')

# 畫單位圓
circle = plt.Circle((0, 0), 1, color='blue', fill=False)
ax.add_artist(circle)

# 原點 O
O = (0, 0)
ax.plot(*O, 'ko')
ax.text(-0.05,-0.05, 'C', fontsize=12, verticalalignment='top', horizontalalignment='right')

# 點 A（60 度）
A = (np.cos(theta), np.sin(theta))
ax.plot(*A, 'ro')
offset_x = 0.08
ax.text(A[0] + offset_x, A[1], r'$A=(\cos\theta,\sin\theta)$', fontsize=12, color='brown')

# 點 B (1, 0)
B = (1, 0)
ax.plot(*B, 'bo')
ax.text(B[0] + 0.05, B[1] + 0.05, r'B=(1,0)', fontsize=12, color='brown')

# OA 向量
arrow = FancyArrowPatch(posA=O, posB=A, arrowstyle='->', color='green', mutation_scale=15, linewidth=2)
ax.add_patch(arrow)

# 圓弧 AB
angle_arc = np.linspace(0, theta, 100)
arc_radius = 1.0
arc_x = arc_radius * np.cos(angle_arc)
arc_y = arc_radius * np.sin(angle_arc)
ax.plot(arc_x, arc_y, 'purple')

# 標記圓弧長 θ
mid_theta = theta / 2
label_radius = 1.05
label_x = label_radius * np.cos(mid_theta)
label_y = label_radius * np.sin(mid_theta)
ax.text(label_x, label_y, r'$\theta$', fontsize=14, color='purple')

# 座標軸設定
ax.set_xlim(-1.2, 1.4)
ax.set_ylim(-0.2, 1.2)
ax.axhline(0, color='black', linewidth=0.5)
ax.axvline(0, color='black', linewidth=0.5)
ax.set_xticks(np.arange(-1, 2, 1))
ax.set_yticks(np.arange(-1, 2, 1))

ax.spines['top'].set_visible(False)
ax.spines['right'].set_visible(False)
ax.spines['left'].set_visible(False)
ax.spines['bottom'].set_visible(False)

# 顯示圖形
plt.savefig("unit_circle_theta.png", dpi=300)
plt.show()

## 正弦定理
- 考慮一個圓內接三角形ABC。則我們可以證明 $$\frac{\overline{BC}}{\sin\angle A}=\text{外接圓的直徑}.$$
- 假設圓半徑為 $r$，角 A 為 $\theta$ 度，邊長BC為 $a$ 且平行於 $y$ 軸。由於圓心角為圓周角的兩倍，且OBC為等腰三角形。因此OB線段的輔角為 $\theta$ 度。此時，$B$ 的座標為 $(r\cos\theta,r\sin \theta)$。也因此 $$\begin{align}a&=2r\sin\theta\\ \frac{a}{\sin\theta}&=2r\end{align}$$


import matplotlib.pyplot as plt
import numpy as np
from matplotlib.patches import Arc

def draw_clean_geometry_diagram():
    # 1. 設定參數
    r = 5  # 圓半徑
    theta_deg = 35  # 設定 theta 為 35 度 (B在第一象限)
    theta_rad = np.radians(theta_deg)
    
    # A點在第二象限
    angle_a_deg = 165 
    angle_a_rad = np.radians(angle_a_deg)

    # 2. 計算座標
    O = np.array([0, 0])
    B = np.array([r * np.cos(theta_rad), r * np.sin(theta_rad)])
    C = np.array([r * np.cos(theta_rad), -r * np.sin(theta_rad)]) # C對稱於B
    A = np.array([r * np.cos(angle_a_rad), r * np.sin(angle_a_rad)])

    # 3. 建立畫布
    fig, ax = plt.subplots(figsize=(8, 8))
    
    # 4. 畫圖形物件
    # 畫圓
    circle = plt.Circle((0, 0), r, color='gray', fill=False, linestyle='--', alpha=0.5)
    ax.add_patch(circle)

    # 畫三角形 ABC
    triangle_x = [A[0], B[0], C[0], A[0]]
    triangle_y = [A[1], B[1], C[1], A[1]]
    ax.plot(triangle_x, triangle_y, color='blue', linewidth=1.5)

    # 畫輔助線 OB 和 OC
    ax.plot([O[0], B[0]], [O[1], B[1]], color='green', linestyle='--', linewidth=1)
    ax.plot([O[0], C[0]], [O[1], C[1]], color='green', linestyle='--', linewidth=1)
    
    # 5. 設定座標軸 (隱藏刻度)
    ax.spines['left'].set_position('center')
    ax.spines['bottom'].set_position('center')
    ax.spines['right'].set_color('none')
    ax.spines['top'].set_color('none')
    
    # --- 關鍵修改：隱藏座標軸上的刻度與數字 ---
    ax.set_xticks([])
    ax.set_yticks([])
    # -------------------------------------
    
    # 設定顯示範圍 (保持正方形比例)
    limit = r + 2
    ax.set_xlim(-limit, limit)
    ax.set_ylim(-limit, limit)
    ax.set_aspect('equal') 

    # 6. 標記文字與符號
    
    # 原點 O
    ax.text(-0.5, -0.5, r'$O$', fontsize=12)
    # 座標軸名稱 (選擇性加入，讓圖更完整)
    ax.text(limit - 0.5, -0.5, r'$x$', fontsize=12)
    ax.text(0.2, limit - 0.5, r'$y$', fontsize=12)
    
    # 頂點 A, B, C
    ax.text(A[0]-0.6, A[1]+0.2, r'$A$', fontsize=12, color='blue')
    ax.text(B[0]+0.2, B[1]+0.2, r'$B=(r\cos\theta,r\sin\theta)$', fontsize=12, color='blue')
    ax.text(C[0]+0.2, C[1]-0.5, r'$C$', fontsize=12, color='blue')

    # B點座標
    #ax.text(B[0]+0.3, B[1], r'$(r\cos \theta, r\sin \theta)$', fontsize=11)

    # 邊長標記 a, b, c
    ax.text(B[0] + 0.3, 0.3, r'$a$', fontsize=12, color='red', va='center') # BC邊
    
    mid_ac = (A + C) / 2
    ax.text(mid_ac[0] - 0.6, mid_ac[1] - 0.6, r'$b$', fontsize=12, color='red') # AC邊
    
    mid_ab = (A + B) / 2
    ax.text(mid_ab[0] - 0.3, mid_ab[1] + 0.4, r'$c$', fontsize=12, color='red') # AB邊

    # 半徑標記 r (OB)
    mid_ob = (O + B) / 2
    ax.text(mid_ob[0]-0.2, mid_ob[1] + 0.4, r'$r$', fontsize=12, color='green')

    # 半徑標記 r (OC)
    mid_oc = (O + C) / 2
    ax.text(mid_oc[0]+0.2, mid_oc[1] + 0.2, r'$r$', fontsize=12, color='green')

    # 7. 標記角度 theta
    
    # (1) 原點處的 theta
    arc_O = Arc((0, 0), 2.5, 2.5, angle=0, theta1=0, theta2=theta_deg,color='purple')
    ax.add_patch(arc_O)
    ax.text(1.5 * np.cos(theta_rad/2), 0.8 * np.sin(theta_rad/2), r'$\theta$', fontsize=11,color='purple')

    # (2) A點處的 theta
    angle_AB = np.degrees(np.arctan2(B[1]-A[1], B[0]-A[0]))
    angle_AC = np.degrees(np.arctan2(C[1]-A[1], C[0]-A[0]))
    
    arc_A = Arc((A[0], A[1]), 2.5, 2.5, angle=0, theta1=angle_AC, theta2=angle_AB, color='purple')
    ax.add_patch(arc_A)
    
    bisector_rad = np.radians((angle_AB + angle_AC) / 2)
    label_dist = 1.8
    ax.text(A[0] + label_dist * np.cos(bisector_rad), 
            A[1] + label_dist * np.sin(bisector_rad), 
            r'$\theta$', fontsize=11, color='purple')

    plt.title('Law of Sines', y=1.02)
    # plt.grid(True) # 隱藏網格讓圖面更乾淨
    plt.show()

if __name__ == "__main__":
    draw_clean_geometry_diagram()

## 餘弦定理
- $a^2+b^2-2ab\cos\theta=c^2$.
- 現在考慮線段BC長度為$a$，線段AC長度為$b$。由於相似三角形的邊長成比例，A點的座標為$(b\cos\theta,b\sin\theta)$，並且B點的座標為$(a,0)$。
- 計算線段AB的長度平方可得
\begin{align}c^2 &=(b\cos\theta-a)^2+b^2\sin^2\theta\\
&=b^2\cos^2\theta-2ab\cos\theta+a^2+b^2\sin^2\theta\\
&=a^2+b^2(\sin^2\theta+\cos^2\theta)-2ab\cos\theta\\
&=a^2+b^2-2ab\cos\theta.
\end{align}

import matplotlib.pyplot as plt
import numpy as np

# 參數設定
a = 5.3
b = 6
theta_deg = 60
theta_rad = np.radians(theta_deg)

# 三點座標
C = np.array([0, 0])
A = np.array([a * np.cos(theta_rad), a * np.sin(theta_rad)])
B = np.array([b, 0])

# 畫圖
fig, ax = plt.subplots()
ax.set_aspect('equal')
ax.set_xlim(-1, b + 1)
ax.set_ylim(-1, a + 1)

# 畫三角形
ax.plot([A[0], B[0], C[0], A[0]], [A[1], B[1], C[1], A[1]], 'k-')

# 畫出 x, y 軸
ax.axhline(0, color='black', linewidth=1.2)
ax.axvline(0, color='black', linewidth=1.2)

# 關閉其他刻度與格線
ax.set_xticks([])
ax.set_yticks([])
ax.grid(False)
ax.spines['top'].set_visible(False)
ax.spines['right'].set_visible(False)
ax.spines['left'].set_visible(False)
ax.spines['bottom'].set_visible(False)

# 標註點
ax.plot(*A, 'ro')
ax.plot(*B, 'ro')
ax.plot(*C, 'ro')
ax.text(A[0],A[1]+0.1, r'$A=(b\cos\theta, b\sin\theta)$', fontsize=12, ha='left', va='bottom')
ax.text(B[0],B[1]-0.1, r'$B=(a,0)$', fontsize=12, ha='left', va='top')
ax.text(C[0]-0.1,C[1]-0.1, '$C$', fontsize=12, ha='right', va='top')

# 標註邊長
ax.text((A[0] + C[0])/2-0.2,(A[1] + C[1])/2+0.2, '$b$', fontsize=12, color='blue')
ax.text((B[0] + C[0])/2,(B[1] + C[1])/2-0.4, '$a$', fontsize=12, color='blue')
ax.text((A[0] + B[0])/2+0.1,(A[1] + B[1])/2+0.1, '$c$', fontsize=12, color='blue')

# 畫圓
circle = plt.Circle((0, 0), 0.8, color='blue', fill=False)
ax.add_artist(circle)

# 畫角 θ 弧線與標註
arc_radius = 0.8
arc = np.linspace(0, theta_rad, 100)
arc_x = arc_radius * np.cos(arc)
arc_y = arc_radius * np.sin(arc)
ax.plot(arc_x, arc_y, 'purple')
ax.text(arc_radius * np.cos(theta_rad / 2)+0.05, arc_radius * np.sin(theta_rad / 2)+0.05, r'$\theta$', fontsize=14, color='purple')

plt.title('Law of Cosines')
plt.savefig("Law of cosines.png", dpi=300)
plt.show()

## 內積公式和柯西不等式(Cauchy Inequality)
- $ab\cos \theta=x_1x_2+y_1y_2$.
- 根據餘弦定理，
\begin{align}
a^2+b^2-2ab\cos\theta&=c^2,\\
x_1^2+y_1^2+x_2^2+y_2^2-2ab\cos\theta&=(x_1-x_2)^2+(y_1-y_2)^2,\\
-2ab\cos\theta&=-2(x_1x_2+y_1y_2),\\
ab\cos\theta&=x_1x_2+y_1y_2.
\end{align}
- 柯西不等式：$(x_1^2+y_1^2)(x_2^2+y_2^2)\geq (x_1y_1+x_2y_2)^2$.
- 因為 $\cos^2\theta\leq\cos^2\theta+\sin^2\theta=1$，根據上面的內積公式可以得到$$(x_1x_2+y_1y_2)^2\leq a^2b^2=(x_1^2+y_1^2)(x_2^2+y_2^2).$$

import numpy as np
import matplotlib.pyplot as plt

# 設定角度（以弧度）
alpha = np.radians(80)
beta = np.radians(20)
theta = alpha - beta

# 向量長度
r = 5

# 三點座標
A = (r * np.cos(alpha), r * np.sin(alpha))
B = ((r+1.3) * np.cos(beta), (r+1.3) * np.sin(beta))
O = (0, 0)

# 建立圖形
fig, ax = plt.subplots()
ax.set_aspect('equal')

# 動態設定邊界，避免遮住文字（特別是 B 點的座標）
x_max = max(A[0], B[0]) + 2.2
y_max = max(A[1], B[1]) + 1.5
ax.set_xlim(-1, x_max)
ax.set_ylim(-1, y_max)

# 畫出 x, y 軸
ax.axhline(0, color='gray', linewidth=1)
ax.axvline(0, color='gray', linewidth=1)

# 畫三角形邊
ax.plot([O[0], A[0]], [O[1], A[1]], 'b')  # OA
ax.plot([O[0], B[0]], [O[1], B[1]], 'b')  # OB
ax.plot([A[0], B[0]], [A[1], B[1]], 'b', linestyle='--', alpha=0.6)  # AB

# 點標記
ax.plot(*A, 'ro')
ax.plot(*B, 'ro')
ax.plot(*O, 'ko')

# 標註點座標
ax.text(A[0] + 0.2, A[1], r'$A=(x_1, y_1)$', fontsize=12)
ax.text(B[0] + 0.2, B[1] + 0.2, r'$B=(x_2, y_2)$', fontsize=12)  # 往右上方偏移避免穿出邊界
ax.text(O[0] - 0.4, O[1] - 0.4, r'$O$', fontsize=12)

# 標註邊 OA, OB, AB 分別為 a, b, c（置中位置）
ax.text((O[0] + A[0]) / 2 - 0.3, (O[1] + A[1]) / 2 + 0.2, r'$a$', fontsize=14, color='black')
ax.text((O[0] + B[0]) / 2 + 0.2, (O[1] + B[1]) / 2-0.25, r'$b$', fontsize=14, color='black')
ax.text((A[0] + B[0]) / 2, (A[1] + B[1]) / 2 + 0.2, r'$c$', fontsize=14, color='black')

# 畫三個角的弧線（依序由內而外：β→α→α-β）
radius_beta = 1.4
radius_alpha = 0.8
radius_theta = 1.4

# β 弧線
beta_arc = np.linspace(0, beta, 100)
ax.plot(radius_beta * np.cos(beta_arc), radius_beta * np.sin(beta_arc), color='blue')
ax.text(radius_beta * np.cos(beta/2) + 0.1,
        radius_beta * np.sin(beta/2),
        r'$\beta$', fontsize=12, color='blue')

# α 弧線
alpha_arc = np.linspace(0, alpha, 100)
ax.plot(radius_alpha * np.cos(alpha_arc), radius_alpha * np.sin(alpha_arc), color='green')
ax.text(radius_alpha * np.cos(alpha/2) + 0.1,
        radius_alpha * np.sin(alpha/2),
        r'$\alpha$', fontsize=12, color='green')

# α - β 弧線
theta_arc = np.linspace(beta, alpha, 100)
ax.plot(radius_theta * np.cos(theta_arc), radius_theta * np.sin(theta_arc), color='red')
ax.text(radius_theta * np.cos((alpha + beta)/2) + 0.1,
        radius_theta * np.sin((alpha + beta)/2),
        r'$\theta=\alpha - \beta$', fontsize=12, color='red')

# 關閉其他刻度與格線
ax.set_xticks([])
ax.set_yticks([])
ax.grid(False)
ax.spines['top'].set_visible(False)
ax.spines['right'].set_visible(False)
ax.spines['left'].set_visible(False)
ax.spines['bottom'].set_visible(False)

# 顯示標題與圖形
plt.title("Inner Prouduct Formula", fontsize=14)
plt.savefig("Inner_product.png", dpi=300)
plt.show()

## 和角公式
- $\cos(\alpha-\beta)=\cos\alpha\cdot\cos\beta+\sin\alpha\cdot \sin \beta$.
- 考慮A, B 兩點都在單位圓上，此時
\begin{align}
a&=b=1,\\
(x_1,y_1)&=(\cos\alpha,\sin\alpha),\\
(x_2,y_2)&=(\cos\beta,\sin\beta).
\end{align}

- 根據內積公式
\begin{align}
ab\cos\theta&=x_1x_2+y_1y_2\\
\cos(\alpha-\beta)&=\cos\alpha\cos\beta+\sin\alpha\sin\beta
\end{align}
- $\cos (\alpha+\beta)=\cos\alpha\cdot \cos \beta-\sin\alpha\cdot\sin \beta$.
- 由於 $\cos(-\beta)=\cos\beta,\sin(-\beta)=-\sin\beta$，因此
\begin{align}
\cos(\alpha+\beta)&=\cos(\alpha-(-\beta))\\
&=\cos\alpha\cos(-\beta)+\sin\alpha\sin(-\beta)\\
&=\cos\alpha\cos\beta-\sin\alpha\sin\beta.
\end{align}

- 使用 $\sin\theta=\cos(\pi/2-\theta)$ 證明下面公式：
- $\sin(\alpha-\beta)=\sin\alpha\cdot\cos\beta -\cos\alpha\cdot\sin\beta$.
- $\sin(\alpha+\beta)=\sin\alpha\cdot\cos\beta+\cos\alpha\cdot\sin\beta$.

import matplotlib.pyplot as plt
import numpy as np
from matplotlib.patches import FancyArrowPatch

# 設定圖形與座標軸
fig, ax = plt.subplots(figsize=(6,6))
ax.set_aspect('equal')
ax.grid(True, which='both')

# 畫單位圓
circle = plt.Circle((0, 0), 1, color='blue', fill=False)
ax.add_artist(circle)

# 原點 O
O = (0, 0)
ax.plot(*O, 'ko')
ax.text(*O, 'O', fontsize=12, verticalalignment='top', horizontalalignment='right')

# A 點（135 度）
alpha = np.pi*3 / 4
A = (np.cos(alpha), np.sin(alpha))
ax.plot(*A, 'ro')

# B 點（45 度）
beta = np.pi / 4
B = (np.cos(beta), np.sin(beta))
ax.plot(*B, 'ro')

# A點向右偏移標籤與座標顯示
offset_x = 0.1
ax.text(A[0] + offset_x, A[1]-0.05, r'$A=(\cos\alpha,\sin\alpha)$', fontsize=12, color='brown')
 

# B點向右偏移標籤與座標顯示
offset_x = 0.05
ax.text(B[0] + offset_x, B[1], r'$B=(\cos\beta,\sin\beta)$', fontsize=12, color='brown')

# 使用 FancyArrowPatch 精準畫 OA 向量
arrow = FancyArrowPatch(posA=O, posB=A, arrowstyle='->', color='green', mutation_scale=15, linewidth=2)
ax.add_patch(arrow)

# 使用 FancyArrowPatch 精準畫 OB 向量
arrow = FancyArrowPatch(posA=O, posB=B, arrowstyle='->', color='green', mutation_scale=15, linewidth=2)
ax.add_patch(arrow)

# 畫角度弧線_A
angle_arc_A = np.linspace(0, alpha, 100)
arc_radius = 0.15
arc_x = arc_radius * np.cos(angle_arc_A)
arc_y = arc_radius * np.sin(angle_arc_A)
ax.plot(arc_x, arc_y, 'purple')
ax.text(arc_radius * np.cos(alpha/2), arc_radius * np.sin(alpha/2), r'$\alpha$', fontsize=14)

# 畫角度弧線_B
angle_arc = np.linspace(0, beta, 100)
arc_radius = 0.3
arc_x = arc_radius * np.cos(angle_arc)
arc_y = arc_radius * np.sin(angle_arc)
ax.plot(arc_x, arc_y, 'green')
ax.text(arc_radius * np.cos(beta/2), arc_radius * np.sin(beta/2), r'$\beta$', fontsize=14)

# 畫角度弧線_AB
angle_arc_AB = np.linspace(beta, alpha, 100)
arc_radius = 0.3
arc_x = arc_radius * np.cos(angle_arc_AB)
arc_y = arc_radius * np.sin(angle_arc_AB)
ax.plot(arc_x, arc_y, 'red')
ax.text(arc_radius * np.cos(alpha*6/7)-0.1, arc_radius * np.sin(alpha*6/7)+0.05, r'$\theta=\alpha-\beta$', fontsize=14)


# 座標軸設定
ax.set_xlim(-1.2, 1.4)
ax.set_ylim(-0.2, 1.2)
ax.set_xticks(np.arange(-1, 2, 1))
ax.set_yticks(np.arange(-1, 2, 1))
ax.axhline(0, color='black', linewidth=0.5)
ax.axvline(0, color='black', linewidth=0.5)

#ax.grid(False)
ax.spines['top'].set_visible(False)
ax.spines['right'].set_visible(False)
ax.spines['left'].set_visible(False)
ax.spines['bottom'].set_visible(False)

plt.title('Trigonometric Additionidentity Formula')
plt.savefig("Trigonometric_Additionidentity.png", dpi=300)
plt.show()

## Cauchy–Schwarz Inequality (廣義柯西不等式)
- $(x_1^2+x_2^2+\cdots+x_n^2)(y_1^2+y_2^2+\cdots+y_n^2)\geq (x_1y_2+x_2y_2+\cdots+x_ny_n)^2$.
- 令 $u=(x_1,\dots,x_n),v=(y_1,\dots,y_n)$，並且定義$n$維向量的內積運算： $$u\cdot v = x_1y_1+\cdots+x_ny_n.$$
- 當 $u\cdot v=0$時，我們可以抽象地說此時$u$和$v$兩向量互相"垂直"！
- 令 $\displaystyle w=\frac{u\cdot v}{v\cdot v}v$.
- 請檢驗 $(u-w)\cdot v = 0$。
- 當我們將$u$分解為$u=(u-w)+w$時，$w$與$v$平行，$u-w$與$v$垂直。
- 請檢驗 $\displaystyle u\cdot u = (u-w)\cdot (u-w)+w\cdot w\geq w\cdot w=\frac{(u\cdot v)^2}{v\cdot v}$，其中\begin{align*}
u\cdot u&=x_1^2+\cdots+x_n^2,\\
v\cdot v&=y_1^2+\cdots+y_n^2,\\
u\cdot v&=x_1y_1+\cdots+x_ny_n.
\end{align*}
- 由上式可得 $(u\cdot u)(v\cdot v)\geq (u\cdot v)^2$。