Axiom Learning·AP PrecalculusChapter 8: Trigonometric Functions
三角函数 — Sine, Cosine, Tangent, and Sinusoidal Models
本章概览
三角函数是描述周期现象（periodic phenomena）的核心工具——从钟摆摆动、声波振动到潮汐变化，都可以用 sine 和 cosine 建模。本章将从单位圆出发，学习特殊角的值、图像变换、正弦型函数建模，以及三角恒等式。
| Term 术语 | Chinese 中文 | Brief meaning |
|---|---|---|
| Periodic function | 周期函数 | Repeats on every interval of length $T$ |
| Period | 周期 | Length of one complete cycle |
| Amplitude | 振幅 | Half the distance from max to min |
| Midline | 中线 | Horizontal center of oscillation |
| Phase shift | 相位平移 | Horizontal shift of sinusoidal graph |
| Radian | 弧度 | Unit of angle: $\pi$ rad $= 180°$ |
| Unit circle | 单位圆 | Circle of radius 1 centered at origin |
| Reference angle | 参考角 | Positive acute angle to nearest x-axis |
| Coterminal angles | 同终边角 | Angles with same terminal side |
| Standard position | 标准位置 | Vertex at origin, initial side on $+x$-axis |
8.1 Periodic Phenomena 周期现象
Periodic Function （周期函数）
A function $f$ is periodic if there exists a smallest positive number $T$ (the period) such that $$f(x+T)=f(x)$$ for every $x$ in the domain. One full repeat of the pattern is called a cycle（循环）.
A function $f$ is periodic if there exists a smallest positive number $T$ (the period) such that $$f(x+T)=f(x)$$ for every $x$ in the domain. One full repeat of the pattern is called a cycle（循环）.
关键观察
- 如果知道函数在任意长度为 $T$ 的区间上的图像，就能推断整个图像。
- 函数的递增/递减区间和凹凸性都以周期 $T$ 为单位重复出现。
- 周期函数没有终点：它在两个方向上都无限延伸。
🤝 一起做 Together T-1 — Identifying Period
A periodic function has the graph shown below over the domain $[-4,4]$. What is its period? If $f(2)=5$, what is $f(10)$?
Matching points repeat every 4 units：相同位置每隔 4 个单位重复。
题意理解
周期函数上的点：只要 $x$ 相差一个周期的整数倍，函数值相同。先读出一个完整循环的长度，就是周期 $T$。
Step 1 — Read period from graph.
The graph completes one full cycle from $x=-4$ to $x=0$ (or from $x=0$ to $x=4$), so the period $T=4$.
The graph completes one full cycle from $x=-4$ to $x=0$ (or from $x=0$ to $x=4$), so the period $T=4$.
Step 2 — Apply periodicity to find $f(10)$.
$f(10) = f(10 - 2\cdot 4) = f(10-8) = f(2) = 5$
$f(10) = f(10 - 2\cdot 4) = f(10-8) = f(2) = 5$
Step 3 — Conclusion.
The period is $T=4$, and $f(10)=\boxed{5}$.
The period is $T=4$, and $f(10)=\boxed{5}$.
📝 练习 E-1 — Using Periodicity
A periodic function has period $6$. If $f(-1)=4$, find $f(17)$ and $f(-13)$.
$17=-1+3(6)$ and $-13=-1-2(6)$, so $f(17)=f(-13)=f(-1)=4$.
中文提示：输入值相差周期 $6$ 的整数倍，输出值相同。
中文提示：输入值相差周期 $6$ 的整数倍，输出值相同。
📝 练习 E-2 — Periodic or Not? Estimating a Period 是否周期？估计周期
(a) Which relationships are periodic (or approximately periodic)? Explain.
(i) the height of a bicycle pedal vs. time, riding at a constant speed; (ii) a city's population vs. year; (iii) the number of cars through an intersection per hour over one month; (iv) the daily high temperature in a city over five years.
(b) The water depth $d$ (m) at a pier is recorded every 2 hours. Estimate the period.
| $t$ (h) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 |
|---|---|---|---|---|---|---|---|---|---|---|
| $d$ (m) | 5 | 7.165 | 7.165 | 5 | 2.835 | 2.835 | 5 | 7.165 | 7.165 | 5 |
(a) (i) periodic: the pedal repeats its position every revolution; (ii) not periodic: population does not repeat a pattern; (iii) approximately periodic with a period of about 1 day (rush hours repeat); (iv) approximately periodic with a period of about 1 year (365 days).
(b) The pattern $5,\ 7.165,\ 7.165,\ 5,\ 2.835,\ 2.835$ starts repeating at $t=12$, so the period is about 12 hours (CED 3.1.B.2).
(b) The pattern $5,\ 7.165,\ 7.165,\ 5,\ 2.835,\ 2.835$ starts repeating at $t=12$, so the period is about 12 hours (CED 3.1.B.2).
8.2 Sine, Cosine, and Tangent 正弦、余弦、正切
Angles in Standard Position 标准位置角
An angle is in standard position（标准位置）if:
- Its vertex is at the origin
- Its initial side lies on the positive $x$-axis
- Its terminal side rotates from the initial side
Degree–Radian Conversion 角度与弧度转换
$$180° = \pi \text{ radians}$$
- Degrees → Radians: multiply by $\dfrac{\pi}{180°}$
- Radians → Degrees: multiply by $\dfrac{180°}{\pi}$
记忆技巧
把 $180°$ 和 $\pi$ 绑定记忆。换算时，想想"乘以一个等于1的分数"：
$90° \times \dfrac{\pi}{180°} = \dfrac{\pi}{2}$（弧度）。
Coterminal Angles 同终边角
Angles are coterminal（同终边角）if they share the same terminal side.
- In degrees: add or subtract $360°$ → $\theta + 360°k$
- In radians: add or subtract $2\pi$ → $\theta + 2\pi k$ ($k$ any integer)
Unit Circle Definitions 单位圆定义
Let $(x,y)$ be the point where the terminal side of angle $\theta$ intersects the unit circle. Then:
$$\cos\theta = x, \qquad \sin\theta = y, \qquad \tan\theta = \frac{y}{x} = \frac{\sin\theta}{\cos\theta} \quad (\cos\theta\neq 0)$$
Unit Circle ↔ Sine/Cosine Wave Tracer · 单位圆—波形联动
One rotating point creates both graphs: its horizontal coordinate is cosine, and its vertical coordinate is sine. 同一个旋转点的横坐标生成 cosine，纵坐标生成 sine。
Drag the angle or start the rotation. 拖动角度，或播放完整的一周。
Quadrant I · 第一象限
Unit Circle · 单位圆
Click the circle to choose an angle. 点击圆周也可选角。
Unwrapped Wave Graphs · 展开的周期图像
sin θ = vertical y-coordinate 纵坐标
cos θ = horizontal x-coordinate 横坐标
current θ 当前角
Angle 角度
45° = π/4
Point on Unit Circle 圆上点
P = (√2/2, √2/2) ≈ (0.707, 0.707)
Signs 符号
Quadrant I: cos +, sin +
Why the waves appear 为什么会出现波形：
as θ increases uniformly, the point's $y$-coordinate rises and falls as $sin\theta$, while its $x$-coordinate moves from right to left and back as $cos\theta$. 角度匀速增大时，圆上点的纵坐标与横坐标分别按 sine 与 cosine 的规律周期变化。
🤝 一起做 Together T-2 — Degree–Radian Conversion
Convert $150°$ to radians.
Multiply by $\dfrac{\pi}{180°}$.
$$150° \times \frac{\pi}{180°} = \frac{150\pi}{180} = \frac{5\pi}{6}$$
$$150° \times \frac{\pi}{180°} = \frac{150\pi}{180} = \frac{5\pi}{6}$$
$150° = \dfrac{5\pi}{6}$ radians
🤝 一起做 Together T-3 — Coterminal Angles
Find two positive and two negative coterminal angles for $\dfrac{\pi}{3}$.
Add / subtract $2\pi$ repeatedly.
Positive: $\dfrac{\pi}{3}+2\pi = \dfrac{7\pi}{3}$, $\dfrac{\pi}{3}+4\pi = \dfrac{13\pi}{3}$
Negative: $\dfrac{\pi}{3}-2\pi = -\dfrac{5\pi}{3}$, $\dfrac{\pi}{3}-4\pi = -\dfrac{11\pi}{3}$
Positive: $\dfrac{\pi}{3}+2\pi = \dfrac{7\pi}{3}$, $\dfrac{\pi}{3}+4\pi = \dfrac{13\pi}{3}$
Negative: $\dfrac{\pi}{3}-2\pi = -\dfrac{5\pi}{3}$, $\dfrac{\pi}{3}-4\pi = -\dfrac{11\pi}{3}$
Positive: $\dfrac{7\pi}{3}$, $\dfrac{13\pi}{3}$ | Negative: $-\dfrac{5\pi}{3}$, $-\dfrac{11\pi}{3}$
📝 练习 E-3 — Angle Representation
(a) Convert $-\dfrac{7\pi}{6}$ to degrees. (b) Give one positive coterminal angle in degrees.
(a) $-\dfrac{7\pi}{6}\cdot\dfrac{180^\circ}{\pi}=-210^\circ$.
(b) $-210^\circ+360^\circ=150^\circ$.
Therefore, $-\dfrac{7\pi}{6}$ and $150^\circ$ share the same terminal side.
(b) $-210^\circ+360^\circ=150^\circ$.
Therefore, $-\dfrac{7\pi}{6}$ and $150^\circ$ share the same terminal side.
Radians, Circles of Any Radius, and Tangent as a Slope 弧度、任意半径的圆、正切 = 斜率
- Radian measure $=\dfrac{\text{arc length}}{\text{radius}}$: $\theta=\dfrac{s}{r}$, so $s=r\theta$. On the unit circle, the radian measure is the arc length (CED 3.2.A.2).
- On a circle of radius $r$ centered at the origin, the terminal ray meets the circle at $P=(r\cos\theta,\ r\sin\theta)$, so $\sin\theta=\dfrac{y}{r}$ and $\cos\theta=\dfrac{x}{r}$ (CED 3.2.A.3–A.4, 3.3.A.1).
- $\tan\theta=\dfrac{y}{x}$ is the slope of the terminal ray (CED 3.2.A.5).
🤝 一起做 Together T-4 — Arc Length, Coordinates, Slope
(a) An angle of $\dfrac{2\pi}{3}$ radians is drawn in a circle of radius 5 cm. Find the arc length it subtends.
(b) Find the point where the terminal ray of $\theta=\dfrac{5\pi}{6}$ meets the circle of radius 4 centered at the origin.
(c) Find the slope of that terminal ray.
(a) $s=r\theta=5\cdot\dfrac{2\pi}{3}=\dfrac{10\pi}{3}\approx10.472$ cm.
(b) $\left(4\cos\dfrac{5\pi}{6},\ 4\sin\dfrac{5\pi}{6}\right)=\left(4\cdot\left(-\dfrac{\sqrt3}{2}\right),\ 4\cdot\dfrac12\right)=(-2\sqrt3,\ 2)$.
(c) slope $=\dfrac{2}{-2\sqrt3}=-\dfrac{\sqrt3}{3}=\tan\dfrac{5\pi}{6}$ ✓
(a) $\frac{10\pi}{3}\approx10.472$ cm · (b) $(-2\sqrt3,\,2)$ · (c) $-\frac{\sqrt3}{3}$
📝 练习 E-4 — A Point on the Terminal Ray 终边上的点
The terminal ray of an angle $\theta$ in standard position passes through $(-3,4)$. Find $\sin\theta$, $\cos\theta$, and $\tan\theta$.
$r=\sqrt{(-3)^2+4^2}=5$. $\sin\theta=\dfrac45$, $\cos\theta=-\dfrac35$, $\tan\theta=-\dfrac43$ (the slope of the ray). $\theta$ is in Quadrant II.
8.3 Sine and Cosine Function Values 正弦与余弦函数值
Special Angles 特殊角
| $\theta$ (rad) | $\theta$ (deg) | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ |
|---|---|---|---|---|
| $0$ | $0°$ | $0$ | $1$ | $0$ |
| $\dfrac{\pi}{6}$ | $30°$ | $\dfrac{1}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{\sqrt{3}}$ |
| $\dfrac{\pi}{4}$ | $45°$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $1$ |
| $\dfrac{\pi}{3}$ | $60°$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{2}$ | $\sqrt{3}$ |
| $\dfrac{\pi}{2}$ | $90°$ | $1$ | $0$ | undefined |
记忆口诀
$\sin$ 值从 $0°$ 到 $90°$ 依次是：$\dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2}$，即 $0, \dfrac{1}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, 1$。$\cos$ 值反过来（倒序）。
Signs by Quadrant 各象限符号
| Quadrant | $\sin$ | $\cos$ | $\tan$ |
|---|---|---|---|
| I | + | + | + |
| II | + | − | − |
| III | − | − | + |
| IV | − | + | − |
记忆口诀 — "ASTC"
All Students Take Calculus：I象限All正，II象限Sine正，III象限Tangent正，IV象限Cosine正。
🤝 一起做 Together T-5 — Values in Other Quadrants
Find the exact value of $\sin\dfrac{5\pi}{6}$ and $\cos\dfrac{5\pi}{6}$.
解题思路
$\dfrac{5\pi}{6}$ 在第 II 象限（大于 $\dfrac{\pi}{2}$，小于 $\pi$）。参考角 = $\pi - \dfrac{5\pi}{6} = \dfrac{\pi}{6}$。
Step 1 — Identify quadrant and reference angle.
$\dfrac{5\pi}{6}$ lies in Quadrant II. Reference angle: $\pi - \dfrac{5\pi}{6} = \dfrac{\pi}{6}$.
$\dfrac{5\pi}{6}$ lies in Quadrant II. Reference angle: $\pi - \dfrac{5\pi}{6} = \dfrac{\pi}{6}$.
Step 2 — Apply quadrant signs.
In QII: $\sin$ is positive, $\cos$ is negative.
$\sin\dfrac{5\pi}{6} = +\sin\dfrac{\pi}{6} = \dfrac{1}{2}$
$\cos\dfrac{5\pi}{6} = -\cos\dfrac{\pi}{6} = -\dfrac{\sqrt{3}}{2}$
In QII: $\sin$ is positive, $\cos$ is negative.
$\sin\dfrac{5\pi}{6} = +\sin\dfrac{\pi}{6} = \dfrac{1}{2}$
$\cos\dfrac{5\pi}{6} = -\cos\dfrac{\pi}{6} = -\dfrac{\sqrt{3}}{2}$
$\sin\dfrac{5\pi}{6} = \dfrac{1}{2}$, $\cos\dfrac{5\pi}{6} = -\dfrac{\sqrt{3}}{2}$
🤝 一起做 Together T-6 — Evaluating Tangent
Find the exact value of $\tan\dfrac{4\pi}{3}$.
Step 1 — Quadrant and reference angle.
$\dfrac{4\pi}{3}$ is in Quadrant III. Reference angle: $\dfrac{4\pi}{3}-\pi = \dfrac{\pi}{3}$.
$\dfrac{4\pi}{3}$ is in Quadrant III. Reference angle: $\dfrac{4\pi}{3}-\pi = \dfrac{\pi}{3}$.
Step 2 — Signs in QIII: $\tan$ is positive.
$\tan\dfrac{4\pi}{3} = +\tan\dfrac{\pi}{3} = \sqrt{3}$
$\tan\dfrac{4\pi}{3} = +\tan\dfrac{\pi}{3} = \sqrt{3}$
$\tan\dfrac{4\pi}{3} = \sqrt{3}$
📝 练习 E-5 — Exact Values from a Reference Angle
Find the exact values of $\sin\dfrac{7\pi}{4}$, $\cos\dfrac{7\pi}{4}$, and $\tan\dfrac{7\pi}{4}$.
$\dfrac{7\pi}{4}$ is in Quadrant IV with reference angle $\dfrac{\pi}{4}$. Thus
$\sin\dfrac{7\pi}{4}=-\dfrac{\sqrt2}{2}$, $\cos\dfrac{7\pi}{4}=\dfrac{\sqrt2}{2}$, and $\tan\dfrac{7\pi}{4}=-1$.
$\sin\dfrac{7\pi}{4}=-\dfrac{\sqrt2}{2}$, $\cos\dfrac{7\pi}{4}=\dfrac{\sqrt2}{2}$, and $\tan\dfrac{7\pi}{4}=-1$.
Where Do the Special Values Come From? 特殊值从哪里来
With hypotenuse 1 (the radius of the unit circle), the legs of the special right triangles are the cosine and sine (CED 3.3.A.2):
45-45-90：两条直角边相等，$\cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{\sqrt2}{2}$。30-60-90 是等边三角形的一半：$\cos\frac{\pi}{6}=\frac{\sqrt3}{2}$，$\sin\frac{\pi}{6}=\frac12$；看 60° 角时两条边的角色互换。再根据象限加上正负号。
8.4 Sine and Cosine Function Graphs 正弦与余弦图像
Parent Functions 母函数
$y = \sin x$
- Domain: $(-\infty,\infty)$
- Range: $[-1,1]$
- Period: $2\pi$
- Amplitude: $1$
- Midline: $y=0$
- Starts at $(0,0)$ — "zero start"
$y = \cos x$
- Domain: $(-\infty,\infty)$
- Range: $[-1,1]$
- Period: $2\pi$
- Amplitude: $1$
- Midline: $y=0$
- Starts at $(0,1)$ — "max start"
$y=\sin x$：从中线向上$y=\cos x$：从最大值出发
Both graphs have amplitude 1, midline $y=0$, and period $2\pi$.
Key Points for One Period 一个周期的关键点
For $y=\sin x$: $(0,0)$, $\left(\dfrac{\pi}{2},1\right)$, $(\pi,0)$, $\left(\dfrac{3\pi}{2},-1\right)$, $(2\pi,0)$
For $y=\cos x$: $(0,1)$, $\left(\dfrac{\pi}{2},0\right)$, $(\pi,-1)$, $\left(\dfrac{3\pi}{2},0\right)$, $(2\pi,1)$
For $y=\cos x$: $(0,1)$, $\left(\dfrac{\pi}{2},0\right)$, $(\pi,-1)$, $\left(\dfrac{3\pi}{2},0\right)$, $(2\pi,1)$
图像规律
将一个周期 $[0, 2\pi]$ 分成 4 等份，每个节点处函数值依次为：最大、零、最小、零（以 $\sin$ 为例，从零出发；以 $\cos$ 为例，从最大值出发）。掌握这5个关键点，就能手绘正弦/余弦图像。
📝 练习 E-6 — Reading a Parent Trig Graph
Use the graph above. On $[0,2\pi]$, list the $x$-values where (a) $\sin x=0$ and (b) $\cos x=-1$.
(a) The sine graph crosses the midline at $x=0,\pi,2\pi$.
(b) The cosine graph reaches its minimum $-1$ at $x=\pi$.
(b) The cosine graph reaches its minimum $-1$ at $x=\pi$.
8.5 Sinusoidal Functions 正弦型函数
General Form 通用形式
$$f(x) = A\sin(B(x-h))+k \qquad \text{or} \qquad f(x) = A\cos(B(x-h))+k$$
| Parameter 参数 | Meaning 含义 | Chinese |
|---|---|---|
| $A$ | Amplitude = $|A|$; reflects over midline if $A<0$ | 振幅（$A<0$ 时翻转） |
| $B$ | Controls period: Period $= \dfrac{2\pi}{|B|}$ | 周期控制参数 |
| $h$ | Phase shift (horizontal shift) | 相位平移（水平移动） |
| $k$ | Vertical shift; midline $y=k$ | 竖直平移；中线 $y=k$ |
Key Formulas 关键公式
$$\text{Amplitude} = |A| = \frac{\max - \min}{2}, \qquad \text{Midline} = k = \frac{\max+\min}{2}$$
$$\text{Period} = \frac{2\pi}{|B|}, \qquad \text{Frequency} = \frac{1}{\text{Period}}$$
🤝 一起做 Together T-7 — Reading Parameters
Find the amplitude, period, midline, and phase shift of $f(x) = 3\sin\!\left(2\!\left(x-\dfrac{\pi}{4}\right)\right)-5$.
Identify $A$, $B$, $h$, $k$.
$A=3$, $B=2$, $h=\dfrac{\pi}{4}$, $k=-5$
$A=3$, $B=2$, $h=\dfrac{\pi}{4}$, $k=-5$
Compute each feature.
Amplitude $= |A| = 3$
Period $= \dfrac{2\pi}{|B|} = \dfrac{2\pi}{2} = \pi$
Midline: $y = -5$
Phase shift: $h = \dfrac{\pi}{4}$ to the right
Amplitude $= |A| = 3$
Period $= \dfrac{2\pi}{|B|} = \dfrac{2\pi}{2} = \pi$
Midline: $y = -5$
Phase shift: $h = \dfrac{\pi}{4}$ to the right
Amplitude $3$, Period $\pi$, Midline $y=-5$, Phase shift $\dfrac{\pi}{4}$ right
Concavity, Symmetry, and Cosine as a Shifted Sine 凹凸性、对称性、余弦 = 平移的正弦
- $\cos\theta=\sin\!\left(\theta+\dfrac{\pi}{2}\right)$: cosine is sine shifted left by $\frac{\pi}{2}$, so every transformation rule works the same for both (CED 3.5.A.1).
- Sinusoidal graphs alternate between concave down and concave up; the points of inflection are on the midline, where the graph is steepest (CED 3.5.A.5).
- $y=\sin\theta$ is symmetric about the origin (odd); $y=\cos\theta$ is symmetric over the $y$-axis (even) (CED 3.5.A.6).
$y=\sin x$ on $[0,2\pi]$: increasing/decreasing (inc/dec) and concave down/up (CD/CU) by quarter period. Red: inflection points on the midline.
📝 练习 E-7 — Rate of Change of a Sinusoid 正弦曲线的变化率
For $y=\cos x$ on $[0,2\pi]$: (a) on which interval is it decreasing and concave down? decreasing and concave up? (b) Where is its rate of change least (most negative)? (c) Is the rate of change of $y=\sin x$ increasing or decreasing on $\left(\frac{\pi}{2},\pi\right)$?
(a) Decreasing and concave down on $\left(0,\frac{\pi}{2}\right)$; decreasing and concave up on $\left(\frac{\pi}{2},\pi\right)$.
(b) At $x=\frac{\pi}{2}$, where the graph crosses the midline going down (a point of inflection).
(c) $\sin x$ is concave down there, so its rate of change is decreasing (it is decreasing faster and faster).
(b) At $x=\frac{\pi}{2}$, where the graph crosses the midline going down (a point of inflection).
(c) $\sin x$ is concave down there, so its rate of change is decreasing (it is decreasing faster and faster).
8.6 Sinusoidal Function Transformations 正弦型函数变换
各参数对图像的影响
- $A$：改变振幅；若 $A<0$，图像关于中线上下翻转。
- $B$：改变周期，$|B|$ 越大，图像越"压缩"（周期越短）。
- $h$：水平平移，向右移 $h$ 个单位（$h>0$）。
- $k$：竖直平移，确定中线位置 $y=k$。
5-Step Graphing Method 五步作图法
- Find amplitude $|A|$ and midline $y=k$.
- Find period $T = \dfrac{2\pi}{|B|}$.
- Divide one period into 4 equal parts (every $T/4$).
- Place key points (zero/max/min) using sine or cosine pattern, starting at $x=h$.
- Apply phase shift $h$ and vertical shift $k$; extend by symmetry.
🤝 一起做 Together T-8 — Graphing a Sinusoid
Graph one period of $y = -2\cos\!\left(\dfrac{1}{2}x\right)+1$ by identifying its key features.
Step 1 — Parameters.
$A=-2$, $B=\dfrac{1}{2}$, $h=0$, $k=1$
$A=-2$, $B=\dfrac{1}{2}$, $h=0$, $k=1$
Step 2 — Amplitude and midline.
Amplitude $= |-2| = 2$. Midline: $y=1$.
Maximum value: $1+2=3$; Minimum value: $1-2=-1$.
Amplitude $= |-2| = 2$. Midline: $y=1$.
Maximum value: $1+2=3$; Minimum value: $1-2=-1$.
Step 3 — Period and quarter-period.
Period $= \dfrac{2\pi}{1/2} = 4\pi$. Quarter period $= \pi$.
Period $= \dfrac{2\pi}{1/2} = 4\pi$. Quarter period $= \pi$.
Step 4 — Key points ($A<0$ so cosine starts at minimum).
$(0,-1)$, $(\pi,1)$, $(2\pi,3)$, $(3\pi,1)$, $(4\pi,-1)$
$(0,-1)$, $(\pi,1)$, $(2\pi,3)$, $(3\pi,1)$, $(4\pi,-1)$
Step 5 — Sketch a smooth curve through these 5 points.
One complete cycle：minimum → midline → maximum → midline → minimum.
📝 练习 E-8 — Constructing a Sinusoidal Equation
A sinusoidal function has maximum value $7$ and minimum value $-3$, and its period is $6\pi$. Write a possible equation in the form $f(x)=A\sin(Bx)+k$.
Amplitude $= \dfrac{7-(-3)}{2}=5$; Midline $= \dfrac{7+(-3)}{2}=2$; $B=\dfrac{2\pi}{6\pi}=\dfrac{1}{3}$
$f(x) = 5\sin\!\left(\dfrac{1}{3}x\right)+2$
$f(x) = 5\sin\!\left(\dfrac{1}{3}x\right)+2$
📝 练习 E-9 — Five Key Points after Transformations
For $g(x)=3\sin\!\left(2\left(x-\dfrac{\pi}{4}\right)\right)-1$, state the amplitude, period, midline, and the five key points of one cycle beginning at $x=\dfrac{\pi}{4}$.
Amplitude $3$; period $\pi$; midline $y=-1$; quarter-period $\pi/4$.
The sine pattern is midline → max → midline → min → midline, so the points are
$\left(\dfrac{\pi}{4},-1\right)$, $\left(\dfrac{\pi}{2},2\right)$, $\left(\dfrac{3\pi}{4},-1\right)$, $(\pi,-4)$, $\left(\dfrac{5\pi}{4},-1\right)$.
The sine pattern is midline → max → midline → min → midline, so the points are
$\left(\dfrac{\pi}{4},-1\right)$, $\left(\dfrac{\pi}{2},2\right)$, $\left(\dfrac{3\pi}{4},-1\right)$, $(\pi,-4)$, $\left(\dfrac{5\pi}{4},-1\right)$.
Factor Out $b$ Before Reading the Phase Shift 先提出 $b$ 再读相位平移
In $y=a\sin(bx+c)+d$ the phase shift is not $-c$. Factor first: $\;a\sin\!\left(b\left(x+\frac{c}{b}\right)\right)+d$, so the phase shift is $-\dfrac{c}{b}$.
例：$\sin(2x-\frac{\pi}{2})=\sin\!\left(2\left(x-\frac{\pi}{4}\right)\right)$，相位平移是向右 $\frac{\pi}{4}$，不是 $\frac{\pi}{2}$。
例：$\sin(2x-\frac{\pi}{2})=\sin\!\left(2\left(x-\frac{\pi}{4}\right)\right)$，相位平移是向右 $\frac{\pi}{4}$，不是 $\frac{\pi}{2}$。
📝 练习 E-10 — Read Every Parameter 读出所有参数
Find the amplitude, period, midline, and phase shift: (a) $y=3\sin\!\left(2x-\frac{\pi}{2}\right)+1$ (b) $y=\cos\!\left(\frac{\pi}{6}x+\frac{\pi}{3}\right)-4$. (c) Show that the function in (a) can also be written as $-3\cos(2x)+1$.
(a) $3\sin\!\left(2\left(x-\frac{\pi}{4}\right)\right)+1$: amplitude 3, period $\pi$, midline $y=1$, phase shift $\frac{\pi}{4}$ right.
(b) $\cos\!\left(\frac{\pi}{6}(x+2)\right)-4$: amplitude 1, period $\frac{2\pi}{\pi/6}=12$, midline $y=-4$, phase shift 2 left.
(c) $\sin\!\left(\theta-\frac{\pi}{2}\right)=-\cos\theta$, so $3\sin\!\left(2x-\frac{\pi}{2}\right)=-3\cos(2x)$. Both start at a minimum at $x=0$.
(b) $\cos\!\left(\frac{\pi}{6}(x+2)\right)-4$: amplitude 1, period $\frac{2\pi}{\pi/6}=12$, midline $y=-4$, phase shift 2 left.
(c) $\sin\!\left(\theta-\frac{\pi}{2}\right)=-\cos\theta$, so $3\sin\!\left(2x-\frac{\pi}{2}\right)=-3\cos(2x)$. Both start at a minimum at $x=0$.
8.7 Sinusoidal Function Context and Data Modeling 正弦型函数情境建模
Modeling Strategy 建模步骤
- Find maximum and minimum values from context.
- Compute amplitude $= \dfrac{\max-\min}{2}$.
- Compute midline $= \dfrac{\max+\min}{2}$.
- Determine period $T$ from the situation.
- Determine phase shift $h$ from starting conditions.
- Choose sine or cosine based on starting value.
选 sine 还是 cosine？
- 若在 $t=0$ 时函数从中线出发（上升），选 $\sin$（无需相位移）。
- 若在 $t=0$ 时函数在最大值，选 $\cos$（无需相位移）。
- 若在 $t=0$ 时函数在最小值，选 $-\cos$（无需相位移）。
🤝 一起做 Together T-9 — Ferris Wheel Model
The height of a point on a Ferris wheel has a maximum of $50$ m and a minimum of $10$ m. One full rotation takes $40$ seconds. At $t=0$, the point is at its minimum height. Write a cosine model.
Step 1 — Amplitude and midline.
Amplitude $= \dfrac{50-10}{2} = 20$ m
Midline $= \dfrac{50+10}{2} = 30$ m
Amplitude $= \dfrac{50-10}{2} = 20$ m
Midline $= \dfrac{50+10}{2} = 30$ m
Step 2 — Period and $B$.
$T = 40$ s, so $B = \dfrac{2\pi}{40} = \dfrac{\pi}{20}$
$T = 40$ s, so $B = \dfrac{2\pi}{40} = \dfrac{\pi}{20}$
Step 3 — Phase shift and sign.
At $t=0$, height is minimum ($10$ m). A cosine starts at its maximum, so we need $-\cos$ (which starts at minimum with no phase shift):
$$h(t) = -20\cos\!\left(\frac{\pi}{20}t\right)+30$$
At $t=0$, height is minimum ($10$ m). A cosine starts at its maximum, so we need $-\cos$ (which starts at minimum with no phase shift):
$$h(t) = -20\cos\!\left(\frac{\pi}{20}t\right)+30$$
Verify: $h(0) = -20(1)+30 = 10$ ✓ $h(20) = -20(-1)+30 = 50$ ✓
$h(t) = -20\cos\!\left(\dfrac{\pi}{20}t\right)+30$
📝 练习 E-11 — Tidal Model and Parameter Interpretation
Ocean tides at a harbor have a high of $8$ m and low of $2$ m. The period is $12$ hours. At $t=0$, the water is at high tide.
(a) Write a cosine model for the water height $h(t)$. (b) Interpret the amplitude and midline in this context.
Amplitude $=3$, Midline $=5$, $B=\dfrac{2\pi}{12}=\dfrac{\pi}{6}$, starts at max $\Rightarrow$ use $+\cos$
(a) $h(t) = 3\cos\!\left(\dfrac{\pi}{6}t\right)+5$.
(b) The amplitude $3$ m is the maximum departure from the average water level. The midline $5$ m is the average of high and low tide.
中文：水位围绕平均高度 $5$ m 上下波动，最大偏离量为 $3$ m。
(a) $h(t) = 3\cos\!\left(\dfrac{\pi}{6}t\right)+5$.
(b) The amplitude $3$ m is the maximum departure from the average water level. The midline $5$ m is the average of high and low tide.
中文：水位围绕平均高度 $5$ m 上下波动，最大偏离量为 $3$ m。
Phase Shift from a Data Point & Analyzing the Model 由数据点定相位平移，并分析模型
- Period: the input distance between consecutive maxima (or consecutive minima); half the period separates a max from the next min (CED 3.7.A.1).
- Phase shift: line up an actual input-output pair with the model. If a maximum occurs at $t=t_0$, use $A\cos\big(B(t-t_0)\big)+k$ (CED 3.7.A.3).
- Analysis (FRQ style): min → midline: increasing, concave up; midline → max: increasing, concave down; max → midline: decreasing, concave down; midline → min: decreasing, concave up.
🤝 一起做 Together T-10 — Greenhouse Temperature (FRQ Style) 温室温度
The temperature in a greenhouse reaches a minimum of 12 °C at 4 AM ($t=4$) and a maximum of 28 °C at 4 PM ($t=16$), where $t$ is hours after midnight. The pattern repeats every 24 hours.
(a) Write a sinusoidal model $T(t)$. (b) Find $T(9)$. (c) On which intervals of $0\le t\le24$ is $T$ increasing at an increasing rate? increasing at a decreasing rate?
(d) Find the average rate of change of $T$ from $t=4$ to $t=10$, with units. (e) During which hours of the day is $T\ge24$ °C?
(a) Amplitude $\frac{28-12}{2}=8$, midline $\frac{28+12}{2}=20$, $B=\frac{2\pi}{24}=\frac{\pi}{12}$. The max is at $t=16$, so
$$T(t)=8\cos\!\left(\frac{\pi}{12}(t-16)\right)+20.$$ Check: $T(4)=8\cos(-\pi)+20=12$ ✓.
(b) $T(9)=8\cos\!\left(-\frac{7\pi}{12}\right)+20\approx17.929$ °C.
(c) The midline is crossed halfway between the min and max, at $t=10$ and $t=22$.
Increasing at an increasing rate (concave up): $(4,10)$. Increasing at a decreasing rate (concave down): $(10,16)$.
Increasing at an increasing rate (concave up): $(4,10)$. Increasing at a decreasing rate (concave down): $(10,16)$.
(d) $\dfrac{T(10)-T(4)}{10-4}=\dfrac{20-12}{6}\approx1.333$ °C per hour.
(e) $8\cos\!\left(\frac{\pi}{12}(t-16)\right)+20\ge24 \Rightarrow \cos\!\left(\frac{\pi}{12}(t-16)\right)\ge\frac12 \Rightarrow -\frac{\pi}{3}\le\frac{\pi}{12}(t-16)\le\frac{\pi}{3}$
$\Rightarrow 12\le t\le20$: from noon to 8 PM. The context ($0\le t\le24$, one day) gives just one interval (CED 3.10.A.3).
$\Rightarrow 12\le t\le20$: from noon to 8 PM. The context ($0\le t\le24$, one day) gives just one interval (CED 3.10.A.3).
$T(t)=8\cos\!\left(\frac{\pi}{12}(t-16)\right)+20$; $T(9)\approx17.929$; CU increasing on $(4,10)$, CD increasing on $(10,16)$; AROC $\approx1.333$ °C/h; $12\le t\le20$.
📝 练习 E-12 — Ferris Wheel Analysis 摩天轮模型分析
Use the Ferris wheel model $h(t)=-20\cos\!\left(\frac{\pi}{20}t\right)+30$ from T-9.
(a) During the first rotation ($0\le t\le40$), when is the rider higher than 40 m?
(b) Find the average rate of change of $h$ on $[0,10]$, with units.
(c) On which interval of the first rotation is $h$ increasing and concave down?
(a) $-20\cos\!\left(\frac{\pi}{20}t\right)+30>40 \Rightarrow \cos\!\left(\frac{\pi}{20}t\right)<-\frac12 \Rightarrow \frac{2\pi}{3}<\frac{\pi}{20}t<\frac{4\pi}{3} \Rightarrow 13.333<t<26.667$ seconds.
(b) $\dfrac{h(10)-h(0)}{10}=\dfrac{30-10}{10}=2$ meters per second.
(c) Min at $t=0$, midline at $t=10$, max at $t=20$: increasing and concave down on $(10,20)$.
(b) $\dfrac{h(10)-h(0)}{10}=\dfrac{30-10}{10}=2$ meters per second.
(c) Min at $t=0$, midline at $t=10$, max at $t=20$: increasing and concave down on $(10,20)$.
Sinusoidal Regression 正弦回归
TI-84 · SinReg
- STAT → 1: Edit: inputs in L1, outputs in L2 (at least 4 points, equally spaced inputs work best).
- STAT → CALC → C: SinReg. Xlist: L1, Ylist: L2, Store RegEQ: Y1.
- The calculator gives $y=a\sin(bx+c)+d$. Period $=\dfrac{2\pi}{b}$; compare it with the context. Use 2nd TRACE (CALC) for values and intersections.
🤝 一起做 Together T-11 — Hours of Daylight 日照时长（calculator 可用）
Average hours of daylight in a city by month ($1=$ January):
| month | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| hours | 9.6 | 10.7 | 12.0 | 13.4 | 14.5 | 15.0 | 14.7 | 13.7 | 12.4 | 11.0 | 9.8 | 9.2 |
(a) Why is a sinusoidal model appropriate? (b) Find the sinusoidal regression. (c) Estimate the daylight in mid-April ($x=4.5$). (d) During which part of the year does the model give more than 14 hours of daylight?
(a) Daylight rises to a single maximum in June, falls to a minimum in December, and repeats every year: periodic with period about 12 months.
(b) SinReg: $y\approx2.892\sin(0.506x-1.526)+12.065$ (your calculator may show slightly different digits). Its period $\frac{2\pi}{0.506}\approx12.4$ months is close to the expected 12 months.
(c) $y(4.5)\approx14.034$ hours.
(d) Intersect with $y=14$: $x\approx4.469$ and $x\approx7.782$, so from about mid-April to late July. The model only makes sense for $1\le x\le12$ in this data's context (CED 3.7.A.5).
$y\approx2.892\sin(0.506x-1.526)+12.065$; about 14.034 h in mid-April; above 14 h for $4.469<x<7.782$.
8.8 The Tangent Function 正切函数
$$y = \tan x = \frac{\sin x}{\cos x}$$
$\tan x$ is undefined when $\cos x = 0$, i.e., at $x = \dfrac{\pi}{2}+k\pi$ (vertical asymptotes).
| Property | $y=\tan x$ |
|---|---|
| Domain | $x \neq \dfrac{\pi}{2}+k\pi$ |
| Range | $(-\infty,\infty)$ |
| Period | $\pi$ (not $2\pi$!) |
| Vertical asymptotes | $x = \dfrac{\pi}{2}+k\pi$ |
| Zeros | $x = k\pi$ |
正切与正弦/余弦的区别
- $\tan x$ 的周期是 $\pi$，而 $\sin x$ 和 $\cos x$ 的周期是 $2\pi$。
- $\tan x$ 的值域是 $(-\infty, \infty)$，没有振幅的概念。
- 每两个相邻竖直渐近线之间有且仅有一个完整的正切图像"枝"。
One branch lies between each pair of adjacent vertical asymptotes；每两个相邻竖直渐近线之间是一整枝。
Transformed Tangent 变换后的正切
$$f(x) = A\tan(B(x-h))+k \qquad \text{Period} = \frac{\pi}{|B|}$$
🤝 一起做 Together T-12 — Tangent Period and Asymptotes
Find the period and vertical asymptotes of $y = \tan(2x)$.
Period.
$B=2$, so Period $= \dfrac{\pi}{|2|} = \dfrac{\pi}{2}$.
$B=2$, so Period $= \dfrac{\pi}{|2|} = \dfrac{\pi}{2}$.
Vertical asymptotes.
Set $2x = \dfrac{\pi}{2}+k\pi$: $$x = \frac{\pi}{4}+\frac{k\pi}{2}, \quad k\in\mathbb{Z}$$ e.g., $x = \ldots,-\dfrac{3\pi}{4},-\dfrac{\pi}{4},\dfrac{\pi}{4},\dfrac{3\pi}{4},\ldots$
Set $2x = \dfrac{\pi}{2}+k\pi$: $$x = \frac{\pi}{4}+\frac{k\pi}{2}, \quad k\in\mathbb{Z}$$ e.g., $x = \ldots,-\dfrac{3\pi}{4},-\dfrac{\pi}{4},\dfrac{\pi}{4},\dfrac{3\pi}{4},\ldots$
Period $= \dfrac{\pi}{2}$; asymptotes at $x = \dfrac{\pi}{4}+\dfrac{k\pi}{2}$
📝 练习 E-13 — Reconstructing a Tangent Function
A function of the form $f(x)=\tan(Bx)$, where $B>0$, has adjacent vertical asymptotes at $x=-\dfrac{\pi}{6}$ and $x=\dfrac{\pi}{6}$. Find $B$, the period, and a formula for $f$.
The distance between adjacent asymptotes equals one tangent period: $T=\dfrac{\pi}{3}$. Since $T=\dfrac{\pi}{|B|}$, $|B|=3$. Taking $B=3$, a possible formula is $f(x)=\tan(3x)$.
Tangent as a Slope, Its Concavity, and $a\tan(b(x+c))+d$ 正切的斜率意义、凹凸性与一般变换
- $\tan\theta$ is the slope of the terminal ray; slopes repeat every half revolution, so the period is $\pi$ (CED 3.8.A.1, B.1).
- Between consecutive asymptotes, $\tan x$ is increasing and changes from concave down to concave up; the point of inflection is at the center of each branch (CED 3.8.B.3).
- $y=a\tan(b(x+c))+d$: vertical dilation by $|a|$ (reflection if $a<0$), period $\dfrac{\pi}{|b|}$, phase shift $-c$, and the line containing the points of inflection moves to $y=d$ (CED 3.8.C).
🤝 一起做 Together T-13 — A Fully Transformed Tangent 一般变换后的正切
For $f(x)=2\tan\!\left(\frac{\pi}{4}(x-1)\right)+3$, find the period, the vertical asymptotes, the point of inflection of the branch through $x=1$, and $f(0)$ and $f(2)$. Describe the concavity on that branch.
Period $=\dfrac{\pi}{\pi/4}=4$.
Asymptotes: $\frac{\pi}{4}(x-1)=\pm\frac{\pi}{2}\Rightarrow x=-1$ and $x=3$; in general $x=3+4k$.
Inflection and values: center $(1,3)$; $f(0)=2\tan\!\left(-\frac{\pi}{4}\right)+3=1$, $f(2)=2\tan\frac{\pi}{4}+3=5$.
Red dashed: asymptotes $x=-1,3,7$. Amber dashed: the line $y=3$ containing the points of inflection.
Concavity: on $(-1,3)$, $f$ is increasing; concave down on $(-1,1)$ and concave up on $(1,3)$.
Period 4; asymptotes $x=-1+4k$; inflection $(1,3)$; $f(0)=1$, $f(2)=5$.
8.9 Inverse Trigonometric Functions 反三角函数
为什么要限制定义域？
$\sin x$、$\cos x$、$\tan x$ 在全域上都不是一一对应（one-to-one）的，因此无法直接求反函数。我们必须将定义域限制在一个区间上（主值分支），使函数单调，才能定义反函数。
Principal Branches 主值范围
| Function | Notation | Input domain | Output range |
|---|---|---|---|
| Inverse sine | $\arcsin x$ or $\sin^{-1}x$ | $[-1,1]$ | $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ |
| Inverse cosine | $\arccos x$ or $\cos^{-1}x$ | $[-1,1]$ | $[0,\pi]$ |
| Inverse tangent | $\arctan x$ or $\tan^{-1}x$ | $(-\infty,\infty)$ | $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ |
⚠ 符号警告
$\sin^{-1}x$ 表示反正弦函数（inverse sine），不是 $\dfrac{1}{\sin x}$（倒数）！$\dfrac{1}{\sin x} = \csc x$（余割函数）。
🤝 一起做 Together T-14 — Evaluating Inverse Trig
(a) $\arcsin\dfrac{1}{2}$ (b) $\arccos\left(-\dfrac{1}{2}\right)$
(a) $\arcsin\dfrac{1}{2}$
Ask: which angle $\theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ has $\sin\theta = \dfrac{1}{2}$?
Answer: $\theta = \dfrac{\pi}{6}$
Ask: which angle $\theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ has $\sin\theta = \dfrac{1}{2}$?
Answer: $\theta = \dfrac{\pi}{6}$
(b) $\arccos\left(-\dfrac{1}{2}\right)$
Ask: which angle $\theta\in[0,\pi]$ has $\cos\theta = -\dfrac{1}{2}$?
Reference angle for $\cos = \dfrac{1}{2}$ is $\dfrac{\pi}{3}$. In $[0,\pi]$, cosine is negative in QII:
$\theta = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$
Ask: which angle $\theta\in[0,\pi]$ has $\cos\theta = -\dfrac{1}{2}$?
Reference angle for $\cos = \dfrac{1}{2}$ is $\dfrac{\pi}{3}$. In $[0,\pi]$, cosine is negative in QII:
$\theta = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$
(a) $\dfrac{\pi}{6}$ (b) $\dfrac{2\pi}{3}$
📝 练习 E-14 — Principal Value of an Inverse Function
Evaluate exactly: $\arcsin\!\left(-\dfrac{\sqrt3}{2}\right)$. Explain why $\dfrac{4\pi}{3}$ is not the output.
The output of $\arcsin$ must lie in $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$. In that interval, the angle whose sine is $-\dfrac{\sqrt3}{2}$ is $-\dfrac{\pi}{3}$. Therefore,
$\arcsin\!\left(-\dfrac{\sqrt3}{2}\right)=-\dfrac{\pi}{3}$. Although $\sin(4\pi/3)=-\sqrt3/2$, $4\pi/3$ is outside the principal range.
$\arcsin\!\left(-\dfrac{\sqrt3}{2}\right)=-\dfrac{\pi}{3}$. Although $\sin(4\pi/3)=-\sqrt3/2$, $4\pi/3$ is outside the principal range.
Graphs of Inverse Trig Functions & Compositions 反三角函数图像与复合
$y=\arcsin x$: domain $[-1,1]$, range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$
$y=\arccos x$: domain $[-1,1]$, range $[0,\pi]$
$y=\arctan x$: all reals → $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, HAs $y=\pm\frac{\pi}{2}$
理解
反三角函数的输入是一个比值（三角函数值），输出是一个角（CED 3.9.A.1）。$\sin(\arcsin x)=x$ 总成立（$-1\le x\le1$），但 $\arcsin(\sin\theta)=\theta$ 只在 $\theta$ 属于主值区间时成立。
🤝 一起做 Together T-15 — Compositions 复合计算
Evaluate exactly: (a) $\arcsin\!\left(\sin\frac{5\pi}{6}\right)$ (b) $\arccos\!\left(\cos\left(-\frac{\pi}{3}\right)\right)$ (c) $\sin\!\left(\arccos\frac35\right)$ (d) Show that $\cos(\arcsin x)=\sqrt{1-x^2}$ for $-1\le x\le1$.
(a) $\sin\frac{5\pi}{6}=\frac12$, and $\arcsin\frac12=\frac{\pi}{6}$ (not $\frac{5\pi}{6}$, which is outside $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$).
(b) $\cos\left(-\frac{\pi}{3}\right)=\frac12$, and $\arccos\frac12=\frac{\pi}{3}$.
(c) Let $\theta=\arccos\frac35\in[0,\pi]$, so $\cos\theta=\frac35$. Then $\sin\theta=\sqrt{1-\frac{9}{25}}=\frac45$ (positive, since $\theta\in[0,\pi]$).
(d) Let $\theta=\arcsin x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Then $\cos\theta\ge0$ and $\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-x^2}$. (So $\arcsin x=\arccos\sqrt{1-x^2}$ when $0\le x\le1$, CED 3.12.A.2.)
(a) $\frac{\pi}{6}$ · (b) $\frac{\pi}{3}$ · (c) $\frac45$
📝 练习 E-15 — More Compositions 复合练习
Evaluate exactly: (a) $\arccos\!\left(\sin\frac{7\pi}{6}\right)$ (b) $\tan\!\left(\arcsin\left(-\frac12\right)\right)$ (c) $\sec(\arctan2)$
(a) $\sin\frac{7\pi}{6}=-\frac12$, $\arccos\left(-\frac12\right)=\frac{2\pi}{3}$. (b) $\arcsin\left(-\frac12\right)=-\frac{\pi}{6}$, $\tan\left(-\frac{\pi}{6}\right)=-\frac{\sqrt3}{3}$.
(c) $\theta=\arctan2$: a right triangle with legs 1 and 2 has hypotenuse $\sqrt5$, so $\cos\theta=\frac{1}{\sqrt5}$ and $\sec\theta=\sqrt5$.
(c) $\theta=\arctan2$: a right triangle with legs 1 and 2 has hypotenuse $\sqrt5$, so $\cos\theta=\frac{1}{\sqrt5}$ and $\sec\theta=\sqrt5$.
8.10 Trigonometric Equations and Inequalities 三角方程与不等式
General Process 解方程步骤
- Isolate the trig expression.
- Find the reference angle using special values or inverse trig.
- Use quadrant signs to find all solutions in the requested interval.
- Add periodicity for the general solution.
Periodic General Solutions 通解形式
- If $\sin\theta = c$: two families, period $2\pi$
- If $\cos\theta = c$: two families, period $2\pi$
- If $\tan\theta = c$: one family, period $\pi$
🤝 一起做 Together T-16 — Solving a Cosine Equation
Find all exact solutions on $[0,2\pi)$: $2\cos x - 1 = 0$.
Step 1 — Isolate.
$2\cos x = 1 \Rightarrow \cos x = \dfrac{1}{2}$
$2\cos x = 1 \Rightarrow \cos x = \dfrac{1}{2}$
Step 2 — Reference angle.
$\cos^{-1}\!\left(\dfrac{1}{2}\right) = \dfrac{\pi}{3}$
$\cos^{-1}\!\left(\dfrac{1}{2}\right) = \dfrac{\pi}{3}$
Step 3 — Find all angles where $\cos$ is positive.
$\cos$ is positive in QI and QIV.
QI: $x = \dfrac{\pi}{3}$ QIV: $x = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}$
$\cos$ is positive in QI and QIV.
QI: $x = \dfrac{\pi}{3}$ QIV: $x = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}$
$x = \dfrac{\pi}{3}$ and $x = \dfrac{5\pi}{3}$
🤝 一起做 Together T-17 — Sine General Solution
Find the general solution: $\sin x = -\dfrac{\sqrt{2}}{2}$.
Reference angle.
$\sin^{-1}\!\left(\dfrac{\sqrt{2}}{2}\right) = \dfrac{\pi}{4}$
$\sin^{-1}\!\left(\dfrac{\sqrt{2}}{2}\right) = \dfrac{\pi}{4}$
$\sin$ is negative in QIII and QIV.
QIII: $x = \pi + \dfrac{\pi}{4} = \dfrac{5\pi}{4}$
QIV: $x = 2\pi - \dfrac{\pi}{4} = \dfrac{7\pi}{4}$
QIII: $x = \pi + \dfrac{\pi}{4} = \dfrac{5\pi}{4}$
QIV: $x = 2\pi - \dfrac{\pi}{4} = \dfrac{7\pi}{4}$
Add period $2\pi$ for general solution.
$x = \dfrac{5\pi}{4}+2k\pi$ or $x = \dfrac{7\pi}{4}+2k\pi$, $k\in\mathbb{Z}$
$x = \dfrac{5\pi}{4}+2k\pi$ or $x = \dfrac{7\pi}{4}+2k\pi$, $k\in\mathbb{Z}$
$x = \dfrac{5\pi}{4}+2k\pi$ or $x = \dfrac{7\pi}{4}+2k\pi$
🤝 一起做 Together T-18 — Trigonometric Inequality
Solve on $[0,2\pi)$: $\cos x > \dfrac{1}{2}$.
Step 1 — Find boundary angles.
$\cos x = \dfrac{1}{2}$ gives $x = \dfrac{\pi}{3}$ (QI) and $x = \dfrac{5\pi}{3}$ (QIV).
$\cos x = \dfrac{1}{2}$ gives $x = \dfrac{\pi}{3}$ (QI) and $x = \dfrac{5\pi}{3}$ (QIV).
Step 2 — Determine where $\cos x > \dfrac{1}{2}$.
$\cos x$ is greater than $\dfrac{1}{2}$ between the two boundary angles — specifically the arc near $x=0$ where cosine is large.
$\cos 0 = 1 > \dfrac{1}{2}$ ✓, so the solution is $x\in\left[0,\dfrac{\pi}{3}\right)\cup\left(\dfrac{5\pi}{3},2\pi\right)$.
$\cos x$ is greater than $\dfrac{1}{2}$ between the two boundary angles — specifically the arc near $x=0$ where cosine is large.
$\cos 0 = 1 > \dfrac{1}{2}$ ✓, so the solution is $x\in\left[0,\dfrac{\pi}{3}\right)\cup\left(\dfrac{5\pi}{3},2\pi\right)$.
$x\in\left[0,\dfrac{\pi}{3}\right)\cup\left(\dfrac{5\pi}{3},2\pi\right)$
📝 练习 E-16 — Tangent Equation
Find all solutions on $[0,\pi)$: $\tan x = \sqrt{3}$.
Reference angle: $\arctan(\sqrt{3}) = \dfrac{\pi}{3}$. $\tan>0$ in QI only on $[0,\pi)$.
$x = \dfrac{\pi}{3}$
$x = \dfrac{\pi}{3}$
More Types of Trig Equations 更多类型的三角方程
- Non-special values: use the inverse function (calculator, radian mode) for one solution, then use symmetry: $\sin x=c \Rightarrow x=\arcsin c$ or $\pi-\arcsin c$; $\cos x=c \Rightarrow x=\pm\arccos c$ (+ $2\pi k$); $\tan x=c \Rightarrow x=\arctan c+\pi k$.
- $\sin(bx)=c$: solve for $bx$ over the stretched interval ($[0,2b\pi)$ if $x\in[0,2\pi)$), then divide by $b$. Expect more solutions.
- Quadratic type: factor like a quadratic in $\sin x$ or $\cos x$. Never divide by $\sin x$ or $\cos x$ (you lose solutions).
🤝 一起做 Together T-19 — Four Kinds of Equations 四种方程
Solve on $[0,2\pi)$: (a) $3\sin x=1$ (b) $\sin2x=\frac12$ (c) $2\sin^2x-\sin x-1=0$ (d) $\cos x=-0.3$
(a) $\sin x=\frac13$: $x=\arcsin\frac13\approx0.340$ and $x=\pi-0.340\approx2.802$ (both where sine is positive).
(b) Let $u=2x\in[0,4\pi)$: $\sin u=\frac12 \Rightarrow u=\frac{\pi}{6},\frac{5\pi}{6},\frac{13\pi}{6},\frac{17\pi}{6}$. So $x=\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}$ (four solutions).
(c) $(2\sin x+1)(\sin x-1)=0$: $\sin x=1\Rightarrow x=\frac{\pi}{2}$; $\sin x=-\frac12\Rightarrow x=\frac{7\pi}{6},\frac{11\pi}{6}$.
(d) $x=\arccos(-0.3)\approx1.875$ and $x=2\pi-1.875\approx4.408$.
(a) $0.340,\ 2.802$ · (b) $\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}$ · (c) $\frac{\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}$ · (d) $1.875,\ 4.408$
📝 练习 E-17 — Equation Practice 方程练习
Solve on $[0,2\pi)$: (a) $\tan x=2.5$ (b) $2\cos^2x=\cos x$ (c) $\cos\!\left(\frac{x}{2}\right)=\frac{\sqrt2}{2}$
(a) $x=\arctan2.5\approx1.190$ and $1.190+\pi\approx4.332$.
(b) $\cos x(2\cos x-1)=0$ (don't divide by $\cos x$!): $\cos x=0\Rightarrow\frac{\pi}{2},\frac{3\pi}{2}$; $\cos x=\frac12\Rightarrow\frac{\pi}{3},\frac{5\pi}{3}$.
(c) $u=\frac{x}{2}\in[0,\pi)$: $\cos u=\frac{\sqrt2}{2}\Rightarrow u=\frac{\pi}{4}$ only, so $x=\frac{\pi}{2}$ (fewer solutions, because the interval for $u$ is shorter).
(b) $\cos x(2\cos x-1)=0$ (don't divide by $\cos x$!): $\cos x=0\Rightarrow\frac{\pi}{2},\frac{3\pi}{2}$; $\cos x=\frac12\Rightarrow\frac{\pi}{3},\frac{5\pi}{3}$.
(c) $u=\frac{x}{2}\in[0,\pi)$: $\cos u=\frac{\sqrt2}{2}\Rightarrow u=\frac{\pi}{4}$ only, so $x=\frac{\pi}{2}$ (fewer solutions, because the interval for $u$ is shorter).
8.11 Secant, Cosecant, and Cotangent 正割、余割、余切
Reciprocal Trigonometric Functions 倒数三角函数
$$\sec\theta = \frac{1}{\cos\theta}, \qquad \csc\theta = \frac{1}{\sin\theta}, \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$$
Domains 定义域
- $\sec\theta$: undefined when $\cos\theta=0$, i.e., $\theta = \dfrac{\pi}{2}+k\pi$
- $\csc\theta$ and $\cot\theta$: undefined when $\sin\theta=0$, i.e., $\theta = k\pi$
名称对应记忆
- secant = second, 对应 cos 的倒数（$\sec=1/\cos$）
- csc = cosecant，对应 sin 的倒数（$\csc=1/\sin$）
- cot = cotangent，对应 tan 的倒数（$\cot=1/\tan$）
🤝 一起做 Together T-20 — Evaluating Reciprocal Functions
Evaluate exactly: $\sec\dfrac{\pi}{3}$, $\csc\dfrac{\pi}{6}$, $\cot\dfrac{\pi}{4}$.
$\sec\dfrac{\pi}{3}$
$\cos\dfrac{\pi}{3} = \dfrac{1}{2}$, so $\sec\dfrac{\pi}{3} = \dfrac{1}{1/2} = 2$
$\cos\dfrac{\pi}{3} = \dfrac{1}{2}$, so $\sec\dfrac{\pi}{3} = \dfrac{1}{1/2} = 2$
$\csc\dfrac{\pi}{6}$
$\sin\dfrac{\pi}{6} = \dfrac{1}{2}$, so $\csc\dfrac{\pi}{6} = \dfrac{1}{1/2} = 2$
$\sin\dfrac{\pi}{6} = \dfrac{1}{2}$, so $\csc\dfrac{\pi}{6} = \dfrac{1}{1/2} = 2$
$\cot\dfrac{\pi}{4}$
$\tan\dfrac{\pi}{4} = 1$, so $\cot\dfrac{\pi}{4} = \dfrac{1}{1} = 1$
$\tan\dfrac{\pi}{4} = 1$, so $\cot\dfrac{\pi}{4} = \dfrac{1}{1} = 1$
$\sec\dfrac{\pi}{3}=2$, $\csc\dfrac{\pi}{6}=2$, $\cot\dfrac{\pi}{4}=1$
📝 练习 E-18 — Reciprocal Values in Other Quadrants
Find $\csc\dfrac{7\pi}{6}$ and $\sec\dfrac{2\pi}{3}$ exactly.
$\sin\dfrac{7\pi}{6} = -\dfrac{1}{2}$ (QIII, ref angle $\dfrac{\pi}{6}$) $\Rightarrow \csc\dfrac{7\pi}{6} = -2$
$\cos\dfrac{2\pi}{3} = -\dfrac{1}{2}$ (QII, ref angle $\dfrac{\pi}{3}$) $\Rightarrow \sec\dfrac{2\pi}{3} = -2$
$\cos\dfrac{2\pi}{3} = -\dfrac{1}{2}$ (QII, ref angle $\dfrac{\pi}{3}$) $\Rightarrow \sec\dfrac{2\pi}{3} = -2$
Graphs of Secant, Cosecant, and Cotangent 正割、余割、余切的图像
$y=\sec x$ (dashed: $\cos x$). VAs where $\cos x=0$
$y=\csc x$ (dashed: $\sin x$). VAs where $\sin x=0$
$y=\cot x$: VAs where $\tan x=0$ ($x=k\pi$)
- $\sec x$ and $\csc x$ have range $(-\infty,-1]\cup[1,\infty)$: where $\cos x$ or $\sin x$ equals $\pm1$, the reciprocal also equals $\pm1$ (a turning point) (CED 3.11.A.3).
- $\cot x$ has period $\pi$, range all reals, and is decreasing between consecutive asymptotes (CED 3.11.A.5).
📝 练习 E-19 — Reading Reciprocal Graphs 读倒数函数图像
(a) List the vertical asymptotes of $y=\csc x$ on $[0,2\pi]$. (b) What is the minimum value of $\sec x$ on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, and where? (c) Solve $\sec x=2$ on $[0,2\pi)$.
(a) $x=0,\ \pi,\ 2\pi$. (b) 1, at $x=0$ (where $\cos x=1$ is largest). (c) $\cos x=\frac12\Rightarrow x=\frac{\pi}{3},\ \frac{5\pi}{3}$.
8.12 Equivalent Representations — Trigonometric Identities 三角恒等式
An identity（恒等式）is an equation that is true for all values in the common domain of the expressions.
Reciprocal 倒数
$\sec\theta = \dfrac{1}{\cos\theta}$$\csc\theta = \dfrac{1}{\sin\theta}$
$\cot\theta = \dfrac{1}{\tan\theta}$
Quotient 商
$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$$\cot\theta = \dfrac{\cos\theta}{\sin\theta}$
Pythagorean 勾股
$\sin^2\theta+\cos^2\theta=1$$1+\tan^2\theta=\sec^2\theta$
$1+\cot^2\theta=\csc^2\theta$
Even–Odd 奇偶性
$\sin(-\theta)=-\sin\theta$ (odd / 奇函数)$\cos(-\theta)=\cos\theta$ (even / 偶函数)
$\tan(-\theta)=-\tan\theta$ (odd / 奇函数)
恒等式证明策略
- 选择较复杂的一边进行化简。
- 将所有函数转化为 $\sin$ 和 $\cos$ 后再化简。
- 寻找可以用勾股恒等式替换的 $\sin^2+\cos^2$、$\tan^2+1$等。
- 两边不能直接做等式变形——只能化简一边使之等于另一边。
🤝 一起做 Together T-21 — Simplifying with Identities
Rewrite in terms of sine and cosine, then simplify: $\sec\theta\cdot\tan\theta\cdot\cos^2\theta$.
Step 1 — Substitute definitions.
$$\sec\theta\cdot\tan\theta\cdot\cos^2\theta = \frac{1}{\cos\theta}\cdot\frac{\sin\theta}{\cos\theta}\cdot\cos^2\theta$$
$$\sec\theta\cdot\tan\theta\cdot\cos^2\theta = \frac{1}{\cos\theta}\cdot\frac{\sin\theta}{\cos\theta}\cdot\cos^2\theta$$
Step 2 — Cancel.
$$= \frac{\sin\theta\cdot\cos^2\theta}{\cos^2\theta} = \sin\theta$$
$$= \frac{\sin\theta\cdot\cos^2\theta}{\cos^2\theta} = \sin\theta$$
$\sin\theta$
🤝 一起做 Together T-22 — Verifying an Identity
Verify the identity: $\dfrac{1-\cos^2\theta}{\sin\theta} = \sin\theta$.
Work on the left side only.
By Pythagorean identity: $1-\cos^2\theta = \sin^2\theta$ $$\frac{1-\cos^2\theta}{\sin\theta} = \frac{\sin^2\theta}{\sin\theta} = \sin\theta \checkmark$$
By Pythagorean identity: $1-\cos^2\theta = \sin^2\theta$ $$\frac{1-\cos^2\theta}{\sin\theta} = \frac{\sin^2\theta}{\sin\theta} = \sin\theta \checkmark$$
LHS $= \sin\theta =$ RHS ✓
📝 练习 E-20 — Identity Practice
Simplify $\dfrac{\tan^2\theta+1}{\sec\theta}$.
Use $\tan^2\theta+1=\sec^2\theta$:
$\dfrac{\sec^2\theta}{\sec\theta} = \sec\theta$
$\dfrac{\sec^2\theta}{\sec\theta} = \sec\theta$
Sum, Difference & Double-Angle Identities 和角、差角、二倍角公式
$$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta \qquad \cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$$
$$\sin2\theta=2\sin\theta\cos\theta \qquad \cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta$$
二倍角公式就是和角公式中令 $\alpha=\beta=\theta$（CED 3.12.B.3）。注意 cos 的和角公式中间是"减号"。选哪种 $\cos2\theta$ 形式，看方程里剩下的是 $\sin$ 还是 $\cos$。
🤝 一起做 Together T-23 — Using the Sum and Double-Angle Identities 使用和角与二倍角公式
(a) Find $\sin75^\circ$ exactly. (b) Find $\cos105^\circ$ exactly. (c) If $\sin\theta=\frac35$ and $\theta$ is in Quadrant II, find $\sin2\theta$ and $\cos2\theta$. (d) Show that $\sin\!\left(\theta+\frac{\pi}{2}\right)=\cos\theta$.
(a) $\sin(45^\circ+30^\circ)=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot\frac12=\dfrac{\sqrt6+\sqrt2}{4}$.
(b) $\cos(60^\circ+45^\circ)=\frac12\cdot\frac{\sqrt2}{2}-\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}=\dfrac{\sqrt2-\sqrt6}{4}$.
(c) In QII, $\cos\theta=-\frac45$. $\sin2\theta=2\cdot\frac35\cdot\left(-\frac45\right)=-\frac{24}{25}$; $\cos2\theta=1-2\left(\frac35\right)^2=\frac{7}{25}$.
(d) $\sin\theta\cos\frac{\pi}{2}+\cos\theta\sin\frac{\pi}{2}=\sin\theta\cdot0+\cos\theta\cdot1=\cos\theta$ ✓ (cosine is a shifted sine).
(a) $\frac{\sqrt6+\sqrt2}{4}$ · (b) $\frac{\sqrt2-\sqrt6}{4}$ · (c) $-\frac{24}{25}$, $\frac{7}{25}$
🤝 一起做 Together T-24 — Solving with an Equivalent Form 用等价形式解方程
Solve on $[0,2\pi)$: (a) $\sin2x=\cos x$ (b) $\cos2x=\sin x$
(a) $2\sin x\cos x-\cos x=0 \Rightarrow \cos x(2\sin x-1)=0$.
$\cos x=0\Rightarrow x=\frac{\pi}{2},\frac{3\pi}{2}$; $\sin x=\frac12\Rightarrow x=\frac{\pi}{6},\frac{5\pi}{6}$.
$\cos x=0\Rightarrow x=\frac{\pi}{2},\frac{3\pi}{2}$; $\sin x=\frac12\Rightarrow x=\frac{\pi}{6},\frac{5\pi}{6}$.
(b) Use $\cos2x=1-2\sin^2x$ (everything in $\sin x$): $2\sin^2x+\sin x-1=0 \Rightarrow (2\sin x-1)(\sin x+1)=0$.
$\sin x=\frac12\Rightarrow\frac{\pi}{6},\frac{5\pi}{6}$; $\sin x=-1\Rightarrow\frac{3\pi}{2}$.
$\sin x=\frac12\Rightarrow\frac{\pi}{6},\frac{5\pi}{6}$; $\sin x=-1\Rightarrow\frac{3\pi}{2}$.
(a) $\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}$ · (b) $\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}$
📝 练习 E-21 — Identity Practice 公式练习
(a) Find $\sin\frac{\pi}{12}$ exactly. (b) Show that $(\sin x+\cos x)^2=1+\sin2x$. (c) Rewrite $f(x)=6\sin x\cos x$ as a single sinusoid and state its amplitude and period. (d) Solve $\cos2x+\cos x=0$ on $[0,2\pi)$.
(a) $\sin\left(\frac{\pi}{3}-\frac{\pi}{4}\right)=\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}-\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6-\sqrt2}{4}$.
(b) $\sin^2x+2\sin x\cos x+\cos^2x=1+\sin2x$ ✓.
(c) $6\sin x\cos x=3\sin2x$: amplitude 3, period $\pi$. The equivalent form reveals information that was hidden (CED 3.12.C.1).
(d) $2\cos^2x-1+\cos x=0 \Rightarrow (2\cos x-1)(\cos x+1)=0$: $x=\frac{\pi}{3},\ \pi,\ \frac{5\pi}{3}$.
(b) $\sin^2x+2\sin x\cos x+\cos^2x=1+\sin2x$ ✓.
(c) $6\sin x\cos x=3\sin2x$: amplitude 3, period $\pi$. The equivalent form reveals information that was hidden (CED 3.12.C.1).
(d) $2\cos^2x-1+\cos x=0 \Rightarrow (2\cos x-1)(\cos x+1)=0$: $x=\frac{\pi}{3},\ \pi,\ \frac{5\pi}{3}$.
Chapter 8 Summary 本章总结
| Topic 主题 | Key Formula / Idea 核心公式 |
|---|---|
| Period | $f(x+T)=f(x)$; read from graph |
| Radian conversion | $180°=\pi$ rad; multiply to convert |
| Unit circle | $\cos\theta=x$, $\sin\theta=y$, $\tan\theta=y/x$ |
| Special angles | $0,\,\pi/6,\,\pi/4,\,\pi/3,\,\pi/2$ and their values |
| Sinusoidal form | $A\sin(B(x-h))+k$; amplitude $|A|$, period $2\pi/|B|$ |
| Tangent | Period $\pi$; asymptotes at $\pi/2+k\pi$ |
| Inverse trig | arcsin: $[-\pi/2,\pi/2]$; arccos: $[0,\pi]$; arctan: $(-\pi/2,\pi/2)$ |
| Pythagorean identity | $\sin^2\theta+\cos^2\theta=1$ → two derived forms |
| Equation strategy | Isolate → reference angle → quadrant signs → add period |
| Radians & circles | $s=r\theta$; $P=(r\cos\theta, r\sin\theta)$; $\tan\theta$ = slope of terminal ray |
| Phase shift | Factor out $b$: $\sin(bx+c)$ shifts by $-\frac{c}{b}$; $\cos\theta=\sin(\theta+\frac{\pi}{2})$ |
| Concavity | Sinusoids: inflection points on the midline; tangent: CD → CU on each branch |
| Modeling | Max at $t_0$ → $A\cos(B(t-t_0))+k$; SinReg; analyze inc/dec and concavity by quarter period |
| Inverse trig | $\arcsin(\sin\theta)=\theta$ only on the principal range; $\cos(\arcsin x)=\sqrt{1-x^2}$ |
| sec / csc / cot | Range of sec, csc: $(-\infty,-1]\cup[1,\infty)$; cot decreasing between asymptotes |
| Sum & double angle | $\sin(\alpha\pm\beta)$, $\cos(\alpha\pm\beta)$, $\sin2\theta=2\sin\theta\cos\theta$, $\cos2\theta=1-2\sin^2\theta$ |
考试重点提示
- AP 考试中三角函数常以建模题形式出现：给出情境，写出 $A\sin(B(x-h))+k$ 或 $A\cos(\cdots)$。
- 必须熟记单位圆特殊值：$30°, 45°, 60°$ 在四个象限的 sin/cos/tan 值。
- 恒等式化简：优先将所有函数化为 $\sin$ 和 $\cos$，再利用勾股恒等式。
- 反三角函数的输出范围一定要记住（arcsin 输出在 $[-\pi/2,\pi/2]$）。
- FRQ 第 3 题常考：由最大值出现的时刻定相位平移，按四分之一周期判断"增/减 × 凹凸"，并在情境定义域内解不等式。
- $\sin(bx)=c$ 先把区间拉伸到 $[0,2b\pi)$ 再求；二次型方程不要约掉 $\sin x$ 或 $\cos x$。
- 和角公式：cos 中间是减号；二倍角 $\cos2\theta$ 有三种形式，按需选择。
Chapter 8 Review 章末习题
Part A选择题 · 不用计算器 Multiple Choice · No Calculator
R-1Coterminal Angles — 同终边角改编自 Practice Exam 2 · MCQ Part A #19
In the $xy$-plane, two different angles $\alpha$ and $\beta$ are in standard position and share a terminal ray. Which of the following gives possible values for $\alpha$ and $\beta$?
(A) $\alpha=\dfrac{2\pi}{3}$ and $\beta=-\dfrac{2\pi}{3}$
(B) $\alpha=\dfrac{5\pi}{6}$ and $\beta=-\dfrac{7\pi}{6}$
(C) $\alpha=\dfrac{\pi}{4}$ and $\beta=\dfrac{3\pi}{4}$
(D) $\alpha=\dfrac{5\pi}{3}$ and $\beta=-\dfrac{5\pi}{3}$
答案：(B) 同终边的角相差 $2\pi$ 的整数倍：$\dfrac{5\pi}{6}-\left(-\dfrac{7\pi}{6}\right)=\dfrac{12\pi}{6}=2\pi$ ✓。
(A) 相差 $\dfrac{4\pi}{3}$——它们关于 $x$ 轴对称，不是同终边；(C) 相差 $\dfrac{\pi}{2}$；(D) 相差 $\dfrac{10\pi}{3}$，不是 $2\pi$ 的整数倍。
R-2A Point on a Circle of Radius 4 — 半径为 4 的圆上的点改编自 2024 AP 真题（亚太卷）· MCQ Part A #5
A circle of radius $4$ is centered at the origin. An angle of measure $\dfrac{5\pi}{6}$ radians is in standard position. What are the coordinates of the point $P$ where the terminal ray of the angle intersects the circle?
(A) $(-2,\ 2\sqrt3)$
(B) $(2\sqrt3,\ -2)$
(C) $\left(-\dfrac{\sqrt3}{2},\ \dfrac12\right)$
(D) $(-2\sqrt3,\ 2)$
答案：(D) 半径为 $r$ 的圆上：$P=(r\cos\theta,\ r\sin\theta)$。$\cos\dfrac{5\pi}{6}=-\dfrac{\sqrt3}{2}$，$\sin\dfrac{5\pi}{6}=\dfrac12$，所以 $P=\left(4\cdot\left(-\dfrac{\sqrt3}2\right),\ 4\cdot\dfrac12\right)=(-2\sqrt3,\ 2)$。
(A) $\cos$、$\sin$ 弄反了；(B) 在第四象限，而 $\dfrac{5\pi}{6}$ 的终边在第二象限；(C) 是单位圆上的点，忘了乘半径 4。
R-3Arc Length and Radius — 弧长与半径改编自 Practice Exam 3 · MCQ Part A #16
An angle of measure $\dfrac{2\pi}{3}$ radians is a central angle of a circle and subtends an arc of length $10$ units. What is the radius of the circle?
(A) $\dfrac{1}{12}$
(B) $\dfrac{\pi}{15}$
(C) $\dfrac{15}{\pi}$
(D) $\dfrac{20\pi}{3}$
答案：(C) 弧度的定义：$\theta=\dfrac sr$，所以 $r=\dfrac s\theta=\dfrac{10}{2\pi/3}=\dfrac{15}{\pi}\approx4.77$。
(A) 用了角度 $120^\circ$：$\dfrac{10}{120}$；(B) 是 $\dfrac\theta s$；(D) 是 $s\cdot\theta$。
R-4Building a Sinusoid from a Description — 由描述写正弦型函数改编自 Practice Exam 3 · MCQ Part A #8
The function $H$ models a periodic phenomenon. The maximum value of $H$ is $14$, which occurs at $t=0$. The minimum value of $H$ is $2$, and the period of $H$ is $8$. Which of the following could be an expression for $H(t)$?
(A) $H(t)=6\cos\left(\dfrac{\pi}{4}t\right)+8$
(B) $H(t)=6\sin\left(\dfrac{\pi}{4}t\right)+8$
(C) $H(t)=6\cos(8t)+8$
(D) $H(t)=12\cos\left(\dfrac{\pi}{4}t\right)+2$
答案：(A) 中线 $d=\dfrac{14+2}{2}=8$，振幅 $a=\dfrac{14-2}{2}=6$，$b=\dfrac{2\pi}{8}=\dfrac\pi4$。$t=0$ 时取最大值，用余弦，因为 $\cos0=1$：$H(t)=6\cos\left(\dfrac\pi4t\right)+8$。
(B) $H(0)=8$，从中线出发；(C) 把周期 8 当成了 $b$；(D) 振幅和中线都不对：最大值 14，最小值 $-10$。
R-5Factor, Don't Divide — 因式分解，别约掉 cosθ改编自 Practice Exam 1 · MCQ Part A #25
What are all values of $\theta$, for $0\le\theta<2\pi$, where $2\cos^2\theta=\cos\theta$?
(A) $\dfrac{\pi}{3}$ and $\dfrac{5\pi}{3}$ only
(B) $\dfrac{\pi}{2}$, $\dfrac{3\pi}{2}$, $\dfrac{\pi}{6}$, and $\dfrac{11\pi}{6}$
(C) $\dfrac{\pi}{3}$, $\dfrac{\pi}{2}$, $\dfrac{3\pi}{2}$, and $\dfrac{5\pi}{3}$
(D) $0$, $\pi$, $\dfrac{\pi}{3}$, and $\dfrac{2\pi}{3}$
答案：(C) 移项因式分解：$2\cos^2\theta-\cos\theta=\cos\theta(2\cos\theta-1)=0$。$\cos\theta=0\Rightarrow\theta=\dfrac\pi2,\dfrac{3\pi}2$；$\cos\theta=\dfrac12\Rightarrow\theta=\dfrac\pi3,\dfrac{5\pi}3$。
(A) 两边同除以 $\cos\theta$，丢掉了 $\cos\theta=0$ 的解；(B) 把 $\cos\theta=\tfrac12$ 和 $\sin\theta=\tfrac12$ 搞混了。
R-6A Difference Identity — 差角公式改编自 2024 AP 真题（亚太卷）· MCQ Part A #23
Which of the following is an equivalent form of the expression $\cos(-2x)\cos x-\sin(2x)\sin(-x)$?
(A) $\cos x$
(B) $\cos(3x)$
(C) $-\cos x$
(D) $\sin(3x)$
答案：(A) 先用奇偶性：$\cos(-2x)=\cos2x$（偶），$\sin(-x)=-\sin x$（奇）。原式 $=\cos2x\cos x+\sin2x\sin x=\cos(2x-x)=\cos x$（差角公式 $\cos(A-B)=\cos A\cos B+\sin A\sin B$）。
(B) 没处理 $\sin(-x)$ 的负号，当成了 $\cos(2x+x)$。
Part B选择题 · 可用计算器 Multiple Choice · Calculator Allowed
R-7Zeros of a Shifted Cosine — 正弦型函数的零点改编自 Practice Exam 1 · MCQ Part B #11
The function $g$ is given by $g(x)=2\cos(1.5x-0.4)+1$. What are all the zeros of $g$ on the interval $0\le x\le\pi$?
(A) $1.130$ and $2.526$
(B) $1.396$ and $2.793$
(C) $1.663$ and $3.059$
(D) $1.663$ only
答案：(C) 用计算器（弧度模式）画 $Y_1=g(x)$，在 $[0,\pi]$ 上找零点：$x\approx1.663$ 和 $x\approx3.059$。也可以代数解：$\cos(1.5x-0.4)=-\tfrac12\Rightarrow1.5x-0.4=\tfrac{2\pi}3$ 或 $\tfrac{4\pi}3$，再解出 $x$。
(A) 把 $-0.4$ 移项时符号错了；(B) 漏了 $-0.4$；(D) 漏掉了第二个零点（$3.059<\pi\approx3.142$）。
R-8An Inverse Tangent Model — 反正切模型改编自 Practice Exam 2 · MCQ Part B #11
The number of people, in thousands, who have entered a festival $t$ hours after it opens is modeled by $N(t)=5.2\arctan(0.6t)$, for $t\ge0$. According to the model, at what time $t$ has the $6000$th person entered the festival?
(A) $0.034$
(B) $1.923$
(C) $2.258$
(D) $3.763$
答案：(D) 解 $5.2\arctan(0.6t)=6$：$\arctan(0.6t)=\dfrac{6}{5.2}$，$0.6t=\tan\left(\dfrac{6}{5.2}\right)\approx2.258$，$t\approx3.763$ 小时。（或用计算器求 $Y_1=N(x)$ 与 $Y_2=6$ 的交点。）
(A) 计算器在角度模式下算 $\tan$ 得到的；(B) 当成了线性方程；(C) 忘了除以 0.6。
R-9Counting Complete Cycles — 完整周期的个数改编自 Practice Exam 1 · MCQ Part B #9
The function $f$ is given by $f(x)=3\sin(6x)-\cos(4x)$. Using the period of $f$, which of the following is the number of complete cycles of the graph of $f$ on the interval $0\le x\le500$?
(A) $159$
(B) $318$
(C) $477$
(D) $954$
答案：(A) $\sin6x$ 的周期是 $\dfrac{2\pi}6=\dfrac\pi3$，$\cos4x$ 的周期是 $\dfrac{2\pi}4=\dfrac\pi2$。$f$ 的周期是两者的最小公倍数 $\pi=3\cdot\tfrac\pi3=2\cdot\tfrac\pi2$。$\dfrac{500}{\pi}\approx159.15$，完整周期 159 个。
(B)、(C) 只用了其中一项的周期 $\tfrac\pi2$、$\tfrac\pi3$。
R-10Finding b and c from Extrema — 由最值点求 b 与 c改编自 Practice Exam 1 · MCQ Part B #6
The function $f$ is given by $f(x)=a\sin\big(b(x+c)\big)+d$, where $a>0$ and $b>0$. The graph of $f$ has a minimum at $(1.5,-3)$, and the next maximum is at $(4,7)$. Which of the following could be the values of $b$ and $c$?
(A) $b=\dfrac{4\pi}{5}$ and $c=-2.75$
(B) $b=\dfrac{2\pi}{5}$ and $c=-2.75$
(C) $b=\dfrac{2\pi}{5}$ and $c=2.75$
(D) $b=5$ and $c=-4$
答案：(B) 最小值到下一个最大值是半个周期：$\dfrac T2=4-1.5=2.5$，$T=5$，$b=\dfrac{2\pi}5$。$\sin$ 在输入为 $\dfrac\pi2$ 时取最大值：$\dfrac{2\pi}5(4+c)=\dfrac\pi2\Rightarrow4+c=1.25\Rightarrow c=-2.75$。检查最小值：$\dfrac{2\pi}5(1.5-2.75)=-\dfrac\pi2$ ✓。另外 $a=5$，$d=2$。
(A) 把 2.5 当成了整个周期；(C) 相位平移方向反了。
Part C简答题 Free Response · Show Your Work
R-11A Waterwheel — 水车上的点改编自 2024 AP 真题（亚太卷）· FRQ #3
A waterwheel has a radius of $3$ meters, and its center $C$ is $1$ meter above the water surface, as shown. The wheel rotates counterclockwise at a constant rate and completes one rotation every $20$ seconds. At time $t=0$, point $W$ on the edge of the wheel is at its highest point. Let $h(t)$ be the height of $W$ above the water surface, in meters (negative values mean $W$ is under water).
(a) Find $h(t)$ for $t=0,\ 5,\ 10,\ 15,$ and $20$.
(b) The function $h$ can be written as $h(t)=a\cos(bt)+d$. Find the values of $a$, $b$, and $d$.
(c) On which interval of $t$ in $[0,20]$ is $h$ decreasing and the rate of change of $h$ increasing? Explain.
(d) A calculator is allowed for this part. During one rotation, for how many seconds is $W$ under water?
(a) $t=0$：最高点 $1+3=4$；$t=5$：转了四分之一圈，与圆心同高，$1$；$t=10$：最低点 $1-3=-2$；$t=15$：$1$；$t=20$：$4$。
(b) 振幅 $a=3$（半径），中线 $d=1$（圆心高度），$b=\dfrac{2\pi}{20}=\dfrac{\pi}{10}$。$t=0$ 在最高点，用 $\cos$ 正好：$h(t)=3\cos\left(\dfrac{\pi}{10}t\right)+1$。
(c) $h$ 在 $0<t<10$ 上递减（从最高点到最低点）。$h$ 的变化率递增 ⇔ 图像上凹，正弦型函数在中线下方上凹，即 $5<t<15$。两者同时成立：$5<t<10$——$W$ 在中线下方继续下降，但下降得越来越慢。
(d) 解 $3\cos\left(\dfrac{\pi}{10}t\right)+1<0$，即 $\cos\left(\dfrac{\pi}{10}t\right)<-\dfrac13$：用计算器求交点 $t\approx6.082$ 和 $t\approx13.918$，所以 $W$ 在水下约 $13.918-6.082\approx7.837$ 秒。
R-12Trig Equations, an Identity, and an Inequality — 三角方程、恒等式与不等式改编自 Practice Exam 1 · FRQ #4
Directions: Solve each part without a calculator. Give exact answers and show the work that leads to your answers.
(a) Solve $2\sin^2x-3\sin x+1=0$ for $0\le x<2\pi$.
(b) Show that $\dfrac{\cos(2x)}{\cos x-\sin x}=\cos x+\sin x$ wherever the left side is defined.
(c) Solve $2\cos x+\sqrt3<0$ for $0\le x<2\pi$.
(d) Find the exact value of $\sec\left(\arctan\dfrac34\right)$.
(a) 看成关于 $\sin x$ 的二次式：$(2\sin x-1)(\sin x-1)=0$，$\sin x=\dfrac12$ 或 $\sin x=1$。$x=\dfrac\pi6,\ \dfrac{5\pi}6,\ \dfrac\pi2$。
(b) 二倍角公式 $\cos2x=\cos^2x-\sin^2x=(\cos x-\sin x)(\cos x+\sin x)$，约去 $\cos x-\sin x$（不为 0）即得 $\cos x+\sin x$。
(c) $\cos x<-\dfrac{\sqrt3}2$。在单位圆上，$\cos x=-\dfrac{\sqrt3}2$ 对应 $x=\dfrac{5\pi}6$ 和 $\dfrac{7\pi}6$，横坐标更小（更靠左）的部分在两者之间：$\dfrac{5\pi}6<x<\dfrac{7\pi}6$。
(d) 设 $\theta=\arctan\dfrac34$，则 $-\dfrac\pi2<\theta<\dfrac\pi2$ 且 $\tan\theta=\dfrac34>0$，$\theta$ 在第一象限。直角三角形对边 3、邻边 4、斜边 5：$\cos\theta=\dfrac45$，$\sec\theta=\dfrac54$。
R-13From a Graph to Sine and Cosine Forms — 由图像写解析式
The graph of the sinusoidal function $g$ is shown. The graph has a maximum at $\left(\dfrac{\pi}{6},5\right)$ and the next minimum at $\left(\dfrac{2\pi}{3},-1\right)$.
(a) Find the amplitude, the equation of the midline, and the period of $g$.
(b) Write an expression for $g(x)$ in the form $a\cos\big(b(x-h)\big)+k$.
(c) Write an expression for $g(x)$ in the form $a\sin\big(b(x-h)\big)+k$.
(d) A student writes $g(x)=3\cos\left(2x-\dfrac{\pi}{3}\right)+2$ and says the graph is the graph of $y=3\cos(2x)+2$ shifted $\dfrac{\pi}{3}$ units to the right. Is the student's expression correct? Is the description of the shift correct? Explain.
(a) 振幅 $\dfrac{5-(-1)}{2}=3$；中线 $y=\dfrac{5+(-1)}{2}=2$；最大到最小是半个周期 $\dfrac{2\pi}3-\dfrac\pi6=\dfrac\pi2$，周期 $\pi$。
(b) $b=\dfrac{2\pi}{\pi}=2$；$\cos$ 从最大值开始，最大值在 $x=\dfrac\pi6$：$g(x)=3\cos\left(2\left(x-\dfrac\pi6\right)\right)+2$。
(c) $\sin$ 在中线上升处开始，比最大值早四分之一周期：$\dfrac\pi6-\dfrac\pi4=-\dfrac\pi{12}$。
$g(x)=3\sin\left(2\left(x+\dfrac\pi{12}\right)\right)+2$
(d) 表达式正确：$3\cos\left(2x-\dfrac\pi3\right)+2=3\cos\left(2\left(x-\dfrac\pi6\right)\right)+2$，和 (b) 一样。但平移量说错了：要先提出 $b=2$，相位平移是向右 $\dfrac\pi6$，不是 $\dfrac\pi3$。（检查：若右移 $\dfrac\pi3$，最大值应在 $x=\dfrac\pi3$，而 $g\left(\dfrac\pi3\right)=3\cos\dfrac\pi3+2=3.5\ne5$。）
R-14A Rotating Lighthouse Beam — 灯塔光束与正切
A lighthouse $L$ stands $2$ kilometers from a straight shoreline. Point $O$ is the point on the shoreline closest to the lighthouse. As the light rotates, its beam makes an angle $\theta$ with segment $LO$ and hits the shoreline at a point $P$ that is $d(\theta)=2\tan\theta$ kilometers from $O$, for $-\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2}$. A calculator is allowed for part (c) only.
(a) Find the exact values of $d\left(\dfrac{\pi}{4}\right)$ and $d\left(\dfrac{\pi}{3}\right)$.
(b) Find the average rate of change of $d$ over $\left[0,\dfrac{\pi}{4}\right]$ and over $\left[\dfrac{\pi}{4},\dfrac{\pi}{3}\right]$. What do these values suggest about the graph of $d$ for $0<\theta<\dfrac{\pi}{2}$?
(c) Find the angle $\theta$ for which the beam hits the shoreline $5$ kilometers from $O$.
(d) Describe the behavior of $d(\theta)$ as $\theta\to\dfrac{\pi}{2}^-$, and explain what this means about the beam.
(a) $d\left(\dfrac\pi4\right)=2\tan\dfrac\pi4=2$ km；$d\left(\dfrac\pi3\right)=2\sqrt3$ km。
(b) $\dfrac{2-0}{\pi/4}=\dfrac8\pi\approx2.546$ km/弧度；$\dfrac{2\sqrt3-2}{\pi/12}=\dfrac{24(\sqrt3-1)}{\pi}\approx5.592$ km/弧度。后一段的平均变化率更大：$d$ 递增且增长越来越快，图像在 $\left(0,\dfrac\pi2\right)$ 上上凹。
(c) $2\tan\theta=5\Rightarrow\theta=\arctan2.5\approx1.190$ 弧度（在 $\left(-\dfrac\pi2,\dfrac\pi2\right)$ 内，$\arctan$ 给出的就是这个角）。
(d) $\displaystyle\lim_{\theta\to\pi/2^-}d(\theta)=\infty$（$\theta=\dfrac\pi2$ 是垂直渐近线）：光束越来越接近与海岸线平行，照到海岸上的点越来越远；$\theta=\dfrac\pi2$ 时光束与海岸平行，照不到海岸。
