Unit B4: Trigonometric
Functions单元 B4：三角函数
The trigonometry of the function family. B4 covers the unit-circle definitions, the Pythagorean and double-angle identities, transformations of $y = a \sin(b(x - c)) + d$, and the systematic solving of trigonometric equations on a stated interval. The HL extension layers on the reciprocal functions ($\sec, \csc, \cot$), the inverse trig functions, and the compound-angle identities used across Papers 1, 2, and 3.本单元聚焦三角函数的函数族视角。B4 涵盖单位圆定义、毕氏与倍角恒等式、$y = a \sin(b(x - c)) + d$ 形式的图像变换，以及在指定区间内系统求解三角方程的方法。HL 拓展加入倒数函数（$\sec, \csc, \cot$）、反三角函数与复角恒等式，对应 Paper 1、Paper 2、Paper 3 三类试题。
How to use this guide本指南使用说明
Trigonometry rewards the student who knows the unit circle cold. Memorise the exact values for the first quadrant, learn the CAST sign rule, and almost every Paper 1 trig problem becomes a sequence of rote moves. B4 is organised so that the cheat sheets at the top of each section give the rote moves; the worked examples drill them on canonical exam problems; the going-deeper blocks earn the last marks.三角学最奖励"单位圆背熟"的学生。第一象限精确值和 CAST 符号规则一旦背牢，几乎所有 Paper 1 三角题都是机械动作。B4 的组织方式如下：每节顶部的速查框给出机械步骤；例题在典型考题上演练这些步骤；末尾的进阶块帮你拿满分。
Memorise the first-quadrant exact values: $\sin 30^\circ = 1/2$, $\sin 45^\circ = \sqrt{2}/2$, $\sin 60^\circ = \sqrt{3}/2$. Memorise the identity $\sin^2 x + \cos^2 x = 1$ and the double-angle formulas. Practise solving $\sin x = c$ on $[0, 2\pi]$ end to end.
背熟第一象限精确值：$\sin 30^\circ = 1/2$、$\sin 45^\circ = \sqrt{2}/2$、$\sin 60^\circ = \sqrt{3}/2$。背熟恒等式 $\sin^2 x + \cos^2 x = 1$ 与倍角公式。完整做一道 $[0, 2\pi]$ 上解 $\sin x = c$ 的题。
Derive the three forms of $\cos 2x$ from $\sin^2 x + \cos^2 x = 1$. Master the compound-angle expansions of $\sin(A \pm B)$, $\cos(A \pm B)$, $\tan(A \pm B)$, and use them to deduce the double-angle formulas as the $A = B$ case. Solve trig equations involving reciprocal and inverse functions cleanly, stating ranges where required.
由 $\sin^2 x + \cos^2 x = 1$ 推导 $\cos 2x$ 的三种形式。熟练 $\sin(A \pm B)$、$\cos(A \pm B)$、$\tan(A \pm B)$ 的复角展开，并把 $A = B$ 作为倍角公式的特例。求解含倒数与反三角函数的方程，必要时写明值域。
Unit Circle and Exact Values单位圆与精确值 SL 3.5
Exact values in the first quadrant.
| $\theta$ | $0$ | $\pi/6$ | $\pi/4$ | $\pi/3$ | $\pi/2$ |
|---|---|---|---|---|---|
| $\sin \theta$ | $0$ | $1/2$ | $\sqrt{2}/2$ | $\sqrt{3}/2$ | $1$ |
| $\cos \theta$ | $1$ | $\sqrt{3}/2$ | $\sqrt{2}/2$ | $1/2$ | $0$ |
| $\tan \theta$ | $0$ | $\sqrt{3}/3$ | $1$ | $\sqrt{3}$ | undef. |
第一象限精确值。
| $\theta$ | $0$ | $\pi/6$ | $\pi/4$ | $\pi/3$ | $\pi/2$ |
|---|---|---|---|---|---|
| $\sin \theta$ | $0$ | $1/2$ | $\sqrt{2}/2$ | $\sqrt{3}/2$ | $1$ |
| $\cos \theta$ | $1$ | $\sqrt{3}/2$ | $\sqrt{2}/2$ | $1/2$ | $0$ |
| $\tan \theta$ | $0$ | $\sqrt{3}/3$ | $1$ | $\sqrt{3}$ | 未定义 |
Find the exact value of $\cos(5\pi/6)$ and $\sin(5\pi/6)$ without a calculator.不使用计算器，求 $\cos(5\pi/6)$ 与 $\sin(5\pi/6)$ 的精确值。
Step 1. Locate the quadrant. $5\pi/6$ lies between $\pi/2$ and $\pi$, so it is in Q2.
第 1 步：定位象限。$5\pi/6$ 介于 $\pi/2$ 与 $\pi$ 之间，故位于 Q2。
Step 2. Reference angle. The acute angle to the $x$-axis is $\pi - 5\pi/6 = \pi/6$.
第 2 步：参考角。与 $x$ 轴所成锐角为 $\pi - 5\pi/6 = \pi/6$。
Step 3. Magnitudes from the first-quadrant table. $\sin(\pi/6) = 1/2$ and $\cos(\pi/6) = \sqrt{3}/2$.
第 3 步：从第一象限表查模长。$\sin(\pi/6) = 1/2$、$\cos(\pi/6) = \sqrt{3}/2$。
Step 4. Signs from CAST. In Q2, $\sin$ is $+$ and $\cos$ is $-$.
第 4 步：用 CAST 定符号。Q2 中 $\sin$ 取 $+$、$\cos$ 取 $-$。
$$ \sin(5\pi/6) \;=\; +\,\tfrac{1}{2}, \qquad \cos(5\pi/6) \;=\; -\,\tfrac{\sqrt{3}}{2}. $$Going deeper: where the exact values come from进阶：精确值的几何来源
The $\pi/4$ values come from an isosceles right triangle with legs $1$ and hypotenuse $\sqrt{2}$. Both legs project onto the unit circle as $\sqrt{2}/2$, so $\sin(\pi/4) = \cos(\pi/4) = \sqrt{2}/2$.
$\pi/4$ 的值源于等腰直角三角形，腰为 $1$，斜边为 $\sqrt{2}$。两腰在单位圆上投影皆为 $\sqrt{2}/2$，故 $\sin(\pi/4) = \cos(\pi/4) = \sqrt{2}/2$。
The $\pi/6$ and $\pi/3$ values come from an equilateral triangle of side $1$. Drop a perpendicular from one vertex to the opposite side; the half-triangle is a right triangle with hypotenuse $1$, short leg $1/2$ (opposite the $\pi/6$ angle) and long leg $\sqrt{3}/2$ (opposite the $\pi/3$ angle). Hence $\sin(\pi/6) = 1/2$, $\cos(\pi/6) = \sqrt{3}/2$, and the $\pi/3$ values are the swap.
$\pi/6$ 与 $\pi/3$ 的值源于边长为 $1$ 的等边三角形。由一顶点向对边作垂线，得到直角三角形：斜边 $1$、短直角边 $1/2$（对 $\pi/6$）、长直角边 $\sqrt{3}/2$（对 $\pi/3$）。故 $\sin(\pi/6) = 1/2$、$\cos(\pi/6) = \sqrt{3}/2$，$\pi/3$ 的值则对调。
Pythagorean and Double-Angle Identities毕氏与倍角恒等式 SL 3.6
Given $\sin^{2} x + \cos^{2} x = 1$, derive $1 + \tan^{2} x = \sec^{2} x$ and state the domain on which the derivation is valid.由 $\sin^{2} x + \cos^{2} x = 1$ 推导 $1 + \tan^{2} x = \sec^{2} x$，并写出推导成立的定义域。
Divide both sides by $\cos^{2} x$.
两边除以 $\cos^{2} x$。
$$ \frac{\sin^{2} x}{\cos^{2} x} + \frac{\cos^{2} x}{\cos^{2} x} \;=\; \frac{1}{\cos^{2} x}. $$Recognise each ratio. $\sin x / \cos x = \tan x$ and $1/\cos x = \sec x$, so
识别各比值。$\sin x / \cos x = \tan x$，$1/\cos x = \sec x$，故
$$ \tan^{2} x + 1 \;=\; \sec^{2} x. $$Domain. Division by $\cos^{2} x$ requires $\cos x \ne 0$, that is $x \ne \pi/2 + k \pi$ for integer $k$. On this set the identity holds. (The companion $1 + \cot^{2} x = \csc^{2} x$ requires $\sin x \ne 0$, that is $x \ne k \pi$.)
定义域。除以 $\cos^{2} x$ 要求 $\cos x \ne 0$，即 $x \ne \pi/2 + k \pi$（$k$ 为整数）。在该集合上恒等式成立。（伴随式 $1 + \cot^{2} x = \csc^{2} x$ 要求 $\sin x \ne 0$，即 $x \ne k \pi$。）
Given $\sin \theta = 3/5$ with $\theta$ in Q2, find $\sin 2\theta$ and $\cos 2\theta$ exactly.已知 $\sin \theta = 3/5$，且 $\theta$ 位于 Q2，精确求 $\sin 2\theta$ 与 $\cos 2\theta$。
Find $\cos \theta$. Pythagoras: $\cos^{2} \theta = 1 - 9/25 = 16/25$, so $\cos \theta = \pm 4/5$. In Q2 $\cos$ is negative, hence $\cos \theta = -4/5$.
求 $\cos \theta$。毕氏：$\cos^{2} \theta = 1 - 9/25 = 16/25$，故 $\cos \theta = \pm 4/5$。Q2 中 $\cos$ 为负，得 $\cos \theta = -4/5$。
Apply the double-angle formulas.
套用倍角公式。
$$ \sin 2\theta \;=\; 2 \sin \theta \cos \theta \;=\; 2 \cdot \tfrac{3}{5} \cdot \bigl(-\tfrac{4}{5}\bigr) \;=\; -\tfrac{24}{25}. $$ $$ \cos 2\theta \;=\; 1 - 2 \sin^{2} \theta \;=\; 1 - 2 \cdot \tfrac{9}{25} \;=\; 1 - \tfrac{18}{25} \;=\; \tfrac{7}{25}. $$Check the quadrant of $2\theta$. Since $\sin 2\theta < 0$ and $\cos 2\theta > 0$, the angle $2\theta$ lies in Q4. (This is consistent with $\theta \in (\pi/2, \pi)$ giving $2\theta \in (\pi, 2\pi)$, where Q4 corresponds to the upper portion.)
检验 $2\theta$ 所在象限。$\sin 2\theta < 0$、$\cos 2\theta > 0$，故 $2\theta$ 位于 Q4。（$\theta \in (\pi/2, \pi)$ 给出 $2\theta \in (\pi, 2\pi)$，其中 Q4 对应靠上半部分，一致。）
Circular Functions and Transformations三角函数与图像变换 SL 3.7
- Amplitude $= |a|$. Half the peak-to-trough distance. The graph oscillates between $d - |a|$ and $d + |a|$.
- Period $= 2\pi / |b|$. The horizontal length of one full cycle.
- Phase shift $= c$. The horizontal translation; the basic sine wave $y = \sin x$ is shifted to the right by $c$ when $c > 0$.
- Vertical shift $= d$. The midline of the wave; the graph oscillates about $y = d$.
The tangent function. $y = \tan x$ has period $\pi$ (not $2\pi$) and vertical asymptotes at $x = \pi/2 + k\pi$, where $\cos x = 0$. It is unbounded between consecutive asymptotes, so it has no amplitude.
- 振幅（amplitude）$= |a|$。峰谷距离的一半。图像在 $d - |a|$ 与 $d + |a|$ 之间振荡。
- 周期（period）$= 2\pi / |b|$。一个完整周期的水平长度。
- 相移（phase shift）$= c$。水平平移；$c > 0$ 时，基本正弦 $y = \sin x$ 向右平移 $c$。
- 纵移（vertical shift）$= d$。波的中线；图像绕 $y = d$ 振荡。
正切函数。$y = \tan x$ 周期为 $\pi$（非 $2\pi$），垂直渐近线位于 $x = \pi/2 + k\pi$（即 $\cos x = 0$ 处）。在两相邻渐近线之间无界，故无振幅。
For the curve $y = 3 \sin\bigl(2(x - \pi/4)\bigr) + 1$, state the amplitude, period, phase shift, vertical shift, and the maximum and minimum values.对曲线 $y = 3 \sin\bigl(2(x - \pi/4)\bigr) + 1$，写出振幅、周期、相移、纵移以及最大值与最小值。
Read the parameters. Here $a = 3$, $b = 2$, $c = \pi/4$, $d = 1$.
读出参数。$a = 3$、$b = 2$、$c = \pi/4$、$d = 1$。
- Amplitude. $|a| = 3$.
- 振幅。$|a| = 3$。
- Period. $2\pi / |b| = 2\pi / 2 = \pi$.
- 周期。$2\pi / |b| = 2\pi / 2 = \pi$。
- Phase shift. $c = \pi/4$ (rightward).
- 相移。$c = \pi/4$（向右）。
- Vertical shift. $d = 1$ (upward).
- 纵移。$d = 1$（向上）。
- Maximum. $d + |a| = 1 + 3 = 4$.
- 最大值。$d + |a| = 1 + 3 = 4$。
- Minimum. $d - |a| = 1 - 3 = -2$.
- 最小值。$d - |a| = 1 - 3 = -2$。
Solving Trigonometric Equations求解三角方程 SL 3.8
- $\sin x = c$: solutions are $x = \alpha$ and $x = \pi - \alpha$. (One in Q1 or Q4, one in Q2 or Q3.)
- $\cos x = c$: solutions are $x = \alpha$ and $x = 2\pi - \alpha$. (One in Q1 or Q2, one in Q3 or Q4.)
- $\tan x = c$: solutions are $x = \alpha$ and $x = \alpha + \pi$. ($\tan$ has period $\pi$, so the second solution is just one period away.)
Always check the stated interval. The number of solutions depends on how many cycles of the trig function fit inside the interval given.
- $\sin x = c$：解为 $x = \alpha$ 与 $x = \pi - \alpha$。（分别在 Q1 或 Q4、Q2 或 Q3。）
- $\cos x = c$：解为 $x = \alpha$ 与 $x = 2\pi - \alpha$。（分别在 Q1 或 Q2、Q3 或 Q4。）
- $\tan x = c$：解为 $x = \alpha$ 与 $x = \alpha + \pi$。（$\tan$ 周期为 $\pi$，故第二解只差一个周期。）
始终校对给定区间。解的个数取决于该区间能容纳几个周期。
Solve $2 \sin x = \sqrt{3}$ on $[0, 2\pi]$.在 $[0, 2\pi]$ 上解 $2 \sin x = \sqrt{3}$。
Isolate $\sin x$. $\sin x = \sqrt{3} / 2$.
分离 $\sin x$。$\sin x = \sqrt{3} / 2$。
Principal solution. From the first-quadrant table, $\sin(\pi/3) = \sqrt{3}/2$, so $\alpha = \pi/3$.
主值。由第一象限表，$\sin(\pi/3) = \sqrt{3}/2$，故 $\alpha = \pi/3$。
Second solution. $\pi - \alpha = \pi - \pi/3 = 2\pi/3$.
第二解。$\pi - \alpha = \pi - \pi/3 = 2\pi/3$。
Check the interval. Both $\pi/3$ and $2\pi/3$ lie in $[0, 2\pi]$, and $\sin$ does not repeat within one $2\pi$ cycle.
校对区间。$\pi/3$ 与 $2\pi/3$ 均在 $[0, 2\pi]$ 内，且 $\sin$ 在一个 $2\pi$ 周期内不会再重复。
$$ x \;=\; \tfrac{\pi}{3} \quad \text{or} \quad x \;=\; \tfrac{2\pi}{3}. $$Going deeper: equations after substitution进阶：代换后的方程
For an equation such as $2 \sin^{2} x - \sin x - 1 = 0$, substitute $u = \sin x$ to convert to a quadratic in $u$: $2 u^{2} - u - 1 = 0$. Factor: $(2 u + 1)(u - 1) = 0$, so $u = -1/2$ or $u = 1$.
如方程 $2 \sin^{2} x - \sin x - 1 = 0$，代 $u = \sin x$ 转化为 $u$ 的二次方程：$2 u^{2} - u - 1 = 0$。因式分解：$(2 u + 1)(u - 1) = 0$，故 $u = -1/2$ 或 $u = 1$。
Then solve $\sin x = -1/2$ and $\sin x = 1$ separately by the standard procedure. The $\sin x = 1$ case has the single solution $x = \pi/2$ on $[0, 2\pi]$; $\sin x = -1/2$ has $x = 7\pi/6$ and $x = 11\pi/6$.
再分别用标准流程解 $\sin x = -1/2$ 与 $\sin x = 1$。$\sin x = 1$ 在 $[0, 2\pi]$ 上只有 $x = \pi/2$；$\sin x = -1/2$ 给出 $x = 7\pi/6$ 与 $x = 11\pi/6$。
Reciprocal and Inverse Trigonometric Functions倒数与反三角函数 HL AHL 3.9
The three inverse functions. To make $\sin, \cos, \tan$ invertible, restrict each to a principal-value range:
| Inverse | Input domain | Output range |
|---|---|---|
| $\arcsin x$ | $[-1, 1]$ | $[-\pi/2, \pi/2]$ |
| $\arccos x$ | $[-1, 1]$ | $[0, \pi]$ |
| $\arctan x$ | $\mathbb{R}$ | $(-\pi/2, \pi/2)$ |
三个反函数。为使 $\sin, \cos, \tan$ 可逆，各限定一段主值域：
| 反函数 | 定义域 | 值域 |
|---|---|---|
| $\arcsin x$ | $[-1, 1]$ | $[-\pi/2, \pi/2]$ |
| $\arccos x$ | $[-1, 1]$ | $[0, \pi]$ |
| $\arctan x$ | $\mathbb{R}$ | $(-\pi/2, \pi/2)$ |
Find the exact value of $\sec(\pi/3)$, $\csc(\pi/4)$, and $\cot(\pi/6)$.求 $\sec(\pi/3)$、$\csc(\pi/4)$、$\cot(\pi/6)$ 的精确值。
Apply the definitions.
套用定义。
$$ \sec(\pi/3) \;=\; \frac{1}{\cos(\pi/3)} \;=\; \frac{1}{1/2} \;=\; 2. $$ $$ \csc(\pi/4) \;=\; \frac{1}{\sin(\pi/4)} \;=\; \frac{1}{\sqrt{2}/2} \;=\; \sqrt{2}. $$ $$ \cot(\pi/6) \;=\; \frac{\cos(\pi/6)}{\sin(\pi/6)} \;=\; \frac{\sqrt{3}/2}{1/2} \;=\; \sqrt{3}. $$Find $\arcsin(-1/2)$ and $\arccos(-1/2)$ exactly.精确求 $\arcsin(-1/2)$ 与 $\arccos(-1/2)$。
$\arcsin(-1/2)$. The output must lie in $[-\pi/2, \pi/2]$. We know $\sin(\pi/6) = 1/2$, so $\sin(-\pi/6) = -1/2$. Hence $\arcsin(-1/2) = -\pi/6$.
$\arcsin(-1/2)$。输出必须落在 $[-\pi/2, \pi/2]$。已知 $\sin(\pi/6) = 1/2$，故 $\sin(-\pi/6) = -1/2$，从而 $\arcsin(-1/2) = -\pi/6$。
$\arccos(-1/2)$. The output must lie in $[0, \pi]$. We need the angle in $[0, \pi]$ whose cosine is $-1/2$. From CAST, $\cos$ is negative in Q2; reference angle $\pi/3$ gives the Q2 angle $\pi - \pi/3 = 2\pi/3$. Hence $\arccos(-1/2) = 2\pi/3$.
$\arccos(-1/2)$。输出必须落在 $[0, \pi]$。求 $[0, \pi]$ 内余弦为 $-1/2$ 的角。CAST 指出 $\cos$ 在 Q2 为负；参考角 $\pi/3$ 对应 Q2 角 $\pi - \pi/3 = 2\pi/3$。故 $\arccos(-1/2) = 2\pi/3$。
Compound-Angle Identities and Trig Symmetries复角恒等式与三角对称性 HL AHL 3.10 · 3.11
Symmetry identities. Sine and tangent are odd functions; cosine is even: $$ \sin(-x) \;=\; -\sin x, \qquad \cos(-x) \;=\; \cos x, \qquad \tan(-x) \;=\; -\tan x. $$ These follow directly from the unit-circle definition: replacing $x$ by $-x$ reflects the point $P$ across the $x$-axis, which negates its $y$-coordinate while leaving the $x$-coordinate fixed.
对称性恒等式。正弦与正切为奇函数，余弦为偶函数： $$ \sin(-x) \;=\; -\sin x, \qquad \cos(-x) \;=\; \cos x, \qquad \tan(-x) \;=\; -\tan x. $$ 这直接由单位圆定义得到：把 $x$ 换为 $-x$ 相当于把点 $P$ 关于 $x$ 轴反射，$y$ 坐标变号、$x$ 坐标不变。
Find the exact value of $\sin(75^\circ)$ by writing $75^\circ = 45^\circ + 30^\circ$.把 $75^\circ$ 写成 $45^\circ + 30^\circ$，求 $\sin(75^\circ)$ 的精确值。
Apply the addition formula.
套用加法公式。
$$ \sin(75^\circ) \;=\; \sin(45^\circ + 30^\circ) \;=\; \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ. $$Substitute exact values. $\sin 45^\circ = \cos 45^\circ = \sqrt{2}/2$, $\sin 30^\circ = 1/2$, $\cos 30^\circ = \sqrt{3}/2$.
代入精确值。$\sin 45^\circ = \cos 45^\circ = \sqrt{2}/2$、$\sin 30^\circ = 1/2$、$\cos 30^\circ = \sqrt{3}/2$。
$$ \sin(75^\circ) \;=\; \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \;=\; \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \;=\; \frac{\sqrt{6} + \sqrt{2}}{4}. $$Going deeper: double-angle from compound-angle进阶：由复角推导倍角
Set $A = B = x$ in the compound-angle formulas. From $\sin(A + B) = \sin A \cos B + \cos A \sin B$,
在复角公式中令 $A = B = x$。由 $\sin(A + B) = \sin A \cos B + \cos A \sin B$，
$$ \sin 2x \;=\; \sin x \cos x + \cos x \sin x \;=\; 2 \sin x \cos x. $$From $\cos(A + B) = \cos A \cos B - \sin A \sin B$,
由 $\cos(A + B) = \cos A \cos B - \sin A \sin B$，
$$ \cos 2x \;=\; \cos^{2} x - \sin^{2} x. $$The other two forms of $\cos 2x$ ($2 \cos^{2} x - 1$ and $1 - 2 \sin^{2} x$) follow by substituting $\sin^{2} x = 1 - \cos^{2} x$ or $\cos^{2} x = 1 - \sin^{2} x$ respectively. The compound-angle formulas are therefore the parent results from which the double-angle formulas descend.
$\cos 2x$ 的另两种形式（$2 \cos^{2} x - 1$ 与 $1 - 2 \sin^{2} x$）分别由代入 $\sin^{2} x = 1 - \cos^{2} x$ 或 $\cos^{2} x = 1 - \sin^{2} x$ 得到。故复角公式是倍角公式之母。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Sketch the unit circle. A quick freehand sketch with the angle marked saves time on quadrant identification.
- 画单位圆草图。手绘标注角位置可加快判别象限。
- Decompose by reference angle. Read the magnitude from the first-quadrant table, then attach the CAST sign for the target quadrant.
- 按参考角分解。从第一象限表读模长，再用 CAST 给目标象限符号。
- Convert to a single function before solving. If an equation mixes $\sin x$ and $\cos x$, replace using $\sin^{2} x + \cos^{2} x = 1$ or a double-angle identity to leave a polynomial in one trig function.
- 解前化为单一函数。若方程中 $\sin x$ 与 $\cos x$ 并存，用 $\sin^{2} x + \cos^{2} x = 1$ 或倍角恒等式替换，化为单个三角函数的多项式。
- State the interval in your answer. Examiners want explicit solutions on the stated interval, not a general formula unless asked.
- 答案中写明区间。阅卷要求给出指定区间上的明确解，除非题目要求通解，否则不要写一般公式。
- Both solutions per cycle. $\sin x = c$ gives two solutions per $2\pi$, $\cos x = c$ gives two per $2\pi$, $\tan x = c$ gives one per $\pi$. Forgetting the second sine or cosine solution is the canonical error.
- 每周期两解。$\sin x = c$ 每 $2\pi$ 内两解，$\cos x = c$ 每 $2\pi$ 内两解，$\tan x = c$ 每 $\pi$ 内一解。漏掉正弦或余弦的第二个解是经典失误。
- Inverse functions return one value, not all. Use the principal range to fix the output; for the other solutions of $\sin x = c$ or $\cos x = c$, use the standard pairs from B4.4.
- 反函数返回单值，并非全部解。用主值域确定输出；对 $\sin x = c$ 或 $\cos x = c$ 的其他解，沿用 B4.4 的标准配对。
- Memorise the compound-angle list. The six identities (three pairs) are the foundation for double-angle, sum-to-product, and product-to-sum transformations. They appear in the formula booklet but recall is faster than lookup.
- 背熟复角公式表。六条恒等式（三对）是倍角、和差化积、积化和差的基础。公式册中虽有，但默写比翻阅更快。
Flashcards闪卡
Unit B4 Practice Quiz单元 B4 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- State the unit-circle definition of $\sin \theta, \cos \theta$, and $\tan \theta$ for any real $\theta$写出任意实数 $\theta$ 下 $\sin \theta, \cos \theta, \tan \theta$ 的单位圆定义
- Recall all first-quadrant exact values at $0, \pi/6, \pi/4, \pi/3, \pi/2$默写 $0, \pi/6, \pi/4, \pi/3, \pi/2$ 处的第一象限精确值
- Apply the CAST sign rule to compute exact values in Q2, Q3, Q4用 CAST 符号规则在 Q2、Q3、Q4 求精确值
- Derive $1 + \tan^{2} x = \sec^{2} x$ and $1 + \cot^{2} x = \csc^{2} x$ from $\sin^{2} x + \cos^{2} x = 1$由 $\sin^{2} x + \cos^{2} x = 1$ 推导 $1 + \tan^{2} x = \sec^{2} x$ 与 $1 + \cot^{2} x = \csc^{2} x$
- State all three forms of $\cos 2x$ and choose the convenient form for a given problem写出 $\cos 2x$ 的三种形式，并按题目选用最合适的
- Identify amplitude, period, phase shift, and vertical shift of $y = a \sin(b(x - c)) + d$辨识 $y = a \sin(b(x - c)) + d$ 的振幅、周期、相移、纵移
- Sketch one cycle of $\tan x$, marking the asymptotes at $x = \pi/2 + k\pi$画出 $\tan x$ 的一个周期，并标注 $x = \pi/2 + k\pi$ 处的渐近线
- Solve $\sin x = c$, $\cos x = c$, $\tan x = c$ on $[0, 2\pi]$ using principal solutions用主值在 $[0, 2\pi]$ 解 $\sin x = c$、$\cos x = c$、$\tan x = c$
- Solve quadratic-style trig equations by substitution $u = \sin x$ or $u = \cos x$用 $u = \sin x$ 或 $u = \cos x$ 代换求解二次型三角方程
- HL Define $\sec, \csc, \cot$ and find exact values at standard angles定义 $\sec, \csc, \cot$ 并求标准角处的精确值
- HL State the principal ranges of $\arcsin, \arccos, \arctan$ and use them to evaluate exact inverses写出 $\arcsin, \arccos, \arctan$ 的主值域并用以求精确反函数值
- HL Recall $\sin(A \pm B)$, $\cos(A \pm B)$, $\tan(A \pm B)$ from memory默写 $\sin(A \pm B)$、$\cos(A \pm B)$、$\tan(A \pm B)$
- HL Use the symmetry identities $\sin(-x) = -\sin x$, $\cos(-x) = \cos x$, $\tan(-x) = -\tan x$ to simplify expressions用对称恒等式 $\sin(-x) = -\sin x$、$\cos(-x) = \cos x$、$\tan(-x) = -\tan x$ 化简表达式
IB Paper-Style PracticeIB 试卷风格练习
B4 Practice and Solutions are on the roadmap. They will ship under Practice Questions/Unit_B4_*.html with the bilingual built-in pattern.
B4 配套的 Practice 与 Solutions 已在排期，上线后位于 Practice Questions/Unit_B4_*.html，采用双语内嵌格式。