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Trigonometric Functions · Solutions三角函数 · 解析

Companion to the IB-Style Practice SetIB 风格练习题的解析配套

EASY MEDIUM HARD Paper 1A Paper 1B Paper 2 Paper 3

Syllabus 3.5, 3.6, 3.7, 3.8, 3.9, 3.10, 3.11考纲 3.5、3.6、3.7、3.8、3.9、3.10、3.11AA HL

PART I  ·  PAPER 1 SECTION A · SOLUTIONS第一部分  ·  第一卷 A 节 · 解析No calculator · 21 marks不可使用计算器 · 21 分

Section A · Worked SolutionsA 节 · 详细解析

Q1EASYPaper 1A3.5 Exact Values (Unit Circle)[4 marks]

Evaluate $\sin 60^\circ + \cos 30^\circ$, $\tan(5\pi/4)$, and $\sin(7\pi/6)$ exactly.精确求 $\sin 60^\circ + \cos 30^\circ$、$\tan(5\pi/4)$、$\sin(7\pi/6)$。

Answers:答案：  (a) $\sqrt{3}$  ·  (b) $1$  ·  (c) $-\tfrac{1}{2}$

(a) $\sin 60^\circ + \cos 30^\circ$ A1·A1

Both are standard first-quadrant exact values: $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$ and $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$ (the complementary-angle identity $\cos\theta = \sin(90^\circ - \theta)$ forces these to agree). Therefore $$ \sin 60^\circ + \cos 30^\circ \;=\; \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \;=\; \sqrt{3}. $$

(b) $\tan(5\pi/4)$ A1

$5\pi/4 = \pi + \pi/4$ lies in quadrant III, where tangent is positive (CAST: in QIII, sin and cos are both negative, so their ratio is positive). The reference angle is $\pi/4$, with $\tan(\pi/4) = 1$. Hence $\tan(5\pi/4) = +1$.

(c) $\sin(7\pi/6)$ A1

$7\pi/6 = \pi + \pi/6$ lies in quadrant III, where sine is negative (CAST). The reference angle is $\pi/6$, with $\sin(\pi/6) = \tfrac{1}{2}$. Therefore $\sin(7\pi/6) = -\tfrac{1}{2}$.
Reference angle then CAST sign — the two-step that never fails. Every Paper 1A exact-value question collapses into the same drill: (i) reduce to the acute reference angle by subtracting the nearest multiple of $\pi$ (or $\pi/2$ if you want the cofunction form); (ii) read the sign from the quadrant via CAST (QI all $+$, QII sin $+$, QIII tan $+$, QIV cos $+$). Skipping step (i) and writing $\sin(7\pi/6) = -\sin(\pi/6) = -\tfrac{1}{2}$ "by feel" works here but breaks for ugly inputs like $\sin(11\pi/6)$ or $\cos(5\pi/3)$. The mechanical two-step also makes part marks easy: an explicit reference angle is the M1, the CAST sign is the A1.

(a) $\sin 60^\circ + \cos 30^\circ$ A1·A1

均为第一象限标准精确值：$\sin 60^\circ = \dfrac{\sqrt{3}}{2}$、$\cos 30^\circ = \dfrac{\sqrt{3}}{2}$（互余恒等式 $\cos\theta = \sin(90^\circ - \theta)$ 保证两者相等）。故 $$ \sin 60^\circ + \cos 30^\circ \;=\; \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \;=\; \sqrt{3}. $$

(b) $\tan(5\pi/4)$ A1

$5\pi/4 = \pi + \pi/4$ 在第三象限，该象限正切为正（CAST：QIII 中 sin、cos 同负，比为正）。参考角 $\pi/4$，$\tan(\pi/4) = 1$。故 $\tan(5\pi/4) = +1$。

(c) $\sin(7\pi/6)$ A1

$7\pi/6 = \pi + \pi/6$ 在第三象限，该象限正弦为负（CAST）。参考角 $\pi/6$，$\sin(\pi/6) = \tfrac{1}{2}$。故 $\sin(7\pi/6) = -\tfrac{1}{2}$。
"参考角 + CAST 定号"——百战不殆的两步走。Paper 1A 任意精确值题最终都归到同一套机械步骤：(i) 减去最近的 $\pi$（或 $\pi/2$ 若要走余函数）化为锐角参考角；(ii) 用 CAST 读所在象限的符号（QI 全正、QII sin 正、QIII tan 正、QIV cos 正）。直接"凭感觉"写 $\sin(7\pi/6) = -\sin(\pi/6) = -\tfrac{1}{2}$ 这里成立，但遇到 $\sin(11\pi/6)$、$\cos(5\pi/3)$ 这种丑角就会翻车。机械两步走也方便分步给分：写出参考角就是 M1，CAST 符号就是 A1。
Q2MEDIUMPaper 1A3.8 Linear Sine Equation[5 marks]

Solve $2 \sin x = \sqrt{3}$ for $x \in [0, 2\pi]$.在 $x \in [0, 2\pi]$ 上解 $2 \sin x = \sqrt{3}$。

Answer:答案：  $x = \dfrac{\pi}{3}$ or或 $x = \dfrac{2\pi}{3}$

(a) Isolate $\sin x$ A1

Divide both sides by $2$: $\sin x = \dfrac{\sqrt{3}}{2}$.

(b) Principal and second solution M1·A1·A1

The principal value is $\alpha = \arcsin\!\bigl(\tfrac{\sqrt{3}}{2}\bigr) = \dfrac{\pi}{3}$. Sine is positive in quadrants I and II; the second QII solution is $$ x \;=\; \pi - \alpha \;=\; \pi - \frac{\pi}{3} \;=\; \frac{2\pi}{3}. $$ Both lie in $[0, 2\pi]$.

(c) No further solutions R1

The next solution would be $\dfrac{\pi}{3} + 2\pi = \dfrac{7\pi}{3} > 2\pi$ (out of range), and similarly $\dfrac{2\pi}{3} + 2\pi > 2\pi$. In quadrants III and IV sine is negative, so it cannot equal $+\tfrac{\sqrt{3}}{2}$ there. Hence the only solutions in $[0, 2\pi]$ are $x = \pi/3$ and $x = 2\pi/3$.
"$\pi - \alpha$" is the QII partner, not $\alpha + \pi$. The most frequent slip on linear sine equations is writing the second solution as $\pi + \alpha$ (which is the QIII partner, where $\sin$ flips sign). For $\sin x = +k$ the two solutions on $[0, 2\pi]$ are $\alpha$ and $\pi - \alpha$; for $\sin x = -k$ they are $\pi + \alpha$ and $2\pi - \alpha$. Memorise the symmetry: sine has axis $x = \pi/2$, so $\sin(\pi/2 + t) = \sin(\pi/2 - t)$, i.e. inputs equidistant from $\pi/2$ have the same sine. That symmetry is what makes $\alpha$ and $\pi - \alpha$ partners, not the period.

(a) 分离 $\sin x$ A1

两边除以 $2$：$\sin x = \dfrac{\sqrt{3}}{2}$。

(b) 主值与第二解 M1·A1·A1

主值 $\alpha = \arcsin\!\bigl(\tfrac{\sqrt{3}}{2}\bigr) = \dfrac{\pi}{3}$。正弦在第一、二象限为正；第二（QII）解为 $$ x \;=\; \pi - \alpha \;=\; \pi - \frac{\pi}{3} \;=\; \frac{2\pi}{3}. $$ 两解皆在 $[0, 2\pi]$ 中。

(c) 无其他解 R1

下一个解为 $\dfrac{\pi}{3} + 2\pi = \dfrac{7\pi}{3} > 2\pi$（超出），同理 $\dfrac{2\pi}{3} + 2\pi > 2\pi$。三、四象限正弦为负，不可能等于 $+\tfrac{\sqrt{3}}{2}$。故 $[0, 2\pi]$ 上仅有 $x = \pi/3$、$x = 2\pi/3$。
"$\pi - \alpha$" 是 QII 伴解，不是 $\alpha + \pi$。线性正弦方程最常见错就是把第二解写成 $\pi + \alpha$（这是 QIII 伴解，$\sin$ 在那里变号）。$\sin x = +k$ 在 $[0, 2\pi]$ 上的两解是 $\alpha$ 与 $\pi - \alpha$；$\sin x = -k$ 的两解是 $\pi + \alpha$ 与 $2\pi - \alpha$。记对称性：正弦关于 $x = \pi/2$ 对称，$\sin(\pi/2 + t) = \sin(\pi/2 - t)$，即与 $\pi/2$ 等距的两输入同正弦值。$\alpha$ 与 $\pi - \alpha$ 是伴解的根源就在这里，不在周期。
Q3MEDIUMPaper 1A3.8 Substitution: $\cos 2x$ Equation[6 marks]

Solve $\cos 2x = \dfrac{1}{2}$ for $x \in [0, 2\pi]$.在 $x \in [0, 2\pi]$ 上解 $\cos 2x = \dfrac{1}{2}$。

Answer:答案：  $x = \dfrac{\pi}{6},\; \dfrac{5\pi}{6},\; \dfrac{7\pi}{6},\; \dfrac{11\pi}{6}$

(a) Substitution range A1

Let $u = 2x$. As $x$ runs through $[0, 2\pi]$, $u$ runs through $[0, 4\pi]$ — two full periods of cosine.

(b) Solve $\cos u = 1/2$ on $[0, 4\pi]$ M1·A1·A1

The principal value is $u = \arccos(1/2) = \pi/3$. Cosine is positive in QI and QIV; on one period $[0, 2\pi]$ the solutions are $u = \pi/3$ and $u = 2\pi - \pi/3 = 5\pi/3$. Add $2\pi$ to each to cover the second period $[2\pi, 4\pi]$: $$ u \;=\; \frac{\pi}{3},\; \frac{5\pi}{3},\; \frac{\pi}{3} + 2\pi = \frac{7\pi}{3},\; \frac{5\pi}{3} + 2\pi = \frac{11\pi}{3}. $$

(c) Back-substitute $x = u/2$ A1·A1

Divide each $u$ by $2$: $$ x \;=\; \frac{\pi}{6},\; \frac{5\pi}{6},\; \frac{7\pi}{6},\; \frac{11\pi}{6}. $$ All four lie in $[0, 2\pi]$. $\checkmark$
Always widen the $u$-interval before solving. Students who forget to expand $u \in [0, 4\pi]$ list only the first two solutions and lose two marks. The rule: if the equation has $\sin(bx)$ or $\cos(bx)$ on the bounded interval $[0, 2\pi]$, then $u = bx$ ranges over $[0, 2 b \pi]$, which contains $b$ full periods — expect $2b$ solutions (for a "generic" right-hand side). Equivalently: a $\cos(bx) = k$ equation on $[0, 2\pi]$ generically has $2b$ solutions. Here $b = 2$ gives $4$ solutions, which is exactly what we found.

(a) 代换区间 A1

令 $u = 2x$。$x$ 取遍 $[0, 2\pi]$ 时，$u$ 取遍 $[0, 4\pi]$ — 即余弦的两个周期。

(b) 在 $[0, 4\pi]$ 上解 $\cos u = 1/2$ M1·A1·A1

主值 $u = \arccos(1/2) = \pi/3$。余弦在 QI、QIV 为正；一个周期 $[0, 2\pi]$ 上的解为 $u = \pi/3$ 与 $u = 2\pi - \pi/3 = 5\pi/3$。各加 $2\pi$ 覆盖第二周期 $[2\pi, 4\pi]$： $$ u \;=\; \frac{\pi}{3},\; \frac{5\pi}{3},\; \frac{\pi}{3} + 2\pi = \frac{7\pi}{3},\; \frac{5\pi}{3} + 2\pi = \frac{11\pi}{3}. $$

(c) 回代 $x = u/2$ A1·A1

每个 $u$ 除以 $2$： $$ x \;=\; \frac{\pi}{6},\; \frac{5\pi}{6},\; \frac{7\pi}{6},\; \frac{11\pi}{6}. $$ 四解皆在 $[0, 2\pi]$。$\checkmark$
解之前先把 $u$ 的区间放大。忘记把 $u \in [0, 4\pi]$ 拉满的同学只列出前两解，直接丢两分。法则：方程含 $\sin(bx)$ 或 $\cos(bx)$ 且 $x \in [0, 2\pi]$，则 $u = bx$ 在 $[0, 2 b \pi]$ 上跑，含 $b$ 个完整周期 — 一般情形有 $2b$ 个解。等价地：$\cos(bx) = k$ 在 $[0, 2\pi]$ 上一般有 $2b$ 解。本题 $b = 2$ 给 $4$ 解，恰好吻合。
Q4HARDPaper 1A3.6 Identity Proof (Double-Angle)[6 marks]

Prove $\dfrac{1 - \cos 2x}{\sin 2x} = \tan x$, and state the excluded $x$-values.证明 $\dfrac{1 - \cos 2x}{\sin 2x} = \tan x$，并写出排除值。

Excluded $x$:$x$ 排除值：  $x \ne \dfrac{k\pi}{2}$ for $k \in \mathbb{Z}$ (i.e. $\sin 2x = 0$ or $\cos x = 0$)（$k \in \mathbb{Z}$；即 $\sin 2x = 0$ 或 $\cos x = 0$）

(a) Rewrite numerator M1·A1

Use the $\sin^{2} x$-form of the double-angle identity: $\cos 2x = 1 - 2 \sin^{2} x$. Then $$ 1 - \cos 2x \;=\; 1 - (1 - 2 \sin^{2} x) \;=\; 2 \sin^{2} x. $$

(b) Rewrite denominator A1

$\sin 2x = 2 \sin x \cos x$ (standard double-angle).

(c) Simplify the ratio M1·A1·A1

$$ \frac{1 - \cos 2x}{\sin 2x} \;=\; \frac{2 \sin^{2} x}{2 \sin x \cos x} \;=\; \frac{\sin x}{\cos x} \;=\; \tan x. \quad \text{AG} $$ The cancellation $\dfrac{2 \sin^{2} x}{2 \sin x} = \sin x$ is valid only when $\sin x \ne 0$; the simplification $\dfrac{\sin x}{\cos x} = \tan x$ requires $\cos x \ne 0$.

(d) Excluded values A1

The LHS requires $\sin 2x \ne 0$, i.e. $2x \ne k\pi$, i.e. $x \ne \dfrac{k\pi}{2}$ ($k \in \mathbb{Z}$). The RHS $\tan x$ requires $\cos x \ne 0$, i.e. $x \ne \dfrac{\pi}{2} + k\pi$. The first set $\{k\pi/2\}$ already contains the second, so the joint exclusion is $x \ne \dfrac{k\pi}{2}$ for $k \in \mathbb{Z}$.
Pick the "right flavour" of $\cos 2x$. The identity $\cos 2x$ has three faces: $\cos^{2} x - \sin^{2} x$, $1 - 2 \sin^{2} x$, $2 \cos^{2} x - 1$. The numerator $1 - \cos 2x$ pleads for the form that kills the $1$: that is $1 - 2 \sin^{2} x$, giving $1 - \cos 2x = 2 \sin^{2} x$. Symmetrically, $1 + \cos 2x = 2 \cos^{2} x$ if you start from $\cos 2x = 2 \cos^{2} x - 1$. Picking the wrong flavour leaves a $1$ to deal with and adds two lines of algebra. Memorise: "minus" goes with sine-squared; "plus" goes with cosine-squared.

(a) 改写分子 M1·A1

用倍角恒等式的 $\sin^{2} x$ 形式：$\cos 2x = 1 - 2 \sin^{2} x$。故 $$ 1 - \cos 2x \;=\; 1 - (1 - 2 \sin^{2} x) \;=\; 2 \sin^{2} x. $$

(b) 改写分母 A1

$\sin 2x = 2 \sin x \cos x$（标准倍角）。

(c) 化简比值 M1·A1·A1

$$ \frac{1 - \cos 2x}{\sin 2x} \;=\; \frac{2 \sin^{2} x}{2 \sin x \cos x} \;=\; \frac{\sin x}{\cos x} \;=\; \tan x. \quad \text{AG} $$ 约分 $\dfrac{2 \sin^{2} x}{2 \sin x} = \sin x$ 需要 $\sin x \ne 0$；$\dfrac{\sin x}{\cos x} = \tan x$ 需要 $\cos x \ne 0$。

(d) 排除值 A1

左边要求 $\sin 2x \ne 0$，即 $2x \ne k\pi$，即 $x \ne \dfrac{k\pi}{2}$（$k \in \mathbb{Z}$）。右边 $\tan x$ 要求 $\cos x \ne 0$，即 $x \ne \dfrac{\pi}{2} + k\pi$。前一集合 $\{k\pi/2\}$ 已包含后者，故联合排除为 $x \ne \dfrac{k\pi}{2}$（$k \in \mathbb{Z}$）。
选对 $\cos 2x$ 的"风味"。$\cos 2x$ 有三张脸：$\cos^{2} x - \sin^{2} x$、$1 - 2 \sin^{2} x$、$2 \cos^{2} x - 1$。分子 $1 - \cos 2x$ 需要消掉那个 $1$：故选 $1 - 2 \sin^{2} x$，得 $1 - \cos 2x = 2 \sin^{2} x$。对称地，$1 + \cos 2x = 2 \cos^{2} x$ 需选 $\cos 2x = 2 \cos^{2} x - 1$。选错风味就留下一个 $1$ 多两行代数。记口诀："减"配 sin 方，"加"配 cos 方。
PART II  ·  PAPER 1 SECTION B · SOLUTIONS第二部分  ·  第一卷 B 节 · 解析No calculator · 11 marks不可使用计算器 · 11 分

Section B · Worked SolutionsB 节 · 详细解析

Q5HARDPaper 1B3.7 Transform Analysis + Solve[11 marks]

For $y = 3 \sin\!\bigl(2(x - \pi/4)\bigr) + 1$ on $[0, \pi]$: state amplitude, period, phase shift, vertical shift; state the range; solve $y = 5/2$.对 $y = 3 \sin\!\bigl(2(x - \pi/4)\bigr) + 1$（$[0, \pi]$）：写振幅、周期、相移、纵移；写值域；解 $y = 5/2$。

Answers:答案：  (a) $A = 3,\; T = \pi,\; \text{phase} = \tfrac{\pi}{4} \text{ right},\; d = 1$  ·  (b) $[-2, 4]$  ·  (c) $x = \dfrac{\pi}{3},\; \dfrac{2\pi}{3}$

(a) Parameters of $a \sin(b(x - c)) + d$ A1·A1·A1·A1

The argument is already factored as $2(x - \pi/4)$, so read directly:
  • Amplitude $|a| = |3| = 3$.
  • Period $T = \dfrac{2\pi}{|b|} = \dfrac{2\pi}{2} = \pi$.
  • Phase shift $c = \dfrac{\pi}{4}$ to the right.
  • Vertical shift $d = 1$ (midline $y = 1$).

(b) Range over $\mathbb{R}$ M1·A1

Since $\sin(\cdot) \in [-1, 1]$, $3 \sin(\cdot) \in [-3, 3]$, so $3 \sin(\cdot) + 1 \in [-2, 4]$. Range $= [-2, 4]$.

(c) Solve $y = 5/2$ M1·A1·A1·A1·A1

Set $3 \sin(2(x - \pi/4)) + 1 = \tfrac{5}{2}$: $$ 3 \sin\!\bigl(2(x - \pi/4)\bigr) \;=\; \tfrac{3}{2} \;\Longrightarrow\; \sin\!\bigl(2(x - \pi/4)\bigr) \;=\; \tfrac{1}{2}. $$ Let $u = 2(x - \pi/4)$. As $x$ runs through $[0, \pi]$, $u$ runs through $[2(0 - \pi/4),\; 2(\pi - \pi/4)] = [-\pi/2,\; 3\pi/2]$ — one full period of $\sin u$. On $[-\pi/2, 3\pi/2]$ the equation $\sin u = 1/2$ has solutions $u = \pi/6$ (QI, principal) and $u = \pi - \pi/6 = 5\pi/6$ (QII). The next $u$ would be $\pi/6 + 2\pi = 13\pi/6 > 3\pi/2$, out of range; and $u = \pi/6 - 2\pi = -11\pi/6 < -\pi/2$, also out of range. So $u \in \{\pi/6,\; 5\pi/6\}$. Back-substitute $x = u/2 + \pi/4$:
  • $u = \pi/6 \Rightarrow x = \pi/12 + \pi/4 = \pi/12 + 3\pi/12 = 4\pi/12 = \pi/3$.
  • $u = 5\pi/6 \Rightarrow x = 5\pi/12 + \pi/4 = 5\pi/12 + 3\pi/12 = 8\pi/12 = 2\pi/3$.
Verify: $x = \pi/3 \Rightarrow 2(\pi/3 - \pi/4) = 2(\pi/12) = \pi/6$; $\sin(\pi/6) = 1/2$ $\checkmark$. $x = 2\pi/3 \Rightarrow 2(2\pi/3 - \pi/4) = 2(5\pi/12) = 5\pi/6$; $\sin(5\pi/6) = 1/2$ $\checkmark$. Hence $x = \dfrac{\pi}{3},\; \dfrac{2\pi}{3}$ in $[0, \pi]$.
Factor before reading phase shift. The single greatest trap on Paper 1B transformations is reading the phase shift from $\sin(2x - \pi/2)$ as "$\pi/2$ right". It is not. Factor: $\sin(2x - \pi/2) = \sin\!\bigl(2(x - \pi/4)\bigr)$, so the phase shift is $\pi/4$. The horizontal scale factor $1/b$ acts on the phase as well as the period: divide $c$ by $b$ to convert a "$-\pi/2$ inside" into the actual rigid shift on the $x$-axis. Mnemonic: "factor first, then read". This is the same trick that converts $f(2x - 6)$ into $f(2(x - 3))$ (shift right $3$, not $6$).

(a) $a \sin(b(x - c)) + d$ 的参数 A1·A1·A1·A1

括号内已是因子形式 $2(x - \pi/4)$，直接读：
  • 振幅 $|a| = |3| = 3$。
  • 周期 $T = \dfrac{2\pi}{|b|} = \dfrac{2\pi}{2} = \pi$。
  • 相移 $c = \dfrac{\pi}{4}$ 向右。
  • 纵移 $d = 1$（中线 $y = 1$）。

(b) 在 $\mathbb{R}$ 上的值域 M1·A1

因 $\sin(\cdot) \in [-1, 1]$，故 $3 \sin(\cdot) \in [-3, 3]$，$3 \sin(\cdot) + 1 \in [-2, 4]$。值域 $= [-2, 4]$。

(c) 解 $y = 5/2$ M1·A1·A1·A1·A1

令 $3 \sin(2(x - \pi/4)) + 1 = \tfrac{5}{2}$： $$ 3 \sin\!\bigl(2(x - \pi/4)\bigr) \;=\; \tfrac{3}{2} \;\Longrightarrow\; \sin\!\bigl(2(x - \pi/4)\bigr) \;=\; \tfrac{1}{2}. $$ 令 $u = 2(x - \pi/4)$。$x$ 取遍 $[0, \pi]$ 时，$u$ 取遍 $[-\pi/2,\; 3\pi/2]$ — 即 $\sin u$ 的一个完整周期。 在 $[-\pi/2, 3\pi/2]$ 上，$\sin u = 1/2$ 的解为 $u = \pi/6$（主值 QI）与 $u = \pi - \pi/6 = 5\pi/6$（QII）。下一个 $u = \pi/6 + 2\pi = 13\pi/6 > 3\pi/2$，超出。 回代 $x = u/2 + \pi/4$：
  • $u = \pi/6 \Rightarrow x = \pi/12 + \pi/4 = \pi/12 + 3\pi/12 = 4\pi/12 = \pi/3$。
  • $u = 5\pi/6 \Rightarrow x = 5\pi/12 + \pi/4 = 5\pi/12 + 3\pi/12 = 8\pi/12 = 2\pi/3$。
验：$x = \pi/3 \Rightarrow 2(\pi/3 - \pi/4) = \pi/6$，$\sin(\pi/6) = 1/2$ $\checkmark$；$x = 2\pi/3 \Rightarrow 2(2\pi/3 - \pi/4) = 5\pi/6$，$\sin(5\pi/6) = 1/2$ $\checkmark$。故 $x = \dfrac{\pi}{3},\; \dfrac{2\pi}{3}$。
读相移前先提因子。Paper 1B 变换题最大陷阱：把 $\sin(2x - \pi/2)$ 的相移读成 "$\pi/2$ 向右"。错。提因子：$\sin(2x - \pi/2) = \sin\!\bigl(2(x - \pi/4)\bigr)$，相移是 $\pi/4$。水平伸缩因子 $1/b$ 同时作用在周期与相移上：要把"括号内的 $-\pi/2$"翻译为 $x$ 轴上的实际刚性平移，必须把 $c$ 除以 $b$。口诀："先提，再读"。同一招也把 $f(2x - 6)$ 转为 $f(2(x - 3))$（右移 $3$ 而非 $6$）。
PART III  ·  PAPER 2 · SOLUTIONS第三部分  ·  第二卷 · 解析Calculator · 16 marks可使用计算器 · 16 分

Paper 2 · Worked Solutions第二卷 · 详细解析

Q6MEDIUMPaper 23.7 Period, Amplitude, Extrema[7 marks]

For $y = -4 \cos(\pi x / 3)$: state amplitude, find period, give max/min and the smallest non-negative $x$ where each occurs.对 $y = -4 \cos(\pi x / 3)$：写振幅、求周期、给出最大值与最小值及各自最小非负 $x$。

Answers:答案：  (a) $|a| = 4$  ·  (b) $T = 6$  ·  (c) $\max = 4$ at于 $x = 3$;  $\min = -4$ at于 $x = 0$

(a) Amplitude A1

Amplitude $= |a| = |-4| = 4$. (The sign of $a$ flips the graph vertically; it does not change the amplitude.)

(b) Period M1·A1

For $\cos(b x)$ with $b = \pi/3$, the period is $$ T \;=\; \frac{2\pi}{|b|} \;=\; \frac{2\pi}{\pi/3} \;=\; 6. $$

(c) Extrema and smallest non-negative $x$ M1·A1·A1·A1

Because of the leading $-4$, the function $y = -4 \cos(\pi x / 3)$ achieves its maximum $+4$ when $\cos(\pi x / 3) = -1$, i.e. $\pi x / 3 = \pi + 2k\pi$, i.e. $x = 3 + 6k$. Smallest non-negative: $x = 3$. It achieves its minimum $-4$ when $\cos(\pi x / 3) = +1$, i.e. $\pi x / 3 = 2k\pi$, i.e. $x = 6k$. Smallest non-negative: $x = 0$. Sanity check. Plot on GDC: $y(0) = -4 \cdot 1 = -4$ (min, $\checkmark$); $y(3) = -4 \cdot \cos\pi = -4 \cdot (-1) = +4$ (max, $\checkmark$); $y(6) = -4 \cdot \cos 2\pi = -4$ (min again, $T = 6$ apart $\checkmark$).
The minus sign flips max and min. Students reflexively read $y = -4 \cos(\pi x / 3)$ as "max at $x = 0$" because $\cos 0 = 1$ is maximal — but the leading $-4$ inverts that: max of $-4 \cos\theta$ occurs where $\cos\theta = -1$, not where $\cos\theta = 1$. The general rule: for $y = a \cos(b(x - c)) + d$ with $a > 0$, max is at $\cos = +1$; with $a < 0$, max is at $\cos = -1$. The midline value is still $d$ in both cases. On Paper 2 always GDC-verify a maximum location after computing it — the sign-flip trap is a $2$-mark loser.

(a) 振幅 A1

振幅 $= |a| = |-4| = 4$。（$a$ 的符号会上下翻转图像，但不改变振幅。）

(b) 周期 M1·A1

$\cos(b x)$ 中 $b = \pi/3$，周期 $$ T \;=\; \frac{2\pi}{|b|} \;=\; \frac{2\pi}{\pi/3} \;=\; 6. $$

(c) 最值及最小非负 $x$ M1·A1·A1·A1

由于前面是 $-4$，$y = -4 \cos(\pi x / 3)$ 取最大值 $+4$ 当 $\cos(\pi x / 3) = -1$，即 $\pi x / 3 = \pi + 2k\pi$，$x = 3 + 6k$。最小非负：$x = 3$。 取最小值 $-4$ 当 $\cos(\pi x / 3) = +1$，即 $\pi x / 3 = 2k\pi$，$x = 6k$。最小非负：$x = 0$。 GDC 验。$y(0) = -4 \cdot 1 = -4$（最小 $\checkmark$）；$y(3) = -4 \cdot (-1) = +4$（最大 $\checkmark$）；$y(6) = -4 \cdot 1 = -4$（再次最小，$T = 6$ 间隔 $\checkmark$）。
负号把最大最小互换。同学们下意识把 $y = -4 \cos(\pi x / 3)$ 读成"$x = 0$ 处最大"，因为 $\cos 0 = 1$ 最大 — 但前面的 $-4$ 把这一点翻转：$-4 \cos\theta$ 的最大值出现在 $\cos\theta = -1$ 处，不是 $\cos\theta = 1$ 处。一般规律：$y = a \cos(b(x - c)) + d$ 中 $a > 0$ 时最大在 $\cos = +1$ 处；$a < 0$ 时最大在 $\cos = -1$ 处。中线值无论怎样都是 $d$。Paper 2 上算出最大点后务必用 GDC 复核 — 这种翻号陷阱稳丢 $2$ 分。
Q7HARDPaper 23.7 Modelling: Tide Height[9 marks]

$H(t) = 5 + 3 \sin(\pi t / 6)$ for $t \in [0, 12]$ hours. (a) Mean; (b) first max time and value; (c) interval where $H > 6$.$H(t) = 5 + 3 \sin(\pi t / 6)$（$t \in [0, 12]$ 小时）。(a) 平均；(b) 首次最大时刻与值；(c) $H > 6$ 时段。

Answers:答案：  (a) $5$ m  ·  (b) $t = 3$ h, $H = 8$ m  ·  (c) $t \in (0.649,\; 5.35)$ h

(a) Mean tide height A1

The model is a sine wave centred on its midline $d = 5$. Mean $H = 5$ m.

(b) First maximum M1·A1·A1

Maximum of $\sin(\pi t / 6)$ is $+1$, attained when $\pi t / 6 = \pi/2 + 2k\pi$, i.e. $t = 3 + 12k$. On $[0, 12]$ the first (and only) maximum is at $t = 3$ hours. Maximum height: $H(3) = 5 + 3 \cdot 1 = 8$ m.

(c) Interval where $H > 6$ M1·A1·A1·A1·A1

Set $5 + 3 \sin(\pi t / 6) > 6$, i.e. $\sin(\pi t / 6) > 1/3$. Let $u = \pi t / 6$, so $u \in [0, 2\pi]$ as $t \in [0, 12]$. Solve $\sin u = 1/3$ on $[0, 2\pi]$: principal value $u_{1} = \arcsin(1/3) \approx 0.3398$ (QI), and $u_{2} = \pi - 0.3398 \approx 2.802$ (QII). Between these, $\sin u > 1/3$; outside (in $[0, u_{1})$ and $(u_{2}, 2\pi]$), $\sin u < 1/3$ (sine is negative on $(\pi, 2\pi)$). Convert back: $t = 6 u / \pi$. So $$ t_{1} \;=\; \frac{6 \cdot 0.3398}{\pi} \;\approx\; 0.6489, \qquad t_{2} \;=\; \frac{6 \cdot 2.802}{\pi} \;\approx\; 5.351. $$ GDC nSolve on $H(t) = 6$ confirms $t_{1} \approx 0.6489$ and $t_{2} \approx 5.351$. Quoting to $3$ s.f.: $t \in (0.649,\; 5.35)$ hours. Endpoint check. The inequality is strict, so the endpoints are excluded. The interval is open: $t \in (0.649,\; 5.35)$.
Convert after solving, not before. Tide modelling problems collapse if you try to "factor the $\pi/6$" out and solve $\sin t > 1/3$ — that is the wrong equation. Always substitute $u$ for the full sine argument, solve the equation in $u$ on its actual range, then convert back to $t$. Second exam-trap: students round intermediate values (e.g. write $\arcsin(1/3) \approx 0.34$) and lose precision, so the final $3$-s.f. answer is wrong in the last digit. Keep $5$+ digits in the GDC store and round only at the very end. Third trap: an interval with strict inequality is open at both ends — closed brackets $[a, b]$ would be a mark loss.

(a) 平均潮位 A1

模型为以中线 $d = 5$ 为轴的正弦波。平均 $H = 5$ 米。

(b) 首次最大 M1·A1·A1

$\sin(\pi t / 6)$ 最大为 $+1$，在 $\pi t / 6 = \pi/2 + 2k\pi$ 处取，即 $t = 3 + 12k$。$[0, 12]$ 上唯一最大点 $t = 3$ 小时。最大高度 $H(3) = 5 + 3 \cdot 1 = 8$ 米。

(c) $H > 6$ 时段 M1·A1·A1·A1·A1

列 $5 + 3 \sin(\pi t / 6) > 6$，即 $\sin(\pi t / 6) > 1/3$。令 $u = \pi t / 6$，$t \in [0, 12]$ 对应 $u \in [0, 2\pi]$。 在 $[0, 2\pi]$ 上解 $\sin u = 1/3$：主值 $u_{1} = \arcsin(1/3) \approx 0.3398$（QI），$u_{2} = \pi - 0.3398 \approx 2.802$（QII）。两者之间 $\sin u > 1/3$；区间外 $\sin u < 1/3$（$(\pi, 2\pi)$ 上 sine 为负）。 回代 $t = 6 u / \pi$： $$ t_{1} \;=\; \frac{6 \cdot 0.3398}{\pi} \;\approx\; 0.6489, \qquad t_{2} \;=\; \frac{6 \cdot 2.802}{\pi} \;\approx\; 5.351. $$ 保留 $3$ 位有效数字（GDC nSolve 解 $H(t) = 6$ 验证）：$t \in (0.649,\; 5.35)$ 小时。 端点核查。不等式严格，端点排除，区间为开区间：$t \in (0.649,\; 5.35)$。
先解再换元，不要先换元再解。潮汐建模题如果硬要"把 $\pi/6$ 提出来"去解 $\sin t > 1/3$，方程就错了。务必把整个 sine 参数代换为 $u$，在 $u$ 的实际区间上解，再回代为 $t$。第二个陷阱：中间舍入（如写 $\arcsin(1/3) \approx 0.34$）使最终 $3$ 位有效数字末位偏差。GDC 至少保留 $5$ 位中间值，仅末步舍入。第三个陷阱：严格不等式对应两端皆开 — 写成闭区间 $[a, b]$ 就要丢分。
PART IV  ·  PAPER 3 · SOLUTIONS第四部分  ·  第三卷 · 解析Calculator · HL extended exploration · 15 marks可使用计算器 · HL 长题探究 · 15 分

Paper 3 · Worked Solutions第三卷 · 详细解析

Q8HARDPaper 33.10/3.11 Triple-Angle & Reciprocal Identities (HL)[15 marks]

(a) Prove $\sin 3x = 3 \sin x - 4 \sin^{3} x$. (b) Prove $\cos 3x = 4 \cos^{3} x - 3 \cos x$. (c) Evaluate $\sin(3\pi/10) - \sin(\pi/10)$. (d) Derive $1 + \tan^{2} x = \sec^{2} x$ and $1 + \cot^{2} x = \csc^{2} x$.(a) 证 $\sin 3x = 3 \sin x - 4 \sin^{3} x$；(b) 证 $\cos 3x = 4 \cos^{3} x - 3 \cos x$；(c) 求 $\sin(3\pi/10) - \sin(\pi/10)$ 精确值；(d) 推导 $1 + \tan^{2} x = \sec^{2} x$、$1 + \cot^{2} x = \csc^{2} x$。

Answers:答案：  (c) $\sin(3\pi/10) - \sin(\pi/10) = \tfrac{1}{2}$  ·  (d) excluded $\cos x = 0$ for sec; $\sin x = 0$ for csc/cotsec 排除 $\cos x = 0$；csc/cot 排除 $\sin x = 0$

(a) Triple-angle for sine M1·A1·A1·A1

Start: $\sin 3x = \sin(2x + x)$. Apply the compound-angle formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$: $$ \sin 3x \;=\; \sin 2x \cos x + \cos 2x \sin x. $$ Substitute $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = 1 - 2 \sin^{2} x$: $$ \sin 3x \;=\; (2 \sin x \cos x)\cos x + (1 - 2 \sin^{2} x)\sin x \;=\; 2 \sin x \cos^{2} x + \sin x - 2 \sin^{3} x. $$ Use $\cos^{2} x = 1 - \sin^{2} x$: $$ \sin 3x \;=\; 2 \sin x (1 - \sin^{2} x) + \sin x - 2 \sin^{3} x \;=\; 2 \sin x - 2 \sin^{3} x + \sin x - 2 \sin^{3} x \;=\; 3 \sin x - 4 \sin^{3} x. \quad \text{AG} $$

(b) Triple-angle for cosine M1·A1·A1·A1

Start: $\cos 3x = \cos(2x + x)$. Apply $\cos(A + B) = \cos A \cos B - \sin A \sin B$: $$ \cos 3x \;=\; \cos 2x \cos x - \sin 2x \sin x. $$ Now choose the $\cos 2x = 2 \cos^{2} x - 1$ flavour (to land on $\cos^{3} x$) and $\sin 2x = 2 \sin x \cos x$: $$ \cos 3x \;=\; (2 \cos^{2} x - 1)\cos x - (2 \sin x \cos x)\sin x \;=\; 2 \cos^{3} x - \cos x - 2 \sin^{2} x \cos x. $$ Use $\sin^{2} x = 1 - \cos^{2} x$: $$ \cos 3x \;=\; 2 \cos^{3} x - \cos x - 2(1 - \cos^{2} x)\cos x \;=\; 2 \cos^{3} x - \cos x - 2 \cos x + 2 \cos^{3} x \;=\; 4 \cos^{3} x - 3 \cos x. \quad \text{AG} $$

(c) Evaluate $\sin(3\pi/10) - \sin(\pi/10)$ M1·M1·A1·A1

Set $x = \pi/10$ in (a): $\sin(3 \cdot \pi/10) = 3 \sin(\pi/10) - 4 \sin^{3}(\pi/10)$, i.e. $$ \sin(3\pi/10) \;=\; 3 \sin(\pi/10) - 4 \sin^{3}(\pi/10). $$ Therefore $$ \sin(3\pi/10) - \sin(\pi/10) \;=\; 2 \sin(\pi/10) - 4 \sin^{3}(\pi/10) \;=\; 2 \sin(\pi/10) \bigl[1 - 2 \sin^{2}(\pi/10)\bigr] \;=\; 2 \sin(\pi/10) \cos(\pi/5), $$ using the double-angle $1 - 2 \sin^{2}\theta = \cos 2\theta$ with $\theta = \pi/10$. Now apply the given identity $4 \cos(\pi/5) \sin(\pi/10) = 1$: $$ 2 \sin(\pi/10) \cos(\pi/5) \;=\; \tfrac{1}{2} \cdot \bigl[4 \sin(\pi/10) \cos(\pi/5)\bigr] \;=\; \tfrac{1}{2} \cdot 1 \;=\; \tfrac{1}{2}. $$ Hence $\sin(3\pi/10) - \sin(\pi/10) = \dfrac{1}{2}$.

(d) Pythagorean reciprocal identities M1·A1·A1

Start from $\sin^{2} x + \cos^{2} x = 1$.
  • Divide by $\cos^{2} x$ (valid when $\cos x \ne 0$, i.e. $x \ne \pi/2 + k\pi$): $\;\tan^{2} x + 1 = \sec^{2} x$, i.e. $1 + \tan^{2} x = \sec^{2} x$.
  • Divide by $\sin^{2} x$ (valid when $\sin x \ne 0$, i.e. $x \ne k\pi$): $\;1 + \cot^{2} x = \csc^{2} x$.
Excluded values: $x \ne \pi/2 + k\pi$ for the sec-identity; $x \ne k\pi$ for the csc/cot-identity.
Triple-angle identities are bridges to exact values of $\pi/10$, $\pi/5$, $3\pi/10$. The values $\sin(\pi/10) = \tfrac{\sqrt{5} - 1}{4}$ and $\cos(\pi/5) = \tfrac{\sqrt{5} + 1}{4}$ come from the triple-angle identity, not from a unit-circle drawing. The trick: set $x = \pi/10$, so $3x = 3\pi/10 = \pi/2 - \pi/5$, hence $\sin 3x = \cos(\pi/5)$. The triple-angle identity then reads $\cos(\pi/5) = 3 \sin(\pi/10) - 4 \sin^{3}(\pi/10)$. Combine with $\cos(\pi/5) = 1 - 2 \sin^{2}(\pi/10)$ (the double-angle $\cos(2 \cdot \pi/10)$) and you get a cubic in $s = \sin(\pi/10)$: $4 s^{3} - 2 s^{2} - 3 s + 1 = 0$, factoring as $(s - 1)(4 s^{2} + 2 s - 1) = 0$. Discard $s = 1$ (out of range), and the quadratic gives $s = \tfrac{-1 + \sqrt{5}}{4}$. Paper 3 problems on the regular pentagon, golden ratio, and Chebyshev polynomials all sit downstream of this single identity.

(a) sine 三倍角 M1·A1·A1·A1

起：$\sin 3x = \sin(2x + x)$。用复角公式 $\sin(A + B) = \sin A \cos B + \cos A \sin B$： $$ \sin 3x \;=\; \sin 2x \cos x + \cos 2x \sin x. $$ 代入 $\sin 2x = 2 \sin x \cos x$、$\cos 2x = 1 - 2 \sin^{2} x$： $$ \sin 3x \;=\; (2 \sin x \cos x)\cos x + (1 - 2 \sin^{2} x)\sin x \;=\; 2 \sin x \cos^{2} x + \sin x - 2 \sin^{3} x. $$ 再用 $\cos^{2} x = 1 - \sin^{2} x$： $$ \sin 3x \;=\; 2 \sin x (1 - \sin^{2} x) + \sin x - 2 \sin^{3} x \;=\; 3 \sin x - 4 \sin^{3} x. \quad \text{AG} $$

(b) cosine 三倍角 M1·A1·A1·A1

起：$\cos 3x = \cos(2x + x)$。用 $\cos(A + B) = \cos A \cos B - \sin A \sin B$： $$ \cos 3x \;=\; \cos 2x \cos x - \sin 2x \sin x. $$ 选 $\cos 2x = 2 \cos^{2} x - 1$ 这一风味（为落到 $\cos^{3} x$）和 $\sin 2x = 2 \sin x \cos x$： $$ \cos 3x \;=\; (2 \cos^{2} x - 1)\cos x - 2 \sin^{2} x \cos x \;=\; 2 \cos^{3} x - \cos x - 2 \sin^{2} x \cos x. $$ 代入 $\sin^{2} x = 1 - \cos^{2} x$： $$ \cos 3x \;=\; 2 \cos^{3} x - \cos x - 2(1 - \cos^{2} x)\cos x \;=\; 4 \cos^{3} x - 3 \cos x. \quad \text{AG} $$

(c) 求 $\sin(3\pi/10) - \sin(\pi/10)$ M1·M1·A1·A1

把 $x = \pi/10$ 代入 (a)： $$ \sin(3\pi/10) \;=\; 3 \sin(\pi/10) - 4 \sin^{3}(\pi/10). $$ 故 $$ \sin(3\pi/10) - \sin(\pi/10) \;=\; 2 \sin(\pi/10) - 4 \sin^{3}(\pi/10) \;=\; 2 \sin(\pi/10) \bigl[1 - 2 \sin^{2}(\pi/10)\bigr] \;=\; 2 \sin(\pi/10) \cos(\pi/5), $$ 其中用了倍角 $1 - 2 \sin^{2}\theta = \cos 2\theta$（$\theta = \pi/10$）。再用题给恒等式 $4 \cos(\pi/5) \sin(\pi/10) = 1$： $$ 2 \sin(\pi/10) \cos(\pi/5) \;=\; \tfrac{1}{2} \cdot 1 \;=\; \tfrac{1}{2}. $$ 故 $\sin(3\pi/10) - \sin(\pi/10) = \dfrac{1}{2}$。

(d) 毕氏倒数形式恒等式 M1·A1·A1

由 $\sin^{2} x + \cos^{2} x = 1$ 出发。
  • 两边除以 $\cos^{2} x$（$\cos x \ne 0$，即 $x \ne \pi/2 + k\pi$）：$\;\tan^{2} x + 1 = \sec^{2} x$，即 $1 + \tan^{2} x = \sec^{2} x$。
  • 两边除以 $\sin^{2} x$（$\sin x \ne 0$，即 $x \ne k\pi$）：$\;1 + \cot^{2} x = \csc^{2} x$。
排除：sec 恒等式 $x \ne \pi/2 + k\pi$；csc/cot 恒等式 $x \ne k\pi$。
三倍角是通向 $\pi/10$、$\pi/5$、$3\pi/10$ 精确值的桥梁。$\sin(\pi/10) = \tfrac{\sqrt{5} - 1}{4}$ 与 $\cos(\pi/5) = \tfrac{\sqrt{5} + 1}{4}$ 来自三倍角恒等式，不是单位圆作图的产物。诀窍：令 $x = \pi/10$，则 $3x = 3\pi/10 = \pi/2 - \pi/5$，故 $\sin 3x = \cos(\pi/5)$。三倍角恒等式于是给 $\cos(\pi/5) = 3 \sin(\pi/10) - 4 \sin^{3}(\pi/10)$。再结合 $\cos(\pi/5) = 1 - 2 \sin^{2}(\pi/10)$（倍角 $\cos(2 \cdot \pi/10)$），得到 $s = \sin(\pi/10)$ 的三次方程：$4 s^{3} - 2 s^{2} - 3 s + 1 = 0$，分解为 $(s - 1)(4 s^{2} + 2 s - 1) = 0$。舍去 $s = 1$，得 $s = \tfrac{-1 + \sqrt{5}}{4}$。正五边形、黄金比、Chebyshev 多项式等 Paper 3 题目都坐在这一个恒等式下游。