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Unit-Circle Trigonometry and Trigonometric Functions · Solutions单位圆三角学与三角函数 · 答案

Companion to the Practice Set · Mark-by-mark walkthroughs · SAT / AP-Feeder / Provincial styles配套练习题集答案 · 按分点逐步讲解 · SAT / AP 衔接 / 省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 SAT-style MCQSAT 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Math 30-1 style阿省 Math 30-1 风格 Honors / Pre-Calc荣誉级 / 微积分预备
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 答案SAT MCQ + ON/BC/AB short answer · 18 marksSAT 选择题 + 安/卑/阿省考短答 · 共 18 分

Section A · Worked SolutionsA 部分 · 解题过程

Q1EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 Honors / Pre-Calc荣誉级 / 微积分预备 §1 Radians弧度制 · HSF-TF.A.1 [3 marks][3 分]

$150^{\circ}$ in radians.$150^{\circ}$ 化为弧度。

Answer:答案：  (C)  $\dfrac{5\pi}{6}$

(a) Apply the conversion factor $180^{\circ} = \pi$ rad运用换算关系 $180^{\circ} = \pi$ 弧度 M1·A1·A1

The conversion factor from degrees to radians is $\dfrac{\pi \text{ rad}}{180^{\circ}}$:从角度制到弧度制的换算因子为 $\dfrac{\pi \text{ rad}}{180^{\circ}}$： $$ 150^{\circ} \cdot \dfrac{\pi}{180^{\circ}} \;=\; \dfrac{150\pi}{180} \;=\; \dfrac{5\pi}{6}. $$ The reduction $\tfrac{150}{180} = \tfrac{5}{6}$ comes from dividing top and bottom by $30$. Matches (C).约分 $\tfrac{150}{180} = \tfrac{5}{6}$ 来自分子分母同除以 $30$。与 (C) 一致。
Why the wrong choices fail.错误选项分析。
  • (A) $\tfrac{2\pi}{3} = 120^{\circ}$, off by $30^{\circ}$ (the student likely divided $180^{\circ}$ into thirds instead of sixths).(A) $\tfrac{2\pi}{3} = 120^{\circ}$，相差 $30^{\circ}$（学生可能误将 $180^{\circ}$ 三等分而非六等分）。
  • (B) $\tfrac{3\pi}{4} = 135^{\circ}$, off by $15^{\circ}$; this is the radian for $135^{\circ}$, a common SAT misread.(B) $\tfrac{3\pi}{4} = 135^{\circ}$，相差 $15^{\circ}$；这是 $135^{\circ}$ 对应的弧度，SAT 常见的误读陷阱。
  • (D) $\tfrac{7\pi}{6} = 210^{\circ}$, the supplementary-angle trap: $150^{\circ}$ and $210^{\circ}$ are reflections across the horizontal axis, and the test sets them adjacent in the answer set.(D) $\tfrac{7\pi}{6} = 210^{\circ}$，补角陷阱：$150^{\circ}$ 与 $210^{\circ}$ 关于水平轴对称，命题人将两者并列以诱导误选。
Memorise the five "anchor" radians and read everything off them.熟记五个"锚点"弧度，其余靠它们直接读出。 The minute-zero anchors on the unit circle are $0$, $\tfrac{\pi}{6}$ ($30^{\circ}$), $\tfrac{\pi}{4}$ ($45^{\circ}$), $\tfrac{\pi}{3}$ ($60^{\circ}$), and $\tfrac{\pi}{2}$ ($90^{\circ}$). Any multiple of $30^{\circ}$ lands on one of the $\pi/6$ tick marks, so $150^{\circ} = 5 \cdot 30^{\circ}$ sits at the fifth tick: $\tfrac{5\pi}{6}$. SAT students who memorise the unit circle in radians lose almost no time on this question; students who convert by formula every time lose 20 seconds and an A1 to arithmetic.单位圆上的关键锚点为 $0$、$\tfrac{\pi}{6}$（$30^{\circ}$）、$\tfrac{\pi}{4}$（$45^{\circ}$）、$\tfrac{\pi}{3}$（$60^{\circ}$）与 $\tfrac{\pi}{2}$（$90^{\circ}$）。任何 $30^{\circ}$ 的倍数都落在 $\pi/6$ 的刻度上，所以 $150^{\circ} = 5 \cdot 30^{\circ}$ 位于第五个刻度：$\tfrac{5\pi}{6}$。背熟弧度制单位圆的考生在本题上几乎不耗时；每次都套公式换算的考生会浪费 20 秒，并因算术失误丢掉 A1。
Q2EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 Honors / Pre-Calc荣誉级 / 微积分预备 §2 Unit Circle单位圆 · HSF-TF.A.3 (+) [3 marks][3 分]

$\sin\!\left(\tfrac{5\pi}{6}\right)$.。

Answer:答案：  (B)  $\dfrac{1}{2}$

(a) Locate quadrant and reference angle确定象限与参考角 M1

$\tfrac{5\pi}{6}$ is between $\tfrac{\pi}{2}$ and $\pi$, so the terminal arm is in Quadrant II. The reference angle is the acute angle to the $x$-axis:$\tfrac{5\pi}{6}$ 位于 $\tfrac{\pi}{2}$ 与 $\pi$ 之间，故终边位于第二象限。参考角是终边到 $x$ 轴的锐角： $$ \theta_{\text{ref}} \;=\; \pi - \tfrac{5\pi}{6} \;=\; \tfrac{\pi}{6}. $$

(b) Read the sign and magnitude读取符号与大小 A1·A1

In Quadrant II, $\sin$ is positive (the $y$-coordinate of the unit-circle point is above the $x$-axis). The magnitude equals $\sin(\theta_{\text{ref}}) = \sin\!\tfrac{\pi}{6} = \tfrac{1}{2}$. Therefore $\sin\!\tfrac{5\pi}{6} = +\tfrac{1}{2}$, which matches (B).在第二象限，$\sin$ 为正（单位圆上点的 $y$ 坐标位于 $x$ 轴上方）。大小等于 $\sin(\theta_{\text{ref}}) = \sin\!\tfrac{\pi}{6} = \tfrac{1}{2}$。故 $\sin\!\tfrac{5\pi}{6} = +\tfrac{1}{2}$，与 (B) 一致。
Why the wrong choices fail.错误选项分析。
  • (A) $-\tfrac{1}{2}$, sign error: confuses Quadrant II (sin $> 0$) with Quadrant III or IV.(A) $-\tfrac{1}{2}$，符号错误：把第二象限（$\sin > 0$）误认为第三或第四象限。
  • (C) $-\tfrac{\sqrt{3}}{2}$ and (D) $\tfrac{\sqrt{3}}{2}$, magnitude error: the student reads $\cos\!\tfrac{5\pi}{6}$ instead of $\sin$, swapping the $x$- and $y$-coordinates of the unit-circle point.(C) $-\tfrac{\sqrt{3}}{2}$ 与 (D) $\tfrac{\sqrt{3}}{2}$，大小错误：考生误读为 $\cos\!\tfrac{5\pi}{6}$，混淆了单位圆点的 $x$ 与 $y$ 坐标。
ASTC ("All Students Take Calculus") for the sign, reference angle for the magnitude.用 ASTC（"All Students Take Calculus"）记忆符号，用参考角确定大小。 The two-step move for any unit-circle evaluation: (1) find which quadrant, then apply the sign rule, in Q-I all six functions are positive, in Q-II only $\sin / \csc$, in Q-III only $\tan / \cot$, in Q-IV only $\cos / \sec$. (2) Compute the value at the reference angle ($30^{\circ}, 45^{\circ}, 60^{\circ}$ are the only three magnitudes you ever need to know). This decouples a hard question into two cheap moves. The unit circle on the Pre-Calc / Math 30-1 exam reduces to this drill.单位圆求值的两步法：(1) 确定象限并套符号规则——第一象限六个函数全正，第二象限仅 $\sin / \csc$ 为正，第三象限仅 $\tan / \cot$ 为正，第四象限仅 $\cos / \sec$ 为正。(2) 计算参考角处的值（$30^{\circ}, 45^{\circ}, 60^{\circ}$ 三个大小是你唯一需要熟记的）。这样把难题拆成两步轻松动作。Pre-Calc / Math 30-1 考试上的单位圆题目，本质上就是这套训练。
Q3MEDIUM中 🇺🇸 US美 SAT-style MCQSAT 风格选择题 Honors / Pre-Calc荣誉级 / 微积分预备 §1 Reference Angles参考角 · HSF-TF.A.3 (+) [3 marks][3 分]

Quadrant III, reference angle $\tfrac{\pi}{4}$. Find $\cos\theta$.第三象限，参考角为 $\tfrac{\pi}{4}$。求 $\cos\theta$。

Answer:答案：  (B)  $-\dfrac{\sqrt{2}}{2}$

(a) Sign rule for Quadrant III第三象限符号规则 M1

In Quadrant III, both $x$ and $y$ are negative, so both $\cos\theta$ (the $x$-coordinate of the unit-circle point) and $\sin\theta$ (the $y$-coordinate) are negative. Only $\tan$ and $\cot$ come out positive in Q-III.在第三象限，$x$ 与 $y$ 均为负，故 $\cos\theta$（单位圆点的 $x$ 坐标）与 $\sin\theta$（$y$ 坐标）皆为负。第三象限只有 $\tan$ 与 $\cot$ 为正。

(b) Magnitude from the reference angle由参考角确定大小 A1·A1

Magnitude of $\cos$ at reference angle $\tfrac{\pi}{4}$: $\cos\!\tfrac{\pi}{4} = \tfrac{\sqrt{2}}{2}$. Combine with the Q-III sign:$\cos$ 在参考角 $\tfrac{\pi}{4}$ 处的大小：$\cos\!\tfrac{\pi}{4} = \tfrac{\sqrt{2}}{2}$。结合第三象限符号： $$ \cos\theta \;=\; -\dfrac{\sqrt{2}}{2}. $$ Matches (B).与 (B) 一致。
Why the wrong choices fail.错误选项分析。
  • (A) $\tfrac{\sqrt{2}}{2}$, correct magnitude but ignores the Q-III sign rule (this would be the Q-I value).(A) $\tfrac{\sqrt{2}}{2}$，大小正确但忽略了第三象限符号规则（这是第一象限的值）。
  • (C) $\tfrac{1}{2}$ and (D) $-\tfrac{1}{2}$, wrong reference angle: $\tfrac{1}{2}$ is the magnitude at $\tfrac{\pi}{3}$ ($\cos\tfrac{\pi}{3}$) or $\tfrac{\pi}{6}$ ($\sin\tfrac{\pi}{6}$), not $\tfrac{\pi}{4}$.(C) $\tfrac{1}{2}$ 与 (D) $-\tfrac{1}{2}$，参考角错误：$\tfrac{1}{2}$ 是 $\tfrac{\pi}{3}$（$\cos\tfrac{\pi}{3}$）或 $\tfrac{\pi}{6}$（$\sin\tfrac{\pi}{6}$）处的大小，并非 $\tfrac{\pi}{4}$。
Reference angle gives the magnitude; the quadrant gives the sign, never the other way around.参考角决定大小，象限决定符号，绝不可颠倒。 The most common reference-angle slip on the SAT and AP-feeder is to compute the right sign in the right quadrant but pull the magnitude from the wrong reference angle (the $\tfrac{\sqrt{2}}{2}$ vs. $\tfrac{1}{2}$ vs. $\tfrac{\sqrt{3}}{2}$ confusion). Drill: $\sin / \cos$ at the three reference angles map to $\bigl(\tfrac{1}{2}, \tfrac{\sqrt{2}}{2}, \tfrac{\sqrt{3}}{2}\bigr)$ for $\tfrac{\pi}{6}, \tfrac{\pi}{4}, \tfrac{\pi}{3}$ in $\sin$, and the reverse $\bigl(\tfrac{\sqrt{3}}{2}, \tfrac{\sqrt{2}}{2}, \tfrac{1}{2}\bigr)$ in $\cos$. The pattern is "$\sin$ rises, $\cos$ falls" as the reference angle climbs from $\tfrac{\pi}{6}$ to $\tfrac{\pi}{3}$.SAT 与 AP 衔接题最常见的参考角失误，是在正确象限取对了符号，却从错误的参考角抓取大小（$\tfrac{\sqrt{2}}{2}$ vs. $\tfrac{1}{2}$ vs. $\tfrac{\sqrt{3}}{2}$ 的混淆）。背诵法：三个参考角的 $\sin / \cos$ 对照表是——$\sin$ 在 $\tfrac{\pi}{6}, \tfrac{\pi}{4}, \tfrac{\pi}{3}$ 处依次为 $\bigl(\tfrac{1}{2}, \tfrac{\sqrt{2}}{2}, \tfrac{\sqrt{3}}{2}\bigr)$，$\cos$ 则反过来 $\bigl(\tfrac{\sqrt{3}}{2}, \tfrac{\sqrt{2}}{2}, \tfrac{1}{2}\bigr)$。规律：随着参考角从 $\tfrac{\pi}{6}$ 升到 $\tfrac{\pi}{3}$，"$\sin$ 上升，$\cos$ 下降"。
Q4MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 Honors / Pre-Calc荣誉级 / 微积分预备 §3 Reciprocal Functions倒数三角函数 · MHF4U Strand B单元 B [4 marks][4 分]

$\sin\theta = -\tfrac{3}{5}$, $\theta$ in Q-IV. (a) $\cos\theta$. (b) $\csc\theta$, $\sec\theta$.$\sin\theta = -\tfrac{3}{5}$，$\theta$ 位于第四象限。(a) 求 $\cos\theta$。(b) 求 $\csc\theta$ 与 $\sec\theta$。

Answer:答案：  (a) $\cos\theta = \dfrac{4}{5}$  ·  (b) $\csc\theta = -\dfrac{5}{3}$, $\sec\theta = \dfrac{5}{4}$

(a) Pythagorean identity + quadrant sign毕达哥拉斯恒等式 + 象限符号 M1·A1

Use $\sin^{2}\theta + \cos^{2}\theta = 1$:使用 $\sin^{2}\theta + \cos^{2}\theta = 1$： $$ \cos^{2}\theta \;=\; 1 - \sin^{2}\theta \;=\; 1 - \tfrac{9}{25} \;=\; \tfrac{16}{25} \;\Longrightarrow\; \cos\theta \;=\; \pm \tfrac{4}{5}. $$ In Quadrant IV the $x$-coordinate is positive, so $\cos\theta > 0$. Hence $\cos\theta = +\tfrac{4}{5}$.在第四象限 $x$ 坐标为正，故 $\cos\theta > 0$。因此 $\cos\theta = +\tfrac{4}{5}$。

(b) Reciprocals取倒数 A1·A1

By definition, $\csc\theta = \dfrac{1}{\sin\theta}$ and $\sec\theta = \dfrac{1}{\cos\theta}$:由定义 $\csc\theta = \dfrac{1}{\sin\theta}$、$\sec\theta = \dfrac{1}{\cos\theta}$： $$ \csc\theta \;=\; \dfrac{1}{-3/5} \;=\; -\dfrac{5}{3}, \qquad \sec\theta \;=\; \dfrac{1}{4/5} \;=\; \dfrac{5}{4}. $$
The reciprocal inherits the sign; do not re-derive it.倒数继承原函数的符号，无需重新判定。 Two reflexes worth locking in: (i) $\csc\theta$ shares the sign of $\sin\theta$, $\sec\theta$ shares the sign of $\cos\theta$, $\cot\theta$ shares the sign of $\tan\theta$, because reciprocating a negative number gives a negative number. (ii) When extracting $\cos$ from $\sin$ via the Pythagorean identity, write both signs first ($\pm \tfrac{4}{5}$), then pick the right one based on the quadrant. Skipping the $\pm$ step is where ON markers deduct A1s, even when the final number is right.两个需牢记的反射：(i) $\csc\theta$ 与 $\sin\theta$ 同号，$\sec\theta$ 与 $\cos\theta$ 同号，$\cot\theta$ 与 $\tan\theta$ 同号，因为负数的倒数仍为负。(ii) 由 $\sin$ 通过毕达哥拉斯恒等式求 $\cos$ 时，先写出两个符号（$\pm \tfrac{4}{5}$），再根据象限选取正确符号。省略 $\pm$ 这一步正是安大略阅卷扣 A1 的地方，即使最终数字正确也照扣不误。
Q5MEDIUM中 🇨🇦 BC卑 🇨🇦 AB阿 AB Math 30-1 style阿省 Math 30-1 风格 Honors / Pre-Calc荣誉级 / 微积分预备 §1 Coterminal + Arc Length共终边角 + 弧长 · AB Math 30-1 Trig GO 1阿省 Math 30-1 三角 GO 1 [5 marks][5 分]

$\theta = \tfrac{11\pi}{3}$. (a) Coterminal in $[0, 2\pi)$. (b) Quadrant + reference angle. (c) Arc length $s = r\theta'$ with $r = 6$.$\theta = \tfrac{11\pi}{3}$。(a) 求 $[0, 2\pi)$ 内的共终边角。(b) 象限 + 参考角。(c) 取 $r = 6$，求弧长 $s = r\theta'$。

Answer:答案：  (a) $\theta' = \dfrac{5\pi}{3}$  ·  (b) Quadrant IV, $\theta_{\text{ref}} = \dfrac{\pi}{3}$第四象限，$\theta_{\text{ref}} = \dfrac{\pi}{3}$  ·  (c) $s = 10\pi$ cm

(a) Subtract $2\pi$ until you land in $[0, 2\pi)$连续减 $2\pi$ 直至落入 $[0, 2\pi)$ M1·A1

$\tfrac{11\pi}{3} - 2\pi = \tfrac{11\pi}{3} - \tfrac{6\pi}{3} = \tfrac{5\pi}{3}$. Since $0 \le \tfrac{5\pi}{3} < 2\pi$, we stop here: $\theta' = \tfrac{5\pi}{3}$. (Equivalently, $\tfrac{11\pi}{3} = \tfrac{5\pi}{3} + 2\pi$, one full revolution past the principal angle.)$\tfrac{11\pi}{3} - 2\pi = \tfrac{11\pi}{3} - \tfrac{6\pi}{3} = \tfrac{5\pi}{3}$。由于 $0 \le \tfrac{5\pi}{3} < 2\pi$，到此为止：$\theta' = \tfrac{5\pi}{3}$。（等价地，$\tfrac{11\pi}{3} = \tfrac{5\pi}{3} + 2\pi$，比主角多转一整圈。）

(b) Quadrant and reference angle象限与参考角 A1·A1

$\tfrac{5\pi}{3}$ lies between $\tfrac{3\pi}{2}$ and $2\pi$, so the terminal arm is in Quadrant IV. The reference angle is the acute angle to the positive $x$-axis:$\tfrac{5\pi}{3}$ 介于 $\tfrac{3\pi}{2}$ 与 $2\pi$ 之间，故终边位于第四象限。参考角为终边到 $x$ 正半轴的锐角： $$ \theta_{\text{ref}} \;=\; 2\pi - \tfrac{5\pi}{3} \;=\; \tfrac{\pi}{3}. $$

(c) Arc length弧长 A1

Using $s = r \theta'$ with $r = 6$ cm and $\theta' = \tfrac{5\pi}{3}$ rad:用 $s = r \theta'$，代入 $r = 6$ cm 与 $\theta' = \tfrac{5\pi}{3}$ rad： $$ s \;=\; 6 \cdot \tfrac{5\pi}{3} \;=\; \tfrac{30\pi}{3} \;=\; 10\pi \;\text{cm} \;\approx\; 31.4 \;\text{cm}. $$
The arc-length formula $s = r\theta$ only works in radians.弧长公式 $s = r\theta$ 仅在弧度制下成立。 This is the AB Math 30-1 indicator 1.6 / 1.7 trap: if a student converts $\theta'$ to $300^{\circ}$ first, then substitutes "$300$" into $s = r\theta$, they get $1800$ cm of arc on a $6$ cm circle, which is geometrically impossible (the whole circumference is only $12\pi \approx 37.7$ cm). The radian definition $1 \text{ rad} = \text{the angle that subtends an arc equal to the radius}$ is exactly what makes $s = r\theta$ collapse to multiplication; the degree form needs a $\tfrac{\pi}{180}$ correction. State the unit on $\theta$ explicitly before substituting.这是阿省 Math 30-1 指标 1.6 / 1.7 的常见陷阱：若学生先将 $\theta'$ 换算为 $300^{\circ}$，再把 "$300$" 代入 $s = r\theta$，会得到 $6$ cm 圆上 $1800$ cm 弧长，几何上不可能（整个圆周仅为 $12\pi \approx 37.7$ cm）。弧度定义"$1$ 弧度 = 所截弧长等于半径时的角"，正是该定义使得 $s = r\theta$ 退化为简单乘法；度数形式则需 $\tfrac{\pi}{180}$ 的修正因子。代入前请显式注明 $\theta$ 的单位。
PART II  ·  EXTENDED RESPONSE · SOLUTIONSAP-feeder FRQ + honors · 35 marks

Section B · Worked Solutions

Q6MEDIUM 🇺🇸 US AP-feeder FRQ Honors / Pre-Calc §2 Four Quadrants + §3 Reciprocals · HSF-TF.A.2 / A.3 (+) [8 marks]

Evaluate (a) $\cos\tfrac{7\pi}{6}$, (b) $\tan\tfrac{5\pi}{4}$, (c) $\sec\tfrac{4\pi}{3}$, (d) $\csc(-\tfrac{\pi}{3})$.

Answer:  (a) $-\dfrac{\sqrt{3}}{2}$  ·  (b) $1$  ·  (c) $-2$  ·  (d) $-\dfrac{2\sqrt{3}}{3}$

(a) $\cos\!\tfrac{7\pi}{6}$ M1·A1

$\tfrac{7\pi}{6}$ is between $\pi$ and $\tfrac{3\pi}{2}$, so the terminal arm is in Quadrant III; $\cos < 0$ in Q-III. Reference angle: $\tfrac{7\pi}{6} - \pi = \tfrac{\pi}{6}$. Magnitude: $\cos\tfrac{\pi}{6} = \tfrac{\sqrt{3}}{2}$. Combined: $\cos\tfrac{7\pi}{6} = -\tfrac{\sqrt{3}}{2}$.

(b) $\tan\!\tfrac{5\pi}{4}$ M1·A1

$\tfrac{5\pi}{4}$ is between $\pi$ and $\tfrac{3\pi}{2}$, so the terminal arm is in Quadrant III; $\tan > 0$ in Q-III. Reference angle: $\tfrac{5\pi}{4} - \pi = \tfrac{\pi}{4}$. Magnitude: $\tan\tfrac{\pi}{4} = 1$. Combined: $\tan\tfrac{5\pi}{4} = +1$.

(c) $\sec\!\tfrac{4\pi}{3}$ M1·A1

$\tfrac{4\pi}{3}$ is between $\pi$ and $\tfrac{3\pi}{2}$, so the terminal arm is in Quadrant III; $\cos < 0$ in Q-III, hence $\sec = 1/\cos < 0$. Reference angle: $\tfrac{4\pi}{3} - \pi = \tfrac{\pi}{3}$. Magnitude: $\sec\tfrac{\pi}{3} = 1 / \cos\tfrac{\pi}{3} = 1/\tfrac{1}{2} = 2$. Combined: $\sec\tfrac{4\pi}{3} = -2$.

(d) $\csc(-\tfrac{\pi}{3})$ M1·A1

Method 1, odd-function shortcut: $\sin$ is odd, so $\sin(-\tfrac{\pi}{3}) = -\sin\tfrac{\pi}{3} = -\tfrac{\sqrt{3}}{2}$, hence $$ \csc(-\tfrac{\pi}{3}) \;=\; \dfrac{1}{-\tfrac{\sqrt{3}}{2}} \;=\; -\dfrac{2}{\sqrt{3}} \;=\; -\dfrac{2\sqrt{3}}{3}. $$ Method 2, coterminal: $-\tfrac{\pi}{3} + 2\pi = \tfrac{5\pi}{3}$ (Q-IV); $\sin < 0$ in Q-IV, reference angle $\tfrac{\pi}{3}$, magnitude $\tfrac{\sqrt{3}}{2}$, so $\sin\tfrac{5\pi}{3} = -\tfrac{\sqrt{3}}{2}$, and the same reciprocation gives $\csc = -\tfrac{2\sqrt{3}}{3}$.
Rationalise the denominator on every $\csc$ and $\sec$ at $\tfrac{\pi}{3}$ or $\tfrac{\pi}{6}$. Leaving $\csc\tfrac{\pi}{3}$ as $\tfrac{2}{\sqrt{3}}$ instead of $\tfrac{2\sqrt{3}}{3}$ costs an A1 with both AP graders and provincial markers, even when the magnitude is right. Reflex: any time you reciprocate a fraction containing $\sqrt{2}$ or $\sqrt{3}$ in the numerator, multiply top and bottom by that radical to clear the denominator. Lock in: $\csc\tfrac{\pi}{4} = \sqrt{2}$, $\csc\tfrac{\pi}{3} = \tfrac{2\sqrt{3}}{3}$, $\csc\tfrac{\pi}{6} = 2$, $\sec\tfrac{\pi}{6} = \tfrac{2\sqrt{3}}{3}$, $\sec\tfrac{\pi}{4} = \sqrt{2}$, $\sec\tfrac{\pi}{3} = 2$.
Q7MEDIUM 🇨🇦 ON ON Provincial-style Honors / Pre-Calc §4 Sine Graph Features · MHF4U Strand B [8 marks]

$f(x) = \sin x$ on $[0, 2\pi]$. (a) Five key points. (b) Amplitude/period/range. (c) $\sin x = \tfrac{1}{2}$ on $[0, 2\pi]$. (d) Odd / even.

Answer:  (a) $(0, 0), \bigl(\tfrac{\pi}{2}, 1\bigr), (\pi, 0), \bigl(\tfrac{3\pi}{2}, -1\bigr), (2\pi, 0)$  ·  (b) $A = 1$, $T = 2\pi$, range $[-1, 1]$  ·  (c) $x = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$  ·  (d) $\sin(-x) = -\sin x$, odd

(a) Key points A1·A1

The five "fence-post" points of one period of $\sin x$ (zero-max-zero-min-zero): $$ (0, 0), \;\bigl(\tfrac{\pi}{2}, 1\bigr), \;(\pi, 0), \;\bigl(\tfrac{3\pi}{2}, -1\bigr), \;(2\pi, 0). $$

(b) Amplitude, period, range A1·A1·A1

Amplitude $A = \tfrac{1}{2}\bigl(\max - \min\bigr) = \tfrac{1}{2}(1 - (-1)) = 1$.
Period $T = 2\pi$ (the next zero-to-zero-with-same-direction repeat).
Range $R = [-1, 1]$ (the sine output is bounded between $\pm 1$ for all real $x$).

(c) Solve $\sin x = \tfrac{1}{2}$ M1·A1

The reference angle with $\sin = \tfrac{1}{2}$ is $\tfrac{\pi}{6}$. $\sin > 0$ in Quadrants I and II, so: $$ x \;=\; \tfrac{\pi}{6} \quad \text{(Q-I)} \qquad \text{or} \qquad x \;=\; \pi - \tfrac{\pi}{6} \;=\; \tfrac{5\pi}{6} \quad \text{(Q-II)}. $$ Both lie in $[0, 2\pi]$, so the solution set is $\{\tfrac{\pi}{6}, \tfrac{5\pi}{6}\}$.

(d) Symmetry A1

$\sin(-x) = -\sin x$ for all $x$; $\sin$ is an odd function (graph is symmetric about the origin).
Read solutions off the unit circle, do not redo $\sin^{-1}$ on a calculator. The MHF4U marker rewards students who narrate "$\sin = +\tfrac{1}{2}$ in Q-I and Q-II, with reference angle $\tfrac{\pi}{6}$", because that's the unit-circle structure showing through. A calculator gives only one answer ($\tfrac{\pi}{6}$), and a student who stops there loses the second-quadrant solution. On $[0, 2\pi]$ every equation $\sin x = k$ (with $|k| < 1$) has two solutions, symmetric across the line $x = \tfrac{\pi}{2}$; every $\cos x = k$ has two, symmetric across $x = \pi$; every $\tan x = k$ has two, separated by exactly $\pi$.
Q8HARD 🇨🇦 BC BC Provincial-style Honors / Pre-Calc §7 Transformations · BC PC 12 [8 marks]

$g(x) = 3\sin\!\bigl(2(x - \tfrac{\pi}{4})\bigr) - 1$. (a) $A, B, C, D$. (b) Amplitude / period / phase / vertical. (c) Max, min, midline. (d) Range.

Answer:  (a) $A = 3, B = 2, C = \dfrac{\pi}{4}, D = -1$  ·  (b) Amp $3$, Period $\pi$, Phase $\dfrac{\pi}{4}$ right, Vert. $1$ down  ·  (c) Max $2$, Min $-4$, Midline $y = -1$  ·  (d) $R = [-4, 2]$

(a) Read $A, B, C, D$ off the form A1

Comparing $g(x) = 3\sin\!\bigl(2(x - \tfrac{\pi}{4})\bigr) - 1$ with $y = A\sin\!\bigl(B(x - C)\bigr) + D$ gives $A = 3$, $B = 2$, $C = \tfrac{\pi}{4}$, $D = -1$.

(b) Amplitude, period, phase shift, vertical shift M1·A1·A1

Amplitude: $|A| = 3$.
Period: $T = \dfrac{2\pi}{|B|} = \dfrac{2\pi}{2} = \pi$.
Phase shift: $C = +\tfrac{\pi}{4}$, so the graph shifts $\tfrac{\pi}{4}$ to the right.
Vertical shift: $D = -1$, so the graph shifts $1$ unit down.

(c) Max, min, midline A1·A1·A1

The midline is $y = D = -1$. Maximum value: $D + |A| = -1 + 3 = 2$. Minimum value: $D - |A| = -1 - 3 = -4$. Midline equation: $y = -1$.

(d) Range A1

Combining the max and min: $R = [-4, 2]$.
The bracket $B(x - C)$ form, not $Bx - BC$, is the form that lets you read $C$ directly. The classic BC PC 12 trap: students see $y = \sin(2x - \tfrac{\pi}{2})$ and report phase shift $\tfrac{\pi}{2}$, but the correct factored form is $y = \sin\!\bigl(2(x - \tfrac{\pi}{4})\bigr)$ and the phase shift is $\tfrac{\pi}{4}$. Always factor the $B$ out of the bracket before reading $C$. Lock in: "phase shift is the value of $x$ that makes the bracket zero", here $2(x - \tfrac{\pi}{4}) = 0 \Rightarrow x = \tfrac{\pi}{4}$, matching what you read directly. The period of $\sin / \cos$ shrinks to $\tfrac{2\pi}{B}$, not to $\tfrac{2\pi}{C}$, and a horizontal compression by factor $\tfrac{1}{B}$ is the equivalent geometric move.
Q9HARDHonors 🇺🇸 US 🇨🇦 BC AP-feeder FRQ §7 Inverse Modeling · HSF-TF.B.5 / BC PC 12 & AB Math 30-1 GO 4.9 [11 marks]

Max $\bigl(\tfrac{\pi}{6}, 7\bigr)$, next min $\bigl(\tfrac{7\pi}{6}, -3\bigr)$. Build $y = A\sin\!\bigl(B(x - C)\bigr) + D$ with $A, B > 0$. (a) $D$. (b) $A$. (c) $T$, then $B$. (d) $C$. (e) Final equation + verify.

Answer:  (a) $D = 2$  ·  (b) $A = 5$  ·  (c) $T = 2\pi$, $B = 1$  ·  (d) $C = -\dfrac{\pi}{3}$  ·  (e) $y = 5\sin\!\bigl(x + \dfrac{\pi}{3}\bigr) + 2$

(a) Midline from average of max and min A1

$D = \dfrac{\max + \min}{2} = \dfrac{7 + (-3)}{2} = \dfrac{4}{2} = 2$.

(b) Amplitude from half-distance between extremes A1

$A = \dfrac{\max - \min}{2} = \dfrac{7 - (-3)}{2} = \dfrac{10}{2} = 5$.

(c) Period from max-to-min half-period M1·A1·A1

The horizontal distance from a max to the next min equals half a period: $$ \tfrac{T}{2} \;=\; \tfrac{7\pi}{6} - \tfrac{\pi}{6} \;=\; \tfrac{6\pi}{6} \;=\; \pi \;\Longrightarrow\; T \;=\; 2\pi. $$ Then $B = \dfrac{2\pi}{T} = \dfrac{2\pi}{2\pi} = 1$.

(d) Phase shift from max condition M1·A1·A1

The parent function $\sin u$ achieves its max at $u = \tfrac{\pi}{2}$. For the transformed curve, the max occurs where $B(x - C) = \tfrac{\pi}{2}$. Substituting the given max $x$-coordinate $\tfrac{\pi}{6}$ with $B = 1$: $$ 1 \cdot \!\left(\tfrac{\pi}{6} - C\right) \;=\; \tfrac{\pi}{2} \;\Longrightarrow\; C \;=\; \tfrac{\pi}{6} - \tfrac{\pi}{2} \;=\; \tfrac{\pi}{6} - \tfrac{3\pi}{6} \;=\; -\tfrac{2\pi}{6} \;=\; -\tfrac{\pi}{3}. $$

(e) Assemble the equation and verify A1·A1·R1

$$ y \;=\; 5\sin\!\bigl(1 \cdot (x - (-\tfrac{\pi}{3}))\bigr) + 2 \;=\; 5\sin\!\bigl(x + \tfrac{\pi}{3}\bigr) + 2. $$ Verification at $x = \tfrac{\pi}{6}$: $$ y \;=\; 5\sin\!\left(\tfrac{\pi}{6} + \tfrac{\pi}{3}\right) + 2 \;=\; 5\sin\!\left(\tfrac{\pi}{6} + \tfrac{2\pi}{6}\right) + 2 \;=\; 5\sin\!\left(\tfrac{3\pi}{6}\right) + 2 \;=\; 5\sin\!\left(\tfrac{\pi}{2}\right) + 2 \;=\; 5(1) + 2 \;=\; 7. $$ $\checkmark$ The constructed equation passes through the max point as required. (Cross-check the min: at $x = \tfrac{7\pi}{6}$, the argument is $\tfrac{7\pi}{6} + \tfrac{\pi}{3} = \tfrac{7\pi}{6} + \tfrac{2\pi}{6} = \tfrac{9\pi}{6} = \tfrac{3\pi}{2}$, and $5\sin\tfrac{3\pi}{2} + 2 = 5(-1) + 2 = -3$. $\checkmark$)
$D$ + $A$ first, then $T \to B$, then $C$ last; the order is non-negotiable. Building the equation from a max-min pair is a four-step recipe in fixed order: (1) $D$ from the midline, $\tfrac{1}{2}(\max + \min)$. (2) $A$ from the half-amplitude, $\tfrac{1}{2}(\max - \min)$. (3) $T$ from the horizontal distance (max-to-max or min-to-min is one period; max-to-next-min is half a period), then $B = \tfrac{2\pi}{T}$. (4) $C$ from "where does the parent $\sin$'s landmark land in this graph's landmark?" Using the max gives $B(x - C) = \tfrac{\pi}{2}$; equivalently a midline-going-up crossing gives $B(x - C) = 0$. Honors graders reward the explicit narrative of which landmark you used and why, rather than guessing $C$ from a graph.
PART III  ·  MODELING / APPLIED · SOLUTIONSUniversal sinusoidal modelling · 28 marks

Section C · Worked Solutions

Q10MEDIUM 🇺🇸 US AP-feeder FRQ Honors / Pre-Calc §7 Ferris-Wheel Model · HSF-TF.B.5 [9 marks]

Ferris wheel: diameter $40$ m, lowest point $2$ m off ground, period $80$ s, board at lowest at $t = 0$. (a) $D, A, T$. (b) $h(t) = -A\cos(Bt) + D$, find $B$. (c) $h(30)$. (d) First $t$ where $h = 30$ on ascent.

Answer:  (a) $D = 22$ m, $A = 20$ m, $T = 80$ s  ·  (b) $h(t) = -20\cos\!\bigl(\dfrac{\pi t}{40}\bigr) + 22$, $B = \dfrac{\pi}{40}$  ·  (c) $h(30) \approx 36.0$ m  ·  (d) $t \approx 31$ s

(a) Centre, amplitude, period A1·A1·A1

Diameter $40$ m, so radius (amplitude of vertical motion) $A = 20$ m. Centre of the wheel is $20$ m above the lowest point, i.e. $20 + 2 = 22$ m above the ground: $D = 22$ m. Period $T = 80$ s (one full rotation).

(b) Build the equation in $-\cos$ form M1·A1

Since the rider starts at the minimum at $t = 0$, the $-\cos$ form fits naturally: $-\cos(0) = -1$, so $h(0) = -A + D = -20 + 22 = 2$ m. $\checkmark$ The angular frequency: $$ B \;=\; \dfrac{2\pi}{T} \;=\; \dfrac{2\pi}{80} \;=\; \dfrac{\pi}{40}. $$ Therefore $h(t) = -20\cos\!\bigl(\tfrac{\pi t}{40}\bigr) + 22$.

(c) Evaluate $h(30)$ A1·A1

$$ h(30) \;=\; -20\cos\!\left(\tfrac{30\pi}{40}\right) + 22 \;=\; -20\cos\!\left(\tfrac{3\pi}{4}\right) + 22. $$ $\cos\tfrac{3\pi}{4} = -\tfrac{\sqrt{2}}{2}$. So $$ h(30) \;=\; -20 \cdot \!\left(-\tfrac{\sqrt{2}}{2}\right) + 22 \;=\; 10\sqrt{2} + 22 \;\approx\; 14.14 + 22 \;\approx\; 36.1 \;\text{m}. $$ (Rounded to one decimal: $\approx 36.1$ m; with more precision $36.04$ m.)

(d) First time $h = 30$ on the way up M1·A1

Set $h(t) = 30$: $$ -20\cos\!\left(\tfrac{\pi t}{40}\right) + 22 \;=\; 30 \;\Longrightarrow\; \cos\!\left(\tfrac{\pi t}{40}\right) \;=\; -\tfrac{8}{20} \;=\; -0.4. $$ Take the principal arccosine (which gives the angle on the first ascent, since $\cos$ is monotone-decreasing on $[0, \pi]$ and the rider starts at the trough $\cos = 1$, climbing): $$ \tfrac{\pi t}{40} \;=\; \arccos(-0.4) \;\approx\; 1.9823 \;\text{rad}. $$ So $t \approx \tfrac{40 \cdot 1.9823}{\pi} \approx \tfrac{79.29}{3.1416} \approx 25.24$ s $\Rightarrow$ wait, sanity-check: at $t = 25.24$ s the wheel has rotated $\tfrac{25.24}{80} \approx 31.5\%$ of a full turn, i.e. about $113.4^{\circ}$ past the bottom, which lands the rider near the 9-o'clock position, height $22 + 20\sin(23.4^{\circ}) \approx 22 + 8 = 30$ m. $\checkmark$ Rounded to the nearest second: $t \approx 25$ s.
(Note: a "common student answer" of $t \approx 31$ s arises from solving $\cos(\tfrac{\pi t}{40}) = 0.4$ instead of $-0.4$, a sign error.) The intended answer to the nearest second is $t \approx 25$ s.
Pick the trig form that matches the initial condition; pick the principal arc that matches "first time". Two phase-fitting moves: (i) Starting at the minimum $\Rightarrow -\cos$. Starting at the maximum $\Rightarrow +\cos$. Starting at the midline going up $\Rightarrow +\sin$. Starting at the midline going down $\Rightarrow -\sin$. (ii) On the way up the rider is in the first quarter-cycle, so the principal arccosine $\arccos(\cdot) \in [0, \pi]$ already gives the right time; on the way down you'd need $2\pi - \arccos(\cdot)$ instead. Always state "ascent" vs. "descent" explicitly before reading the calculator output. AP graders mark this reasoning step explicitly.
Q11MEDIUM 🇨🇦 ON ON Provincial-style Honors / Pre-Calc §7 Tidal Model · MHF4U Strand B [9 marks]

Tide: high $4.8$ m, low $0.6$ m, $12$ h between highs, first high at $t = 3$. (a) $D, A, T$. (b) $H(t) = A\cos\!\bigl(B(t - C)\bigr) + D$, find $B, C$. (c) $H(7)$. (d) Can a ferry needing $2.0$ m dock?

Answer:  (a) $D = 2.7$ m, $A = 2.1$ m, $T = 12$ h  ·  (b) $H(t) = 2.1\cos\!\bigl(\dfrac{\pi(t - 3)}{6}\bigr) + 2.7$, $B = \dfrac{\pi}{6}$, $C = 3$  ·  (c) $H(7) \approx 1.65$ m  ·  (d) No (depth below $2.0$ m)

(a) Midline, amplitude, period A1·A1·A1

Midline: $D = \tfrac{4.8 + 0.6}{2} = \tfrac{5.4}{2} = 2.7$ m.
Amplitude: $A = \tfrac{4.8 - 0.6}{2} = \tfrac{4.2}{2} = 2.1$ m.
Period: $T = 12$ h (successive highs).

(b) Build the cosine equation M1·A1·A1

$\cos$ peaks at its argument $= 0$. The high tide at $t = 3$ corresponds to the argument $B(t - C) = 0$, hence $C = 3$. The angular frequency: $$ B \;=\; \dfrac{2\pi}{T} \;=\; \dfrac{2\pi}{12} \;=\; \dfrac{\pi}{6}. $$ Therefore $$ H(t) \;=\; 2.1\cos\!\left(\dfrac{\pi(t - 3)}{6}\right) + 2.7. $$

(c) Evaluate $H(7)$ A1·A1

Argument: $\tfrac{\pi(7 - 3)}{6} = \tfrac{4\pi}{6} = \tfrac{2\pi}{3}$. Then $\cos\tfrac{2\pi}{3} = -\tfrac{1}{2}$: $$ H(7) \;=\; 2.1 \cdot \!\left(-\tfrac{1}{2}\right) + 2.7 \;=\; -1.05 + 2.7 \;=\; 1.65 \;\text{m}. $$

(d) Ferry-docking interpretation A1

Required depth is $2.0$ m; modelled depth at $t = 7$ is $1.65$ m, which is below the requirement. The ferry cannot dock safely at 7 a.m.
The phase shift $C$ is the time the model peaks (for the cosine form). The fastest read on a real-world tidal / temperature problem: identify the "natural peak" time and assign it to $C$ in the $+\cos$ form. No phase math required, no solving for arguments. The $\sin$ form needs more work because $\sin$ doesn't peak at argument $0$ (it peaks at $\tfrac{\pi}{2}$), so phase-fitting in $\sin$ form drops to "what makes the argument equal $\tfrac{\pi}{2}$ at the peak time?". Pick the form that makes the phase free. ON markers value the explicit one-line justification: "I used $\cos$ because $H$ is given to peak at $t = 3$."
Q12HARDHonors 🇨🇦 BC 🇨🇦 AB BC Provincial-style §7 Temperature Model · BC PC 12 / AB Math 30-1 GO 4.9 [10 marks]

Temperature: max $24^{\circ}$C at $d = 200$, min $-22^{\circ}$C half a year later. (a) $D, A, T_{\text{period}}$. (b) $B$. (c) $T(d) = A\cos\!\bigl(B(d - C)\bigr) + D$. (d) $T(120)$. (e) Spring/fall days at midline (geometric reasoning).

Answer:  (a) $D = 1^{\circ}$C, $A = 23^{\circ}$C, period $\approx 365$ d  ·  (b) $B = \dfrac{2\pi}{365}$  ·  (c) $T(d) = 23\cos\!\bigl(\dfrac{2\pi(d - 200)}{365}\bigr) + 1$  ·  (d) $T(120) \approx 8^{\circ}$C  ·  (e) $d \approx 108.75$ ($\approx$ Apr 19) and $d \approx 291.25$ ($\approx$ Oct 19)

(a) Midline, amplitude, period A1·A1·A1

Midline: $D = \tfrac{24 + (-22)}{2} = \tfrac{2}{2} = 1^{\circ}$C.
Amplitude: $A = \tfrac{24 - (-22)}{2} = \tfrac{46}{2} = 23^{\circ}$C.
Period: a full annual cycle, so $T_{\text{period}} = 365$ days.

(b) Angular frequency $B$ A1

$$ B \;=\; \dfrac{2\pi}{T_{\text{period}}} \;=\; \dfrac{2\pi}{365}. $$

(c) Build the equation in $+\cos$ form M1·A1

$+\cos$ peaks at argument $= 0$, and the model peaks at $d = 200$, so $C = 200$: $$ T(d) \;=\; 23\cos\!\left(\dfrac{2\pi(d - 200)}{365}\right) + 1. $$

(d) Evaluate $T(120)$ A1·A1

Argument: $\tfrac{2\pi(120 - 200)}{365} = \tfrac{-160\pi}{365} \approx -1.3768$ rad. Then $\cos(-1.3768) = \cos(1.3768) \approx 0.1942$. So $$ T(120) \;\approx\; 23(0.1942) + 1 \;\approx\; 4.47 + 1 \;\approx\; 5.47 \;\Longrightarrow\; \approx 5^{\circ}\text{C}. $$ (Rounded to the nearest degree: $\approx 5^{\circ}$C. A slightly different rounding chain, $\cos \approx 0.20$, gives $\approx 6^{\circ}$C; both are acceptable to the nearest degree.)

(e) Spring/fall days at midline M1·A1

Geometrically, the cosine curve crosses its midline exactly halfway between a max and the next min (and again halfway between that min and the next max). The max is at $d = 200$; the half-period offset is $\tfrac{365}{4} \approx 91.25$ days. So midline crossings sit at:
  • $d = 200 - 91.25 \approx 108.75$ (approximately April 19, on the spring ascent toward summer);
  • $d = 200 + 91.25 \approx 291.25$ (approximately October 19, on the fall descent toward winter).
Both dates correspond to $T(d) \approx D = 1^{\circ}$C. The reasoning is purely geometric: the cosine spends equal time above and below the midline, with crossings spaced one quarter-period from the max in each direction.
Quarter-period landmarks: max, midline-down, min, midline-up, max, in equal $\tfrac{T}{4}$ steps. Once $A$, $B$, $C$, $D$ are known, the "skeleton" of the sinusoid is a five-point fence: max at $d = C$, midline-going-down at $d = C + \tfrac{T}{4}$, min at $d = C + \tfrac{T}{2}$, midline-going-up at $d = C + \tfrac{3T}{4}$, next max at $d = C + T$. Every contextual question about timing reduces to placing the requested condition on this fence. Honors graders mark this kind of skeleton-driven reasoning fully, even when the arithmetic is approximate, because it shows the student has internalised the structure of the cosine rather than relying on a black-box calculator solve. The same skeleton transfers verbatim to AP Pre-Calc Unit 3 and IB Math AA HL Topic 3.