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Trigonometric Identities and Equations · Solutions三角恒等式与三角方程 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · SAT / AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · SAT / AP 衔接 / 安 / 卑 / 艾省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB艾 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 Provincial-style艾伯塔省考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解SAT MCQ + ON/BC/AB short answer · 18 marksSAT 选择题 + 安/卑/艾省考短答 · 共 18 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §1 Pythagorean Identity勾股恒等式 · HSF-TF.C.8 [3 marks][3 分]

$\sin \theta = \tfrac{3}{5}$, $\theta$ in Quadrant II. Find $\cos \theta$.$\sin \theta = \tfrac{3}{5}$，$\theta$ 在第二象限。求 $\cos \theta$。

Answer:答案：  (B)  $\cos \theta = -\tfrac{4}{5}$

(a) Apply $\sin^{2} \theta + \cos^{2} \theta = 1$套用勾股恒等式 $\sin^{2} \theta + \cos^{2} \theta = 1$ M1·A1

$\cos^{2} \theta = 1 - \sin^{2} \theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}$, so $\cos \theta = \pm \tfrac{4}{5}$.故 $\cos \theta = \pm \tfrac{4}{5}$。

(b) Use the quadrant to fix the sign用象限判定符号 R1

In Quadrant II, $\cos \theta < 0$ (the $x$-coordinate on the unit circle is negative there). Hence $\cos \theta = -\tfrac{4}{5}$, matching option (B).第二象限内 $\cos \theta < 0$（单位圆上该处的 $x$ 坐标为负）。故 $\cos \theta = -\tfrac{4}{5}$，对应选项 (B)。
Why the wrong choices fail.干扰项分析。
  • (A) $\tfrac{4}{5}$, correct magnitude but wrong sign: this is the Quadrant I answer; the student forgot the quadrant constraint.绝对值正确但符号错误：这是第一象限的答案，学生忽略了象限约束。
  • (C) $\tfrac{3}{4}$, this is $\sin / \cos$ structure applied incorrectly (looks like a tangent-style fraction from the $3$ and $4$ visible in the working).误用 $\sin / \cos$ 结构（从演算中可见的 $3$ 与 $4$ 拼出一个看似正切的分数）。
  • (D) $-\tfrac{3}{4}$, same fraction-confusion as (C) with a sign flip; both (C) and (D) confuse magnitudes from the $3$-$4$-$5$ triangle.与 (C) 同样的分数混淆，再翻一次号；(C) 与 (D) 都是把 $3$-$4$-$5$ 三角形的边长混在一起。
The "ASTC" rule (All-Sin-Tan-Cos) decides every sign, but always cross-check the unit-circle picture."ASTC"（All-Sin-Tan-Cos）口诀决定每个符号，但永远要回到单位圆图像核对。 A faster diagnostic than ASTC: just remember $\cos = x$-coordinate, $\sin = y$-coordinate on the unit circle. Quadrant II is upper-left, so $x < 0$ (cos negative) and $y > 0$ (sin positive). That single fact handles every "which sign?" question in one second, with no mnemonic needed. SAT loves $3$-$4$-$5$ and $5$-$12$-$13$ triples precisely because the magnitudes are recognisable and the only real test is the sign.比 ASTC 更快的诊断方法是：记住单位圆上 $\cos$ 就是 $x$ 坐标、$\sin$ 就是 $y$ 坐标。第二象限位于左上方，所以 $x < 0$（cos 为负）、$y > 0$（sin 为正）。这一条规律一秒就能搞定任何"符号怎么定"的题目，根本不用口诀。SAT 偏爱 $3$-$4$-$5$ 与 $5$-$12$-$13$ 三元组，正是因为绝对值一目了然，真正考的是符号。
Q2EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §2 Quotient / Reciprocal商 / 倒数恒等式 · HSF-TF.A.2 [3 marks][3 分]

Simplify $\dfrac{1}{\cos \theta} \cdot \sin \theta$.化简 $\dfrac{1}{\cos \theta} \cdot \sin \theta$。

Answer:答案：  (C)  $\tan \theta$

(a) Combine the product into one fraction把乘积合并为一个分式 M1·A1·A1

$$ \frac{1}{\cos \theta} \cdot \sin \theta \;=\; \frac{\sin \theta}{\cos \theta} \;=\; \tan \theta. $$ This is the quotient identity, $\tan \theta \;=\; \dfrac{\sin \theta}{\cos \theta}$, read backwards. Matches option (C).这正是商恒等式 $\tan \theta \;=\; \dfrac{\sin \theta}{\cos \theta}$ 反向使用。对应选项 (C)。
Why the wrong choices fail.干扰项分析。
  • (A) $\cot \theta = \cos / \sin$, this is the reciprocal of the correct answer; obtained by swapping numerator and denominator.这是正确答案的倒数，由颠倒分子分母得到。
  • (B) $\sec \theta = 1 / \cos$, this is what you get if you forget to multiply by $\sin \theta$; only half the expression is simplified.若忘记乘以 $\sin \theta$ 就会得到这个，只化简了一半表达式。
  • (D) $\csc \theta = 1 / \sin$, this would arise if the student mistook $\sin \theta$ in the numerator as $1 / \sin \theta$ in the denominator (a confused "reciprocal-of-numerator" error).若把分子的 $\sin \theta$ 误认为分母上的 $1 / \sin \theta$ 就会出现这个（"把分子当倒数"的混乱错误）。
Six trig ratios, two relationships: quotient and reciprocal.六个三角比，两种关系：商与倒数。 Lock the six ratios into one map: $\sin$ and $\csc$ are reciprocals; $\cos$ and $\sec$ are reciprocals; $\tan$ and $\cot$ are reciprocals. Inside that, $\tan = \sin/\cos$ and $\cot = \cos/\sin$ are the quotient identities. Every "simplify" SAT item is either a quotient compression (this one), a Pythagorean conversion ($1 - \sin^{2} = \cos^{2}$), or a reciprocal substitution ($\sec = 1/\cos$). Tag each problem to one of those three moves before doing any algebra.把六个三角比整合成一张关系图：$\sin$ 与 $\csc$ 互为倒数，$\cos$ 与 $\sec$ 互为倒数，$\tan$ 与 $\cot$ 互为倒数。在这之内，$\tan = \sin/\cos$ 与 $\cot = \cos/\sin$ 是商恒等式。SAT 上每一道"化简"题不外乎商压缩（本题）、勾股替换（$1 - \sin^{2} = \cos^{2}$）或倒数替换（$\sec = 1/\cos$）。动笔前先把题目归到这三招中的某一招。
Q3MEDIUM中 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §7 Linear Trig Equation一次三角方程 · HSF-TF.B.7 (+) [3 marks][3 分]

Number of solutions of $2 \sin x - 1 = 0$ on $0 \le x < 2\pi$.$2 \sin x - 1 = 0$ 在 $0 \le x < 2\pi$ 上的解的个数。

Answer:答案：  (B)  Exactly two solutions恰有两个解

(a) Isolate $\sin x$分离 $\sin x$ M1

$2 \sin x - 1 = 0 \Longrightarrow \sin x = \tfrac{1}{2}$.

(b) Use the unit circle on $[0, 2\pi)$在 $[0, 2\pi)$ 上扫单位圆 A1·A1

$\sin x = \tfrac{1}{2}$ at $x = \tfrac{\pi}{6}$ (reference angle) and at $x = \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$ (the other angle in the same row of the unit circle). Both lie in $[0, 2\pi)$. So there are exactly two solutions: $x \in \{\tfrac{\pi}{6}, \tfrac{5\pi}{6}\}$, matching option (B).$\sin x = \tfrac{1}{2}$ 在 $x = \tfrac{\pi}{6}$（参考角）以及 $x = \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$（单位圆上同一行的另一角）处取得。两者都在 $[0, 2\pi)$ 内。故恰有两个解：$x \in \{\tfrac{\pi}{6}, \tfrac{5\pi}{6}\}$，对应选项 (B)。
Why the wrong choices fail.干扰项分析。
  • (A) Exactly one恰有一个, the trap for students who stop at the principal value $x = \arcsin(\tfrac{1}{2}) = \tfrac{\pi}{6}$ and forget that sine takes every value in $(-1, 1)$ twice per period.这是只写主值 $x = \arcsin(\tfrac{1}{2}) = \tfrac{\pi}{6}$ 就停笔的陷阱，忘了 sine 在 $(-1, 1)$ 内的每个值在一个周期内会取两次。
  • (C) Exactly three恰有三个, would require a non-linear equation (e.g. $\sin x = c$ for $|c| < 1$ on an interval longer than $2\pi$). On exactly one period there can be at most two.需要非一次方程或区间长于 $2\pi$ 才会出现；恰好一个周期内最多两个解。
  • (D) No solutions无解, would require $|\sin x| > 1$; $\tfrac{1}{2}$ is comfortably inside $[-1, 1]$, so the equation is solvable.这需要 $|\sin x| > 1$；而 $\tfrac{1}{2}$ 稳稳落在 $[-1, 1]$ 内，方程显然有解。
"Two solutions per period" is the default for $\sin x = c$ and $\cos x = c$ with $|c| < 1$.当 $|c| < 1$ 时，$\sin x = c$ 与 $\cos x = c$ 的默认情形是"一个周期两个解"。 The exceptions are the boundary cases $c = \pm 1$ (one solution per period, the peak or trough) and $|c| > 1$ (no solutions). For tangent the default flips to one solution per period because $\tan$ has period $\pi$ instead of $2\pi$. Counting solutions on a restricted interval is mostly bookkeeping: principal value $+$ unit-circle reflection partner, then check both lie in the stated interval.例外是边界情形 $c = \pm 1$（一个周期一个解，即峰或谷）和 $|c| > 1$（无解）。tangent 的默认翻转为一个周期一个解，因为 $\tan$ 的周期是 $\pi$ 而非 $2\pi$。限定区间上的"数解题"基本上就是记账：主值加上单位圆对称伙伴，再核对两者是否都落在题目给定的区间内。
Q4MEDIUM 🇨🇦 ON ON Provincial-style §1–§2 All Six Ratios · MHF4U Trig Functions [4 marks]

$\cos \theta = -\tfrac{5}{13}$, $\theta$ in Quadrant III. Find the other five ratios.

Answer:  $\sin = -\tfrac{12}{13}$ · $\tan = \tfrac{12}{5}$ · $\csc = -\tfrac{13}{12}$ · $\sec = -\tfrac{13}{5}$ · $\cot = \tfrac{5}{12}$

(a) Find $\sin \theta$ via the Pythagorean identity M1·A1

$\sin^{2} \theta = 1 - \cos^{2} \theta = 1 - \tfrac{25}{169} = \tfrac{144}{169}$, so $\sin \theta = \pm \tfrac{12}{13}$. In Quadrant III, $\sin \theta < 0$, hence $\sin \theta = -\tfrac{12}{13}$.

(b) Tangent via the quotient identity A1

$\tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{-12/13}{-5/13} = \tfrac{12}{5}$ (positive in Q3, as expected from ASTC).

(c) Three reciprocals A1

$\csc \theta = \dfrac{1}{\sin \theta} = -\tfrac{13}{12}$,   $\sec \theta = \dfrac{1}{\cos \theta} = -\tfrac{13}{5}$,   $\cot \theta = \dfrac{1}{\tan \theta} = \tfrac{5}{12}$.
Quadrant-III sign audit: $\sin, \cos, \csc, \sec$ all negative; $\tan, \cot$ both positive. $\checkmark$
Reach the other five from one ratio in three moves: Pythagorean $\to$ quotient $\to$ reciprocals. The order matters. Step 1: get the missing one of $\{\sin, \cos\}$ from $\sin^{2} + \cos^{2} = 1$. Step 2: get $\tan$ from $\sin / \cos$. Step 3: flip each of the three ratios to get the three reciprocals $\csc, \sec, \cot$. ON markers split the four marks across exactly these stages: M1 for invoking Pythagorean, A1 for the missing primary, A1 for the quotient, A1 for the three reciprocals as a single line. Skipping the explicit quadrant justification costs the R-mark on the next-tier item.
Q5MEDIUM 🇨🇦 AB AB Provincial-style §7 Linear Equation · Math 30-1 GO 5 (5.2) [5 marks]

$\sqrt{3} \tan \theta + 1 = 0$. (a) Solutions on $[0, 2\pi)$. (b) General solution.

Answer:  (a) $\theta = \tfrac{5\pi}{6}, \tfrac{11\pi}{6}$  ·  (b) $\theta = \tfrac{5\pi}{6} + k\pi$, $k \in \mathbb{Z}$

(a) Isolate $\tan \theta$ and find solutions on $[0, 2\pi)$ M1·A1·A1

$\sqrt{3} \tan \theta + 1 = 0 \Longrightarrow \tan \theta = -\dfrac{1}{\sqrt{3}}$. Reference angle: $\tan^{-1}(1/\sqrt{3}) = \tfrac{\pi}{6}$. Tangent is negative in Quadrants II and IV.
  • Q2 solution: $\theta = \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$.
  • Q4 solution: $\theta = 2\pi - \tfrac{\pi}{6} = \tfrac{11\pi}{6}$.
Both lie in $[0, 2\pi)$. Solution set: $\bigl\{\tfrac{5\pi}{6}, \tfrac{11\pi}{6}\bigr\}$.

(b) General solution using period $\pi$ A1·A1

Tangent has period $\pi$, so consecutive solutions differ by $\pi$. Notice $\tfrac{11\pi}{6} = \tfrac{5\pi}{6} + \pi$, confirming they are one period apart. The general solution collapses to a single family: $$ \theta \;=\; \tfrac{5\pi}{6} + k\pi, \quad k \in \mathbb{Z}. $$
Tangent is the single-family equation; sine and cosine are two-family equations. Because $\tan$ has period $\pi$ and is one-to-one on each period, every "$\tan \theta = c$" equation has its general solution in a single family $\theta = \theta_{0} + k\pi$. Sine and cosine, with period $2\pi$, almost always need two families ($\theta_{0} + 2k\pi$ and $\pi - \theta_{0} + 2k\pi$ for sine; $\pm \theta_{0} + 2k\pi$ for cosine). Math 30-1 GO 5.2 explicitly tests whether you recognise tangent's single-family shortcut, students who write two families for tangent leave A1s on the table.
PART II  ·  EXTENDED RESPONSE · SOLUTIONSAP-feeder FRQ + ON / BC / honors · 35 marks

Section B · Worked Solutions

Q6MEDIUMHonors 🇺🇸 US AP-feeder FRQ §3 Sum & Difference · HSF-TF.C.9 (+) [8 marks]

(a) $\cos 15^{\circ}$ exactly. (b) $\sin(7\pi/12)$ exactly. (c) Verify $\sin(\alpha + \beta) + \sin(\alpha - \beta) = 2 \sin \alpha \cos \beta$.

Answer:  (a) $\cos 15^{\circ} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$  ·  (b) $\sin\!\left(\tfrac{7\pi}{12}\right) = \dfrac{\sqrt{6} + \sqrt{2}}{4}$  ·  (c) verified AG

(a) Difference formula on $\cos(45^{\circ} - 30^{\circ})$ M1·A1·A1

Using $\cos(A - B) = \cos A \cos B + \sin A \sin B$: $$ \cos 15^{\circ} \;=\; \cos(45^{\circ} - 30^{\circ}) \;=\; \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ}. $$ Substitute the special-angle values: $$ \cos 15^{\circ} \;=\; \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \;=\; \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \;=\; \frac{\sqrt{6} + \sqrt{2}}{4}. $$

(b) Sum formula on $\sin(\pi/3 + \pi/4)$ M1·A1·A1

Using $\sin(A + B) = \sin A \cos B + \cos A \sin B$: $$ \sin\!\left(\tfrac{7\pi}{12}\right) \;=\; \sin\!\left(\tfrac{\pi}{3} + \tfrac{\pi}{4}\right) \;=\; \sin \tfrac{\pi}{3} \cos \tfrac{\pi}{4} + \cos \tfrac{\pi}{3} \sin \tfrac{\pi}{4}. $$ $$ \;=\; \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} \;=\; \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \;=\; \frac{\sqrt{6} + \sqrt{2}}{4}. $$ (Notice this equals $\cos 15^{\circ}$, because $\sin(105^{\circ}) = \cos(15^{\circ})$ via the co-function identity. $\checkmark$)

(c) Verify $\sin(\alpha + \beta) + \sin(\alpha - \beta) = 2 \sin \alpha \cos \beta$ M1·A1

Expand both terms on the LHS: $$ \sin(\alpha + \beta) \;=\; \sin \alpha \cos \beta + \cos \alpha \sin \beta, $$ $$ \sin(\alpha - \beta) \;=\; \sin \alpha \cos \beta - \cos \alpha \sin \beta. $$ Add. The $\cos \alpha \sin \beta$ terms cancel; the $\sin \alpha \cos \beta$ terms double: $$ \sin(\alpha + \beta) + \sin(\alpha - \beta) \;=\; 2 \sin \alpha \cos \beta. \quad \mathrm{AG} $$
The sum-plus-difference cancellation is the seed of the product-to-sum identities. What part (c) just verified is one of four "sum-to-product / product-to-sum" identities that show up in AP Calc BC and IB AA HL Topic 3.7. The pattern: sum of sines with arguments $\alpha \pm \beta$ collapses to $2 \sin \alpha \cos \beta$ because the mixed terms have opposite signs and cancel. Same trick gives $\cos(\alpha + \beta) + \cos(\alpha - \beta) = 2 \cos \alpha \cos \beta$ and the two subtractive analogues. AP Calc uses these to integrate products like $\int \sin(3x) \cos(2x) \, dx$, recognising the sum-difference pattern saves five lines of substitution.
Q7HARDHonors 🇨🇦 ON ON Provincial-style §4 Double-Angle · MHF4U Trig Functions [9 marks]

$\sin \theta = \tfrac{4}{5}$, $\theta$ in Quadrant II. (a) $\cos \theta$. (b) $\sin 2\theta$. (c) $\cos 2\theta$. (d) Quadrant of $2\theta$ + sign audit.

Answer:  (a) $\cos \theta = -\tfrac{3}{5}$  ·  (b) $\sin 2\theta = -\tfrac{24}{25}$  ·  (c) $\cos 2\theta = -\tfrac{7}{25}$  ·  (d) $2\theta$ in Quadrant III, signs consistent

(a) $\cos \theta$ via Pythagorean identity M1·A1

$\cos^{2} \theta = 1 - \tfrac{16}{25} = \tfrac{9}{25}$, so $\cos \theta = \pm \tfrac{3}{5}$. In Q2, $\cos \theta < 0$, hence $\cos \theta = -\tfrac{3}{5}$.

(b) $\sin 2\theta = 2 \sin \theta \cos \theta$ M1·A1

$$ \sin 2\theta \;=\; 2 \cdot \tfrac{4}{5} \cdot \bigl(-\tfrac{3}{5}\bigr) \;=\; -\tfrac{24}{25}. $$

(c) Choose a form of $\cos 2\theta$ M1·A1·A1

The three equivalent forms are $\cos 2\theta = \cos^{2} \theta - \sin^{2} \theta = 1 - 2 \sin^{2} \theta = 2 \cos^{2} \theta - 1$. Since the prompt gave $\sin \theta$ directly, the cleanest choice is the all-sine form $1 - 2 \sin^{2} \theta$ (no need to invoke $\cos \theta$ from (a)): $$ \cos 2\theta \;=\; 1 - 2 \sin^{2} \theta \;=\; 1 - 2 \cdot \tfrac{16}{25} \;=\; 1 - \tfrac{32}{25} \;=\; -\tfrac{7}{25}. $$

(d) Quadrant of $2\theta$ and sign audit A1·A1

$\theta \in (\tfrac{\pi}{2}, \pi)$, so $2\theta \in (\pi, 2\pi)$. With $\sin 2\theta = -\tfrac{24}{25} < 0$ and $\cos 2\theta = -\tfrac{7}{25} < 0$, both coordinates are negative, the unit-circle position is in Quadrant III, where $\pi < 2\theta < \tfrac{3\pi}{2}$. The signs of parts (b) and (c) are exactly what Q3 demands.
Pick the form of $\cos 2\theta$ that uses the variable you already have. $\cos 2\theta$ has three faces because students sometimes have $\sin$ alone, sometimes $\cos$ alone, sometimes both. If you only have $\sin$, use $1 - 2 \sin^{2}$, no quadrant decision needed for $\cos \theta$. If you only have $\cos$, use $2 \cos^{2} - 1$. The $\cos^{2} - \sin^{2}$ form is the only one that needs both and is therefore the most error-prone. ON markers split (c) into M1 (correct form chosen) + A1 (substitution) + A1 (final), the M1 rewards the strategic choice, not the answer.
Q8HARDHonors 🇨🇦 AB AB Provincial-style §6 Verifying Identities · Math 30-1 GO 6 (6.4) [9 marks]

(a) $\tan + \cot = \sec \csc$. (b) $(1-\cos)/\sin = \sin/(1+\cos)$. (c) $(\sin^{2}-\cos^{2})/(\sin-\cos) = \sin+\cos$.

Answer:  all three identities verified · LHS reductions reach RHS in each case (AG)

(a) Reduce to $\sin / \cos$ form, common denominator M1·A1·A1

LHS: $$ \tan \theta + \cot \theta \;=\; \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \;=\; \frac{\sin^{2} \theta + \cos^{2} \theta}{\sin \theta \cos \theta} \;=\; \frac{1}{\sin \theta \cos \theta}. $$ Using the Pythagorean identity $\sin^{2} + \cos^{2} = 1$ in the numerator. Then read the result via reciprocals: $\dfrac{1}{\sin \theta \cos \theta} = \dfrac{1}{\sin \theta} \cdot \dfrac{1}{\cos \theta} = \csc \theta \sec \theta = \sec \theta \csc \theta = $ RHS. $\;\mathrm{AG}$

(b) Conjugate trick: multiply by $(1 + \cos \theta)/(1 + \cos \theta)$ M1·A1·A1

LHS: $$ \frac{1 - \cos \theta}{\sin \theta} \;=\; \frac{(1 - \cos \theta)(1 + \cos \theta)}{\sin \theta \, (1 + \cos \theta)} \;=\; \frac{1 - \cos^{2} \theta}{\sin \theta \, (1 + \cos \theta)}. $$ Apply $1 - \cos^{2} \theta = \sin^{2} \theta$: $$ \;=\; \frac{\sin^{2} \theta}{\sin \theta \, (1 + \cos \theta)} \;=\; \frac{\sin \theta}{1 + \cos \theta} \;=\; \text{RHS}. \quad \mathrm{AG} $$

(c) Factor the difference of squares in the numerator M1·A1·A1

The numerator factors as $\sin^{2} \theta - \cos^{2} \theta = (\sin \theta - \cos \theta)(\sin \theta + \cos \theta)$. The shared factor cancels (valid because $\sin \theta \ne \cos \theta$): $$ \frac{(\sin \theta - \cos \theta)(\sin \theta + \cos \theta)}{\sin \theta - \cos \theta} \;=\; \sin \theta + \cos \theta \;=\; \text{RHS}. \quad \mathrm{AG} $$
Three strategies in one question: reduce-to-sin/cos, conjugate, factor. Math 30-1 GO 6 indicator 6.4 explicitly names these as the verification toolkit. The strategy follows the structure of the LHS, not the RHS: (a) had two trig ratios so we unified them via $\sin / \cos$; (b) had a $1 - \cos$ that screamed "conjugate pair" since $(1-\cos)(1+\cos) = \sin^{2}$ is a built-in Pythagorean factoring; (c) had a difference-of-squares pattern in the numerator. Train the eye to read the LHS structure first: if you see "$1 \pm$ something" reach for the conjugate; if you see a difference of squared trig ratios reach for the factoring; if everything is jumbled, fall back on the universal reducer "convert to sin and cos." AB markers also reward students who explicitly name the strategy in one sentence before the algebra.
Q9HARDHonors 🇺🇸 US 🇨🇦 BC AP-feeder FRQ §5 Half-Angle · HSF-TF.C.9 (+) / BC PC 12 enrichment [9 marks]

(a) $\cos 22.5^{\circ}$ via half-angle. (b) $\sin(\theta/2)$ given $\cos \theta = -\tfrac{7}{25}$, $\theta \in $ Q2. (c) Derive $\cos$ half-angle from $\cos 2x = 2 \cos^{2} x - 1$.

Answer:  (a) $\cos 22.5^{\circ} = \dfrac{\sqrt{2 + \sqrt{2}}}{2}$  ·  (b) $\sin(\theta/2) = \tfrac{4}{5}$  ·  (c) half-angle formula derived AG

(a) Half-angle with $\theta = 45^{\circ}$ M1·A1·A1

$22.5^{\circ} = 45^{\circ} / 2$. Since $22.5^{\circ}$ is in Quadrant I, $\cos 22.5^{\circ} > 0$, take the positive root: $$ \cos 22.5^{\circ} \;=\; \sqrt{\dfrac{1 + \cos 45^{\circ}}{2}} \;=\; \sqrt{\dfrac{1 + \tfrac{\sqrt{2}}{2}}{2}} \;=\; \sqrt{\dfrac{2 + \sqrt{2}}{4}} \;=\; \dfrac{\sqrt{2 + \sqrt{2}}}{2}. $$

(b) Quadrant of $\theta/2$, then half-angle for sin M1·R1·A1·A1

$\theta \in (\tfrac{\pi}{2}, \pi)$ (Q2), so $\theta/2 \in (\tfrac{\pi}{4}, \tfrac{\pi}{2})$ — Q1. Hence $\sin(\theta/2) > 0$, take the positive root. $$ \sin\!\left(\tfrac{\theta}{2}\right) \;=\; \sqrt{\dfrac{1 - \cos \theta}{2}} \;=\; \sqrt{\dfrac{1 - (-\tfrac{7}{25})}{2}} \;=\; \sqrt{\dfrac{\tfrac{32}{25}}{2}} \;=\; \sqrt{\dfrac{16}{25}} \;=\; \tfrac{4}{5}. $$

(c) Derive the $\cos$ half-angle formula M1·A1

Start from $\cos 2x = 2 \cos^{2} x - 1$ and substitute $x = \theta/2$: $$ \cos \theta \;=\; 2 \cos^{2}\!\bigl(\tfrac{\theta}{2}\bigr) - 1. $$ Solve for $\cos^{2}(\theta/2)$: $$ \cos^{2}\!\bigl(\tfrac{\theta}{2}\bigr) \;=\; \dfrac{1 + \cos \theta}{2}. $$ Take square roots, attaching the quadrant-determined sign: $$ \cos\!\bigl(\tfrac{\theta}{2}\bigr) \;=\; \pm \sqrt{\dfrac{1 + \cos \theta}{2}}. \quad \mathrm{AG} $$
The $\pm$ in the half-angle formula is not optional, it is a quadrant question disguised as a sign. Every half-angle problem decomposes into two steps: (i) compute the radicand from the given $\cos \theta$, (ii) figure out which quadrant $\theta/2$ lives in and pick the sign from there. The half-step from $\theta$ to $\theta/2$ usually changes the quadrant (if $\theta$ is in Q2, $\theta/2$ is in Q1; if $\theta$ is in Q3, $\theta/2$ is in Q2; the halving compresses the angle into a different region of the unit circle). AP graders explicitly mark down students who forget the quadrant step and write the formula with both signs in the final answer, the unique-value question is what the prompt asked for.
PART III  ·  MODELING / APPLIED · SOLUTIONSUniversal · 28 marks

Section C · Worked Solutions

Q10MEDIUM 🇨🇦 BC BC Provincial-style §7 Quadratic Trig Equation · PC 12 Trig [9 marks]

(a) $2 \sin^{2} x - \sin x - 1 = 0$ on $[0, 2\pi)$. (b) $2 \cos^{2} x = 1$ on $[0, 2\pi)$. (c) General solution to (b).

Answer:  (a) $x = \tfrac{\pi}{2}, \tfrac{7\pi}{6}, \tfrac{11\pi}{6}$  ·  (b) $x = \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4}$  ·  (c) $x = \tfrac{\pi}{4} + \tfrac{k\pi}{2}$, $k \in \mathbb{Z}$

(a) Substitute $u = \sin x$, factor the quadratic M1·A1·A1·A1

Let $u = \sin x$. Then $2 u^{2} - u - 1 = 0$. Factor (AC-method: $AC = -2$, integers $(-2, 1)$): $$ 2 u^{2} - 2 u + u - 1 \;=\; 2u(u - 1) + (u - 1) \;=\; (2u + 1)(u - 1) \;=\; 0. $$ So $u = 1$ or $u = -\tfrac{1}{2}$, i.e. $\sin x = 1$ or $\sin x = -\tfrac{1}{2}$.
  • $\sin x = 1 \Rightarrow x = \tfrac{\pi}{2}$ (one solution, the peak of sine on $[0, 2\pi)$).
  • $\sin x = -\tfrac{1}{2} \Rightarrow$ reference angle $\tfrac{\pi}{6}$; sine negative in Q3 and Q4: $x = \pi + \tfrac{\pi}{6} = \tfrac{7\pi}{6}$ and $x = 2\pi - \tfrac{\pi}{6} = \tfrac{11\pi}{6}$.
Three solutions total: $x \in \bigl\{\tfrac{\pi}{2}, \tfrac{7\pi}{6}, \tfrac{11\pi}{6}\bigr\}$.

(b) Take square roots, then sweep the unit circle M1·A1·A1

$\cos^{2} x = \tfrac{1}{2}$, so $\cos x = \pm \tfrac{\sqrt{2}}{2}$. Reference angle $\tfrac{\pi}{4}$. Both signs of cosine appear — positive in Q1 and Q4, negative in Q2 and Q3 — so the four solutions are: $$ x \;\in\; \bigl\{ \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4} \bigr\}. $$ (Equivalently: $\cos^{2} x = \tfrac{1}{2}$ is the same as $\cos 2x = 0$ via the double-angle identity, which has period $\pi/2$ on $x$, immediately predicting four solutions on a $2\pi$ interval.)

(c) Single-family general solution A1·A1

The four solutions in (b) form an arithmetic progression with common difference $\tfrac{\pi}{2}$. Hence $$ x \;=\; \tfrac{\pi}{4} + \tfrac{k\pi}{2}, \quad k \in \mathbb{Z}. $$
Always substitute before factoring trig quadratics; the algebra has nothing to do with trigonometry. Part (a) was a quadratic in $u = \sin x$ wearing a trig costume. The substitution makes the structure obvious and prevents the most common BC error, treating $\sin^{2} x$ as $\sin (x^{2})$ or "$2$ sinning $x$." The back-substitution step is where trig reappears: every value of $u$ in $(-1, 1)$ corresponds to two angles per period; every value of $u$ at $\pm 1$ corresponds to one angle per period. Counting solutions before sweeping the unit circle is a useful sanity check, here we expected one solution from $u = 1$ and two from $u = -\tfrac{1}{2}$, totaling three. $\checkmark$
Q11HARDHonors 🇨🇦 AB AB Provincial-style §7 Identity-Required Equation · Math 30-1 GO 5 (5.6) [9 marks]

(a) $\cos 2x + \sin x = 0$ on $[0, 2\pi)$. (b) $\sin x \tan x = \sin x$ on $[0, 2\pi)$, with domain restriction. (c) Why dividing by $\sin x$ in (b) loses roots.

Answer:  (a) $x = \tfrac{\pi}{2}, \tfrac{7\pi}{6}, \tfrac{11\pi}{6}$  ·  (b) $x = 0, \tfrac{\pi}{4}, \pi, \tfrac{5\pi}{4}$ (with $x \ne \tfrac{\pi}{2}, \tfrac{3\pi}{2}$)  ·  (c) dividing by $\sin x$ deletes the $\sin x = 0$ root family

(a) Convert $\cos 2x$ to all-sin form, then factor M1·A1·A1·A1

Use the double-angle identity $\cos 2x = 1 - 2 \sin^{2} x$: $$ (1 - 2 \sin^{2} x) + \sin x \;=\; 0 \;\Longleftrightarrow\; -2 \sin^{2} x + \sin x + 1 \;=\; 0 \;\Longleftrightarrow\; 2 \sin^{2} x - \sin x - 1 \;=\; 0. $$ This is the same factoring as Q10(a): $(2 \sin x + 1)(\sin x - 1) = 0$, giving $\sin x = 1$ or $\sin x = -\tfrac{1}{2}$. Solutions on $[0, 2\pi)$: $x \in \bigl\{\tfrac{\pi}{2}, \tfrac{7\pi}{6}, \tfrac{11\pi}{6}\bigr\}$.

(b) Factor, do not divide M1·A1·A1

Move everything to one side and factor: $$ \sin x \tan x - \sin x \;=\; 0 \;\Longleftrightarrow\; \sin x \, (\tan x - 1) \;=\; 0. $$ Two cases:
  • $\sin x = 0 \Rightarrow x = 0$ or $x = \pi$ on $[0, 2\pi)$.
  • $\tan x = 1 \Rightarrow$ reference angle $\tfrac{\pi}{4}$, tangent positive in Q1 and Q3: $x = \tfrac{\pi}{4}$ or $x = \pi + \tfrac{\pi}{4} = \tfrac{5\pi}{4}$.
Domain restriction: $\tan x$ is undefined at $x = \tfrac{\pi}{2}$ and $x = \tfrac{3\pi}{2}$, so those values are excluded from the original equation's domain. None of the four candidate solutions lies at the forbidden values, so all four are valid: $x \in \bigl\{0, \tfrac{\pi}{4}, \pi, \tfrac{5\pi}{4}\bigr\}$.

(c) The "divide-by-sin" mistake R1·A1

Dividing both sides by $\sin x$ requires $\sin x \ne 0$, which excludes the two roots $x = 0$ and $x = \pi$. The leftover equation $\tan x = 1$ then gives only $x = \tfrac{\pi}{4}, \tfrac{5\pi}{4}$ — a four-root problem dropped to two roots. The algebraic reason: factoring an expression to $0$ uses the zero-product property, $AB = 0 \Leftrightarrow A = 0 \text{ or } B = 0$, both factors must be tested. Dividing is a non-invertible operation when the divisor can be zero, so it deletes the corresponding root family without warning.
"Never divide by something that could be zero" is the single most useful trig-equation rule. Math 30-1 GO 5.6 explicitly tests this awareness, and it generalises far beyond trigonometry. The lossless analogue of "divide by $X$" is "factor out $X$ and use the zero-product property," which preserves the $X = 0$ root family explicitly. The diagnostic for whether you have lost roots: count the degree of the original equation in $\sin x$ (here, the LHS has $\sin x \tan x = \sin^{2}x / \cos x$, a degree-2 expression in $\sin$). A degree-2 equation on $[0, 2\pi)$ should produce at most four roots; if you find only two, ask which family you divided away.
Q12HARD 🇺🇸 US 🇨🇦 BC AP-feeder FRQ §7 Modeling with Trig Equations · HSF-TF.B.7 (+) [10 marks]

Ferris wheel: $h(t) = 12 - 10 \cos(\pi t / 20)$. (a) $h(0) = 2$, $h_{\max}$ at $t = 20$ s. (b) Times $h = 17$ on $[0, 40)$. (c) General "$h = 17$" times. (d) Why every intermediate height is hit twice per revolution.

Answer:  (a) $h(0) = 2$, $h_{\max}(20) = 22$ m  ·  (b) $t = \tfrac{40}{3}$ s and $t = \tfrac{80}{3}$ s  ·  (c) $t = \tfrac{40}{3} + 40 k$ or $t = \tfrac{80}{3} + 40 k$, $k \in \mathbb{Z}_{\ge 0}$  ·  (d) $\cos$ is monotone on each half-period

(a) Boundary checks A1·A1

$h(0) = 12 - 10 \cos 0 = 12 - 10 = 2$ m. $\checkmark$ Maximum height: $\cos$ attains $-1$ at argument $\pi$, so $\pi t / 20 = \pi \Rightarrow t = 20$ s. Then $h(20) = 12 - 10(-1) = 22$ m. The peak is $10$ m above the centre (radius), consistent with the geometry.

(b) Solve $h(t) = 17$ on $[0, 40)$ M1·A1·A1·A1

$$ 12 - 10 \cos\!\left(\tfrac{\pi t}{20}\right) = 17 \;\Longleftrightarrow\; \cos\!\left(\tfrac{\pi t}{20}\right) = -\tfrac{1}{2}. $$ Reference angle for $\cos = -\tfrac{1}{2}$: $\arccos(\tfrac{1}{2}) = \tfrac{\pi}{3}$. Cosine is negative in Q2 and Q3, so on one period of the argument $\tfrac{\pi t}{20} \in [0, 2\pi)$: $$ \tfrac{\pi t}{20} = \pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3} \quad \text{or} \quad \tfrac{\pi t}{20} = \pi + \tfrac{\pi}{3} = \tfrac{4\pi}{3}. $$ Solve for $t$: $$ t = \tfrac{20}{\pi} \cdot \tfrac{2\pi}{3} = \tfrac{40}{3} \;\text{s} \quad \text{or} \quad t = \tfrac{20}{\pi} \cdot \tfrac{4\pi}{3} = \tfrac{80}{3} \;\text{s}. $$ Both lie in $[0, 40)$, since $\tfrac{40}{3} \approx 13.3$ and $\tfrac{80}{3} \approx 26.7$. Two valid times.

(c) General solution across rotations A1·A1

The period of $h$ is $40$ s, so successive crossings of any horizontal line $h = c$ ($2 < c < 22$) repeat every $40$ s: $$ t \;=\; \tfrac{40}{3} + 40 k \quad \text{or} \quad t \;=\; \tfrac{80}{3} + 40 k, \qquad k \in \mathbb{Z}_{\ge 0}. $$

(d) Twice-per-revolution argument A1·A1

On each half-period, $h(t)$ is monotone: ascending on $[0, 20]$ (rider rising), descending on $[20, 40]$ (rider falling). Each intermediate height $c$ with $2 < c < 22$ therefore corresponds to exactly one time on the ascent and exactly one on the descent — two times per revolution. The boundary heights $c = 2$ (boarding) and $c = 22$ (peak) are hit exactly once per revolution, at $t = 0$ (or $t = 40$, same point) and $t = 20$ respectively.
Modeling-context trig equations are "$\cos = $ number" wearing a different costume. The recipe is always the same: (i) isolate the trig ratio, (ii) compute the reference angle, (iii) sweep the unit circle for sign-consistent quadrants, (iv) un-do the argument's linear transformation to recover $t$. HSF-TF.B.7 (+) is exactly this workflow with the "interpret in context" wrapper. AP graders explicitly reward students who state the period of the modeled function before writing the general solution — the period is what bridges the principal solutions to the family. Common loss-of-mark: forgetting that the radian-to-second conversion is $t = (T / 2\pi) \cdot \theta$ where $T$ is the period of $h$; here $T = 40$ s gives $t = (20/\pi) \theta$, which is exactly what part (b) used.