PART I · SHORT RESPONSE第一部分 · 短答题SAT-style MCQ + ON/BC/AB short answer · 18 marksSAT 风格选择题 + 安/卑/艾省考短答 · 共 18 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. For trig equations on a restricted interval, state every solution explicitly (do not write only the principal value). Calculator not required on Q1–Q4; permitted on Q5 only for arithmetic checks.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写下足以让阅卷人核对的过程。对限定区间上的三角方程，必须明确列出每一个解（不能只写主值）。Q1–Q4 不可使用计算器；Q5 仅可用于算术核对。
If $\sin \theta = \tfrac{3}{5}$ and $\theta$ lies in Quadrant II, what is $\cos \theta$?若 $\sin \theta = \tfrac{3}{5}$ 且 $\theta$ 位于第二象限，则 $\cos \theta$ 是多少？
Which expression is equivalent to $\dfrac{1}{\cos \theta} \cdot \sin \theta$ for every $\theta$ in the natural domain?在自然定义域内，下列哪个表达式与 $\dfrac{1}{\cos \theta} \cdot \sin \theta$ 对每一个 $\theta$ 都相等？
How many solutions does the equation $2 \sin x - 1 = 0$ have on the interval $0 \le x < 2\pi$?方程 $2 \sin x - 1 = 0$ 在区间 $0 \le x < 2\pi$ 上有多少个解？
(A)Exactly one solution恰有一个解
(B)Exactly two solutions恰有两个解
(C)Exactly three solutions恰有三个解
(D)No solutions无解
Q4MEDIUM中🇨🇦 ON安ON Provincial-style安大略省考风格§1–§2 All Six Ratios六个三角比 · MHF4U Trig Functions三角函数[4 marks][4 分]
Given $\cos \theta = -\tfrac{5}{13}$ with $\theta$ in Quadrant III, determine the exact values of the other five trigonometric ratios: $\sin \theta$, $\tan \theta$, $\csc \theta$, $\sec \theta$, $\cot \theta$. State the quadrant-based sign for each.已知 $\cos \theta = -\tfrac{5}{13}$，且 $\theta$ 位于第三象限。求其余五个三角比 $\sin \theta$、$\tan \theta$、$\csc \theta$、$\sec \theta$、$\cot \theta$ 的精确值，并根据象限判定每一个的正负号。
Q5MEDIUM中🇨🇦 AB艾AB Provincial-style艾伯塔省考风格§7 Linear Equation一次方程 · Math 30-1 GO 5 (5.2)[5 marks][5 分]
(a)State all solutions on the restricted interval $0 \le \theta < 2\pi$.列出限定区间 $0 \le \theta < 2\pi$ 上的全部解。[3]
(b)State the general solution in radians, using the parameter $k \in \mathbb{Z}$. Use the fact that $\tan$ has period $\pi$.用弧度写出通解，参数 $k \in \mathbb{Z}$。利用 $\tan$ 的周期为 $\pi$ 这一事实。[2]
PART II · EXTENDED RESPONSE第二部分 · 解答题AP-feeder FRQ + ON / BC / honors · 35 marksAP 衔接简答 + 安 / 卑 / 荣誉级 · 共 35 分
Section B · Extended ResponseB 部分 · 解答题
Show every algebraic step. For identity verifications, work down a single column (LHS only or RHS only); never write LHS = RHS on the first line. Cite each identity by name (Pythagorean, quotient, reciprocal, sum, double-angle, half-angle) as you invoke it. Calculator not permitted on Q6–Q9.写出每一步代数运算。证明恒等式时只能沿单侧（只动 LHS 或只动 RHS）化简，第一行绝不能写 LHS = RHS。每次使用恒等式都要点名（勾股、商、倒数、和差、二倍角、半角）。Q6–Q9 不可使用计算器。
Use the sum and difference formulas to evaluate exact values of trigonometric expressions, expressing each answer as a single simplified radical fraction (no decimals).利用和差公式求三角表达式的精确值，每个答案写成最简的单一根式分数（不可用小数）。
(a)Compute $\cos 15^{\circ}$ exactly by writing $15^{\circ} = 45^{\circ} - 30^{\circ}$. State the formula used.将 $15^{\circ} = 45^{\circ} - 30^{\circ}$，精确求出 $\cos 15^{\circ}$，并写明所用公式。[3]
Suppose $\sin \theta = \tfrac{4}{5}$ with $\theta$ in Quadrant II.设 $\sin \theta = \tfrac{4}{5}$，且 $\theta$ 位于第二象限。
(a)Determine the exact value of $\cos \theta$ using the Pythagorean identity. Justify the sign from the quadrant.利用勾股恒等式求 $\cos \theta$ 的精确值，并根据象限说明符号。[2]
(b)Use the double-angle formula $\sin 2\theta = 2 \sin \theta \cos \theta$ to compute $\sin 2\theta$ exactly.用二倍角公式 $\sin 2\theta = 2 \sin \theta \cos \theta$ 精确求出 $\sin 2\theta$。[2]
(c)Use a suitable form of $\cos 2\theta$ to compute $\cos 2\theta$ exactly. State which of the three equivalent forms you chose and why.选用 $\cos 2\theta$ 的合适形式精确求出 $\cos 2\theta$，并说明你在三个等价形式中选了哪一个、为何这样选。[3]
(d)Determine the quadrant in which the angle $2\theta$ terminates, and confirm your answers in (b) and (c) carry the right signs for that quadrant.判断角 $2\theta$ 终边所在象限，并核对 (b)(c) 的结果是否与该象限的符号一致。[2]
Q8HARD难Honors荣誉级🇨🇦 AB艾AB Provincial-style艾伯塔省考风格§6 Verifying Identities证明恒等式 · Math 30-1 GO 6 (6.4)[9 marks][9 分]
Verify each identity by working from a single side. Document the strategy chosen (reduce-to-sin-cos, common denominator, factoring, or conjugate). Conclude with AG (answer given) once both sides match.仅从单侧化简来证明每个恒等式。注明你选用的策略（化为 sin/cos、通分、因式分解或共轭）。两侧化为相同形式后，以 AG（answer given，原题已给）收束。
(a)Verify $\;\tan \theta + \cot \theta \;=\; \sec \theta \csc \theta\;$ by reducing the LHS to $\sin / \cos$ form and combining with a common denominator.把 LHS 化为 $\sin / \cos$ 形式并通分，证明 $\;\tan \theta + \cot \theta \;=\; \sec \theta \csc \theta$。[3]
(b)Verify $\;\dfrac{1 - \cos \theta}{\sin \theta} \;=\; \dfrac{\sin \theta}{1 + \cos \theta}\;$ by multiplying the LHS numerator and denominator by the conjugate $1 + \cos \theta$, then applying the Pythagorean identity.将 LHS 的分子分母同乘以共轭 $1 + \cos \theta$，再用勾股恒等式，证明 $\;\dfrac{1 - \cos \theta}{\sin \theta} \;=\; \dfrac{\sin \theta}{1 + \cos \theta}$。[3]
Q9HARD难Honors荣誉级🇺🇸 US美🇨🇦 BC卑AP-feeder FRQAP 衔接简答题§5 Half-Angle半角公式 · HSF-TF.C.9 (+) / BC PC 12 enrichment卑诗 PC 12 拓展[9 marks][9 分]
The half-angle formulas state $\cos\!\left(\dfrac{\theta}{2}\right) = \pm \sqrt{\dfrac{1 + \cos \theta}{2}}$ and $\sin\!\left(\dfrac{\theta}{2}\right) = \pm \sqrt{\dfrac{1 - \cos \theta}{2}}$, with the sign determined by the quadrant of $\theta/2$.半角公式为 $\cos\!\left(\dfrac{\theta}{2}\right) = \pm \sqrt{\dfrac{1 + \cos \theta}{2}}$ 与 $\sin\!\left(\dfrac{\theta}{2}\right) = \pm \sqrt{\dfrac{1 - \cos \theta}{2}}$，符号由 $\theta/2$ 所在象限确定。
(a)Use the half-angle formula to compute $\cos 22.5^{\circ}$ exactly. Justify the choice of sign from the quadrant of $22.5^{\circ}$.用半角公式精确求 $\cos 22.5^{\circ}$，并根据 $22.5^{\circ}$ 所在象限说明符号选择。[3]
(b)Suppose $\cos \theta = -\tfrac{7}{25}$ with $\theta$ in Quadrant II. Determine the quadrant of $\theta/2$, then compute the exact value of $\sin(\theta/2)$.设 $\cos \theta = -\tfrac{7}{25}$ 且 $\theta$ 位于第二象限。先判定 $\theta/2$ 所在象限，再精确求出 $\sin(\theta/2)$。[4]
(c)Derive the half-angle formula for $\cos$ from the double-angle identity $\cos 2x = 2 \cos^{2} x - 1$ by substituting $x = \theta/2$. State each algebraic step.在二倍角恒等式 $\cos 2x = 2 \cos^{2} x - 1$ 中代入 $x = \theta/2$，推导 $\cos$ 的半角公式。写出每一步代数运算。[2]
PART III · MODELING / APPLIED第三部分 · 建模 / 应用题Universal · 28 marks通用 · 共 28 分
Section C · Equations and ApplicationsC 部分 · 方程与应用
Solve each trigonometric equation on the stated interval. Where general solutions are requested, use $+ 2k\pi$ (or $+ k\pi$ for tangent) with $k \in \mathbb{Z}$. Always verify candidate solutions against domain restrictions on $\sec$, $\csc$, $\cot$. Calculator permitted only for arithmetic checks; exact-value answers are required.在题目给定区间内求解每一个三角方程。若题目要求通解，正余弦用 $+ 2k\pi$、正切用 $+ k\pi$，$k \in \mathbb{Z}$。务必将候选解与 $\sec$、$\csc$、$\cot$ 的定义域限制对照，剔除增根。计算器只可用于算术核对；最终答案必须为精确值。
Solve each second-degree trigonometric equation on the restricted interval $0 \le x < 2\pi$. Show the substitution (e.g. let $u = \sin x$), factor or apply the quadratic formula, then back-substitute and list every solution.在限定区间 $0 \le x < 2\pi$ 上求解下列二次三角方程。请写出换元（如令 $u = \sin x$），因式分解或用求根公式，再回代写出全部解。
(a) $2 \sin^{2} x - \sin x - 1 = 0$. [4]
(b) $2 \cos^{2} x = 1$. State the four solutions.写出四个解。[3]
(c)State the general solution to (b) in radians, using parameters $k \in \mathbb{Z}$.用弧度写出 (b) 的通解，参数 $k \in \mathbb{Z}$。[2]
Q11HARD难Honors荣誉级🇨🇦 AB艾AB Provincial-style艾伯塔省考风格§7 Identity-Required Equation须用恒等式的方程 · Math 30-1 GO 5 (5.6)[9 marks][9 分]
Some trigonometric equations cannot be solved until a Pythagorean or double-angle identity converts the expression to a single trig ratio. Solve each equation on $0 \le x < 2\pi$, citing the identity used at the conversion step.有些三角方程必须先用勾股或二倍角恒等式把表达式化为单一三角比才能求解。在 $0 \le x < 2\pi$ 上求解下列方程，并在转化步骤处注明所用恒等式。
(a) $\cos 2x + \sin x = 0$. (Hint: replace $\cos 2x$ with the form $1 - 2 \sin^{2} x$, then factor.)（提示：将 $\cos 2x$ 换成 $1 - 2 \sin^{2} x$ 的形式，再因式分解。）[4]
(b) $\sin x \tan x = \sin x$. (Hint: do not divide both sides by $\sin x$, factor instead, or you will lose roots.) State the domain restriction that excludes any extraneous solution.（提示：不要两边同除 $\sin x$，要因式分解，否则会丢根。）写出排除增根的定义域限制。[3]
(c)Briefly explain why dividing both sides of (b) by $\sin x$ is a mistake: name the lost roots and the algebraic reason.简要说明在 (b) 中两边同除 $\sin x$ 为何错误：指出丢掉的根以及代数上的原因。[2]
A Ferris wheel of radius $10$ m has its centre at a height of $12$ m above the ground. It rotates at a constant rate, completing one full revolution every $40$ seconds. A rider boards at the lowest point at time $t = 0$ s. The rider's height above the ground, in metres, after $t$ seconds is modeled by一架半径 $10$ m 的摩天轮，圆心距地面 $12$ m，以恒定速率旋转，每 $40$ 秒一圈。乘客在 $t = 0$ s 从最低点上车。$t$ 秒后乘客距地面的高度（米）由下式给出：
$$ h(t) \;=\; 12 \;-\; 10 \cos\!\left(\dfrac{\pi t}{20}\right). $$
(a)Verify that $h(0) = 2$ m (the boarding platform height) and that the maximum height occurs at $t = 20$ s. State the maximum height.验证 $h(0) = 2$ m（上车平台高度）且最高点出现在 $t = 20$ s。写出最高点高度。[2]
(b)Determine all times $t$ in the first revolution ($0 \le t < 40$) at which the rider is exactly $17$ m above the ground. Express each answer in seconds, exactly.求第一圈内（$0 \le t < 40$）乘客恰好距地面 $17$ m 的所有时刻 $t$。每个答案以秒为单位，精确给出。[4]
(c)State the general solution for "rider at height $17$ m" across all rotations, using the period of $h$. Use $k \in \mathbb{Z}_{\ge 0}$.利用 $h$ 的周期，写出"乘客高度为 $17$ m"在所有圈数下的通解，参数 $k \in \mathbb{Z}_{\ge 0}$。[2]
(d)Explain in one sentence why the rider passes through every height between $2$ m and $22$ m exactly twice per revolution, except for the boarding height and the peak.用一句话解释：除上车高度与最高点外，乘客每圈为何恰好两次经过 $2$ m 到 $22$ m 之间的任一高度。[2]
🇺🇸 US Common Core美国共同核心HSF-TF.C.8 · HSF-TF.C.9 (+) · HSF-TF.B.7 (+) · HSF-TF.A.2
🇨🇦 Alberta艾伯塔Math 30-1 Trig GO 5 (equations, indicators 5.1–5.6) + GO 6 (identities, indicators 6.1–6.7)Math 30-1 三角 GO 5（方程，指标 5.1–5.6）+ GO 6（恒等式，指标 6.1–6.7）
Full 4-column Syllabus Map lives in ../Study Guides/Unit_9_Trigonometric_Identities_and_Equations.html.完整的四列大纲对照见 ../Study Guides/Unit_9_Trigonometric_Identities_and_Equations.html。