📚 Review Set 9C: Trigonometric Equations & Identities | 复习集 9C：三角方程与恒等式
Review set 9C in IB Mathematics: Analysis and Approaches SL consolidates the essential techniques for solving trigonometric equations and manipulating trigonometric identities. This article revisits the core concepts, provides strategic approaches, and highlights typical pitfalls to help you master this topic confidently.
IB数学分析与方法SL课程的复习集9C巩固了解三角方程和三角恒等式变形的基本技巧。本文重温核心概念，提供解题策略，并指出常见陷阱，帮助你自信掌握这一主题。
1. Overview of Review Set 9C | 复习集9C概述
ur command of fundamental identities like sin²θ + cos²θ = 1 and the double-angle formulas.
本复习集涵盖从 sin x = 1/2 这样直接的问题，到需要因式分解、二次方程技巧或多倍角处理的复杂方程。它还严格测试你对 sin²θ + cos²θ = 1 和二倍角公式等基本恒等式的掌握程度。
The questions are designed to mirror IB exam style, where exact solutions within a specified domain (usually in radians) are expected and a deep understanding of periodic behaviour is essential.
题目设计模拟IB考试风格，要求在指定区间（通常为弧度制）内给出精确解，并且深刻理解周期行为至关重要。
2. Key Trigonometric Identities | 关键三角恒等式
iding it by cos²θ yields 1 + tan²θ = sec²θ, and dividing by sin²θ gives 1 + cot²θ = csc²θ.
勾股恒等式是基石：sin²θ + cos²θ = 1。将其除以 cos²θ 得到 1 + tan²θ = sec²θ，除以 sin²θ 得到 1 + cot²θ = csc²θ。
The double-angle formulas are heavily tested: sin 2θ = 2 sinθ cosθ; cos 2θ = cos²θ – sin²θ = 2cos²θ – 1 = 1 – 2sin²θ. You must be able to choose the most useful form depending on the equation.
二倍角公式是重点考查内容：sin 2θ = 2 sinθ cosθ；cos 2θ = cos²θ – sin²θ = 2cos²θ – 1 = 1 – 2sin²θ。你必须能够根据方程选择最便于使用的形式。
its reciprocal relations secθ = 1/cosθ, cscθ = 1/sinθ, cotθ = 1/tanθ complete the toolkit.
正切恒等式 tanθ = sinθ / cosθ 及其倒数关系 secθ = 1/cosθ、cscθ = 1/sinθ、cotθ = 1/tanθ 构成了完整的工具包。
3. Solving Basic Trigonometric Equations | 解基本三角方程
isolating the trigonometric function. For sin x = k, find the reference angle α = arcsin(|k|) in radians. Then use the sign of k to determine the quadrants.
首先分离三角函数。对于 sin x = k，求出以弧度表示的参考角 α = arcsin(|k|)。然后根据 k 的符号确定象限。
General solutions: for sine, x = α + 2πn or x = π – α + 2πn; for cosine, x = ±α + 2πn; for tangent, x = α + πn, where n ∈ ℤ Finally, list the specific solutions that lie within the given interval, such as [0, 2π].
通解：正弦为 x = α + 2πn 或 x = π – α + 2πn；余弦为 x = ±α + 2πn；正切为 x = α + πn，其中 n ∈ ℤ。最后列出落在指定区间（如 [0, 2π]）内的具体解。
4. Equations with Multiple Angles | 多倍角方程
or 2x = 5π/6 + 2πn. Then divide all terms by 2 to obtain x = π/12 + πn or x = 5π/12 + πn.
解如 sin(2x) = 1/2 的方程时，首先将 2x 当作变量处理：2x = π/6 + 2πn 或 2x = 5π/6 + 2πn。然后将每一项除以2，得到 x = π/12 + πn 或 x = 5π/12 + πn。
Remember to generate successive integer values for n until all solutions within the required domain are collected. Because the period is halved, you will typically obtain twice as many solutions as in the single-angle case.
记住用连续的整数值代入 n，直到收集到所有在要求范围内的解。因为周期减半，你通常会得到单倍角情形两倍数量的解。
5. Using the Unit Circle and CAST | 使用单位圆和CAST规则
The unit circle gives a visual interpretation: sinθ is the y-coordinate, cosθ is the x-coordinate. The CAST diagram indicates which trigonometric functions are positive in each quadrant (C: cos positive, A: all positive, S: sin positive, T: tan positive).
单位圆提供了直观的几何解释：sinθ 是 y 坐标，cosθ 是 x 坐标。CAST图指出各个象限中哪些三角函数为正（C：cos为正，A：所有为正，S：sin为正，T：tan为正）。
Using CAST, you can quickly determine the quadrants for the solutions without rote formula memorisation. Draw the reference angle from the horizontal axis and read off the angles in the correct quadrants.
利用CAST，你可以快速确定解的象限，无需死记公式。从水平轴画出参考角，并在正确象限内读出角度。
6. Equations Involving Factoring | 含因式分解的方程
Factoring is a powerful technique. For instance, consider 2 sin x cos x – sin x = 0. Factor out sin x to get sin x (2 cos x – 1) = 0. Then set each factor equal to zero.
因式分解是一种强有效的方法。例如，考虑 2 sin x cos x – sin x = 0。提取 sin x 得到 sin x (2 cos x – 1) = 0，然后分别令每个因子等于零。
Solve sin x = 0, giving x = πn, and 2 cos x – vide both sides by a trigonometric expression that could be zero, as this loses solutions.
解 sin x = 0，得出 x = πn，再解 2 cos x – 1 = 0，即 cos x = 1/2。找出指定区间内的全部解。切勿草率地将两边同时除以可能为零的三角函数表达式，那样会丢失解。
7. Quadratic Trigonometric Equations | 二次三角方程
olve the quadratic, discarding any u-values outside the range [-1, 1] for sine and cosine.
形如 a sin²x + b sin x + c = 0 的方程可通过代换求解。设 u = sin x，得到 a u² + b u + c = 0。解该二次方程，对于正弦和余弦，舍去任何超出 [-1, 1] 范围的 u 值。
After finding valid u, back-substitute and solve the basic trigonometric equation. The same method applies to equations in cos x or tan x, remembering that tangent has no range restriction.
找到有效的 u 后，回代并解基本的三角方程。同样的方法适用于关于 cos x 或 tan x 的方程，记住正切没有值域限制。
8. Equations with Reciprocal Functions | 涉及倒数函数的方程
moves the difficulty. For example, csc x = 2 is equivalent to 1/sin x = 2, so sin x = 1/2.
倒数函数 csc、sec 和 cot 常令学生心生畏惧，但简单的改写即可消除困难。例如，csc x = 2 等价于 1/sin x = 2，故 sin x = 1/2。
s x = 0 for sec and tan) and exclude those x-values from your solution set if they appear.
注意这些函数无定义的位置（对于 csc 和 cot，sin x = 0；对于 sec 和 tan，cos x = 0），如果解中出现这些 x 值，需将它们排除。
9. Identities in Proofs | 恒等式证明
Proof questions ask you to show that one side of an equation is identical to the other. Start with the more complicated side and transform it step by step using known identities. Writing everything in terms of sine and cosine is a reliable first step.
证明题要求展示等式的一边恒等于另一边。从较为复杂的一边入手，利用已知恒等式逐步变形。将每一个函数都化为正弦和余弦是一个可靠的起步策略。
Avoid the temptation to move terms across the equals sign unless you are certain both sides are non-zero. Always work on one side only and state which identity you are applying at each stage.
避免在等号两边移动项，除非你确定两边均非零。始终只处理一边，并说明每一步用到了哪个恒等式。
10. Applications: Modeling with Trig Equations | 应用：三角方程建模
Real-world problems often model periodic phenomena with an equation like h(t) = A sin(B(t – cific value, set h(t) equal to that number and solve the resulting trigonometric equation.
现实问题常用形如 h(t) = A sin(B(t – C)) + D 的方程描述周期现象。要确定何时高度达到某个特定值，令 h(t) 等于该数值，并解所得的三角方程。
cal restrictions on the variable t, such as t ≥ 0. Only discrete values may be meaningful.
记得在情境中解释解，考虑周期以及变量 t 的物理限制，例如 t ≥ 0。通常只有离散值是有意义的。
11. Common Mistakes and Tips | 常见错误与提示
g sin(2x) = k. Many students write the general solution as if the coefficient of x were 1.
在解 sin(2x) = k 时忘记调整周期。许多学生写出的通解就好像 x 的系数是 1 一样。
Working in degrees when the domain is given in radians. Always check whether the question specifies radians or degrees, but IB papers default to radians unless explicitly stated.
在给出弧度区间时却使用角度制。务必检查题目要求的是弧度还是角度，但除非明确说明，IB试卷默认使用弧度。
lose all solutions where sin x = 0. Instead, factor and apply the zero product principle.
除以可能为零的三角函数表达式。如果你从两边约去 sin x，将丢失所有 sin x = 0 的解。正确的做法是进行因式分解并应用零乘积原理。
Not testing whether solutions satisfy the original equation, especially after squaring steps or manipulating reciprocal functions, which may introduce extraneous solutions.
在平方步骤或处理倒数函数后，未检验解是否满足原方程。这些操作可能引入增根。
12. Practice Questions and Solutions | 练习题与解答
Example 1: Solve 2 sin x = √3 for 0 ≤ x ≤ 2π.
例1：解方程 2 sin x = √3，0 ≤ x ≤ 2π。
Solution: sin x = √3/2, reference angle = π/3. Sine is positive in quadrants I and II, so x = π/3, 2π/3.
解：sin x = √3/2，参考角 = π/3。正弦在第一、二象限为正，故 x = π/3, 2π/3。
Example 2: Solve 2 cos² x – cos x – 1 = 0 for 0 ≤ x < 2π.
例2：解方程 2 cos² x – cos x – 1 = 0，0 ≤ x < 2π。
Solution: Let u = cos x, then 2u² – u – 1 = 0 → (2u + 1)(u – 1) = 0 → u = –1/2 or u = 1. For cos x = –1/2, x = 2π/3, 4π/3; for cos x = 1, x = 0. Thus the solutions are x = 0, 2π/3, 4π/3.
解：令 u = cos x，则 2u² – u – 1 = 0 → (2u + 1)(u – 1) = 0 → u = –1/2 或 u = 1。由
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