📚 Trigonometric Identities and Equations | 三角恒等式与方程
Trigonometric identities and equations form the backbone of A-Level Pure Mathematics. Mastering them allows you to simplify complex expressions, prove relationships, and solve equations that model wave behaviour, oscillations, and geometrical problems. In this article, we cover everything Edexcel students need: fundamental identities, reciprocal ratios, double-angle and compound-angle formulas, harmonic form, and systematic strategies for solving equations within given intervals.
三角恒等式与方程是 A-Level 纯数学的核心支柱。掌握它们能够帮助你化简复杂表达式、证明三角关系，并求解用于描述波动、振动与几何问题的方程。本文将全面覆盖 Edexcel 学生需要掌握的内容：基本恒等式、倒数比、倍角与和差角公式、辅助角形式，以及在给定区间内解方程的系统策略。
1. Trigonometric Ratios and the Unit Circle | 三角比与单位圆
a point (x, y) on the circumference gives cos θ = x, sin θ = y, and tan θ = y/x where x ≠ 0. The signs of these ratios in the four quadrants are memorised using 'All Students Take Calculus' – All positive in Quadrant I, Sine positive in II, Tangent in III, Cosine in IV.
三个基本的三角比是正弦、余弦和正切，它们在单位圆上定义。对于从正 x 轴量起的任意角度 θ，圆周上的点 (x, y) 满足 cos θ = x，sin θ = y，tan θ = y/x（x ≠ 0）。四个象限中三角比的符号可用 'All Students Take Calculus' 记忆：第一象限全正，第二象限正弦正，第三象限正切正，第四象限余弦正。
Radian measure is essential for calculus and equation solving. 180° equals π rad, so to convert degrees to radians multiply by π/180. Working in radians often simplifies differentiation and integration later, and Edexcel equations are frequently set in radians unless specified otherwise.
弧度制对微积分和解方程至关重要。180° 等于 π rad，因此度数乘以 π/180 即可化为弧度。使用弧度常会简化后续的微分与积分运算，Edexcel 考试中的方程如果没有特别说明，通常也要求在弧度下求解。
2. Fundamental Pythagorean and Quotient Identities | 基本勾股恒等式与商数恒等式
The two most important identities derived from the unit circle are: sin²θ + cos²θ ≡ 1 and tan θ ≡ sin θ / cos θ, cos θ ≠ 0. These 'must-know' relationships allow you to replace one ratio with another and underpin all further manipulation.
由单位圆导出的两个最重要的恒等式是：sin²θ + cos²θ ≡ 1 和 tan θ ≡ sin θ / cos θ（cos θ ≠ 0）。这两个必背关系允许你在三角比之间进行替换，是所有后续变形的基础。
sin²θ + cos²θ ≡ 1
tan θ ≡ sin θ / cos θ
From the first identity you can derive forms like sin²θ ≡ 1 – cos²θ and cos²θ ≡ 1 – sin²θ. These are extremely useful when solving quadratic trigonometric equations where you need to write everything in terms of a single trigonometric function.
从第一个恒等式可以推导出 sin²θ ≡ 1 – cos²θ 和 cos²θ ≡ 1 – sin²θ。当你需要将二次三角方程写成单一三角函数的表达式时，这些形式极其有用。
3. Reciprocal and Derived Identities | 倒数比与引申恒等式
Edexcel also introduces the reciprocal trigonometric functions: sec θ ≡ 1/cos θ, cosec θ ≡ 1/sin θ, and cot θ ≡ 1/tan θ ≡ cos θ/sin θ. Using the Pythagorean identity you obtain further relations: 1 + tan²θ ≡ sec²θ and 1 + cot²θ ≡ cosec²θ.
Edexcel 也引入了倒数三角函数：sec θ ≡ 1/cos θ，cosec θ ≡ 1/sin θ，以及 cot θ ≡ 1/tan θ ≡ cos θ/sin θ。利用勾股恒等式可以进一步得到：1 + tan²θ ≡ sec²θ 和 1 + cot²θ ≡ cosec²θ。
These derived identities are often tested in proof questions or when simplifying expressions that involve sec, cosec and cot. Being able to rewrite everything in terms of sine and cosine is a safe first step in any identity proof.
这些引申恒等式常出现在证明题中，或在化简含有 sec、cosec 和 cot 的表达式时考查。在证明恒等式时，首先将所有东西写成正弦和余弦的形式通常是一个稳妥的策略。
4. Solving Basic Trigonometric Equations | 解基本三角方程
n equation like sin θ = 0.5 for 0 ≤ θ < 2π, first find the principal value in the range – ymmetry and periodicity: sin θ is also positive in Quadrant II, giving a second solution π – π/6 = 5π/6. Always sketch the graph or use the unit circle to avoid missing solutions.
求解如 sin θ = 0.5 在 0 ≤ θ < 2π 内的方程，先在 –π/2 到 π/2 范围内找到主值：arcsin(0.5) = π/6。再利用对称性与周期性：sin θ 在第二象限也为正，得到第二个解 π – π/6 = 5π/6。应始终画图或使用单位圆，避免遗漏解。
For cos θ = –√2/2, principal value is 3π/4. Cosine is also negative in Quadrant III, giving 2π – lue is π/4, and because tan has period π, the second solution in [0, 2π) is 5π/4. Remember to list all solutions within the specified interval.
对于 cos θ = –√2/2，主值为 3π/4。余弦在第三象限也为负，得到 2π – 3π/4 = 5π/4。对于 tan θ = 1，主值为 π/4，由于正切周期为 π，[0, 2π) 内的第二个解是 5π/4。务必列出指定区间内的所有解。
5. Quadratic Trigonometric Equations | 二次型三角方程
a term like sin²θ or cos²θ, it often becomes a quadratic in disguise. For example, 2 sin²θ – sin θ – 1 = 0 for 0 ≤ θ ≤ 2π. Let x = sin θ, giving 2x² – x – 1 = 0, which factorises to (2x + 1)(x – 1) = 0. Thus sin θ = –½ or sin θ = 1. Solve each basic equation separately within the interval.
当方程包含 sin²θ 或 cos²θ 这样的项时，通常会化成一个隐藏的二次方程。例如在 0 ≤ θ ≤ 2π 内解 2 sin²θ – sin θ – 1 = 0。令 x = sin θ，得到 2x² – x – 1 = 0，分解为 (2x + 1)(x – 1) = 0。于是 sin θ = –½ 或 sin θ = 1，再在区间内分别求解。
If the equation mixes sine and cosine, use sin²θ + cos²θ ≡ 1 to replace one ratio. For 3 cos²θ + 5 sin θ = 1, substitute cos²θ = 1 – sin²θ to get a quadratic in sin θ: 3(1 – sin²θ) + 5 sin θ = 1 → –3 sin²θ + 5 sin θ + 2 = 0 → 3 sin²θ – 5 sin θ – 2 = 0, which can be solved by factorisation or formula.
如果方程混合了正弦和余弦，利用 sin²θ + cos²θ ≡ 1 替换其中一个比。对于 3 cos²θ + 5 sin θ = 1，代入 cos²θ = 1 – sin²θ，得到关于 sin θ 的二次方程：3(1 – sin²θ) + 5 sin θ = 1 → –3 sin²θ + 5 sin θ + 2 = 0 → 3 sin²θ – 5 sin θ – 2 = 0，随后通过因式分解或求根公式求解。
6. Using Identities to Solve More Complex Equations | 运用恒等式解较复杂方程
like 2θ or half-angles, identities are vital. For instance, cos 2θ + sin θ = 0. Use cos 2θ ≡ 1 – 2 sin²θ to obtain 1 – 2 sin²θ + sin θ = 0 → 2 sin²θ – sin θ – The identity turned the double-angle equation into a quadratic, which can then be solved.
当方程涉及不同参数时，如 2θ 或半角，恒等式至关重要。例如 cos 2θ + sin θ = 0。利用 cos 2θ ≡ 1 – 2 sin²θ 得到 1 – 2 sin²θ + sin θ = 0 → 2 sin²θ – sin θ – 1 = 0。这个恒等式将倍角方程转化为二次方程，可随即求解。
y by sin θ (provided sin θ ≠ 0) and bring terms together: cos θ = 2 cos θ sin θ → cos θ (1 – This gives cos θ = 0 or sin θ = ½, which are then solved and checked to ensure sin θ ≠ 0.
类似地，含有 sec、cosec 或 cot 的方程应首先改写为正弦和余弦。对于 cot θ = 2 cos θ，写成 cos θ/sin θ = 2 cos θ，两边乘以 sin θ（确保 sin θ ≠ 0）并移动项：cos θ = 2 cos θ sin θ → cos θ (1 – 2 sin θ) = 0。得到 cos θ = 0 或 sin θ = ½，求解后再验证 sin θ ≠ 0。
7. Double-Angle Formulas and Their Use in Equations | 倍角公式及其在方程中的应用
The three double-angle forms are essential: sin 2θ ≡ 2 sin θ cos θ; cos 2θ ≡ cos²θ – sin²θ ≡ 2 cos²θ – 1 ≡ 1 – 2 sin²θ; and tan 2θ ≡ 2 tan θ / (1 – tan²θ). You must be able to choose the most useful form of cos 2θ depending on whether you want to express it in terms of cos θ or sin θ alone.
三个倍角公式必不可少：sin 2θ ≡ 2 sin θ cos θ；cos 2θ ≡ cos²θ – sin²θ ≡ 2 cos²θ – 1 ≡ 1 – 2 sin²θ；tan 2θ ≡ 2 tan θ / (1 – tan²θ)。你必须能够根据想把 cos 2θ 表示成单一的 cos θ 还是 sin θ，来选择最合适的形式。
sin 2θ ≡ 2 sin θ cos θ
cos 2θ ≡ cos²θ – sin²θ ≡ 2 cos²θ – 1 ≡ 1 – 2 sin²θ
relies on replacing sin 2θ with 2 sin θ cos θ. Then 2 sin θ cos θ = sin θ → sin θ (2 cos θ – doubling the angle means you must find solutions for θ in the original range, not for 2θ.
求解 sin 2θ = sin θ 这样的方程，需要将 sin 2θ 替换为 2 sin θ cos θ，得到 2 sin θ cos θ = sin θ → sin θ (2 cos θ – 1) = 0，从而 sin θ = 0 或 cos θ = ½。务必注意区间调整：若 0 ≤ θ ≤ 2π，则寻找的是原区间内的 θ 解，而不是 2θ 的解。
8. Compound-Angle Formulas and Proofs | 和差角公式与证明
The addition formulas for sine and cosine are: sin(A ± B) ≡ sin A cos B ± cos A sin B; cos(A ± B) ≡ cos A cos B ∓ sin A sin B. For tangent, tan(A ± B) ≡ (tan A ± tan B) / (1 ∓ tan A tan B). These are tested both in equation solving and in 'show that' style proof questions.
正弦和余弦的和差角公式为：sin(A ± B) ≡ sin A cos B ± cos A sin B；cos(A ± B) ≡ cos A cos B ∓ sin A sin B。正切公式为：tan(A ± B) ≡ (tan A ± tan B) / (1 ∓ tan A tan B)。这些公式既有可能出现在解方程中，也有可能出现在“证明”类题目里。
To prove an identity like sin(45° + θ) – sin(45° – θ) = √2 g the addition formulas, then simplify. The cos terms cancel, leaving 2 cos 45° sin θ = 2(√2/2) sin θ = √2 sin θ, as required. Clear layout and showing all expansions step by step is crucial for full marks.
要证明如 sin(45° + θ) – sin(45° – θ) = √2 sin θ 这样的恒等式，先分别展开两个正弦和差角，再化简。cos 项抵消后剩下 2 cos 45° sin θ = 2(√2/2) sin θ = √2 sin θ，得证。清晰的布局和逐步展示展开过程是获得满分的关鍵。
9. Harmonic Form: a sin θ ± b cos θ | 辅助角形式：a sin θ ± b cos θ
and R sin(θ + α) = R sin θ cos α + R cos θ sin α, then equate coefficients: a = R cos α, b = R sin α. Hence R = √a² + b² and tan α = b/a.
将 a sin θ + b cos θ 表达为 R sin(θ ± α) 或 R cos(θ ± α) 的形式，是解方程和求极值的利器。展开 R sin(θ + α) = R sin θ cos α + R cos θ sin α，比较系数得：a = R cos α，b = R sin α。于是 R = √a² + b² 且 tan α = b/a。
a sin θ + b cos θ ≡ R sin(θ + α), where R = √a² + b², tan α = b/a
To solve 3 sin θ + 4 cos θ = 2, write the left side as R sin(θ + α). Here R = √3² + 4² = 5, and tan α = 4/3 ⇒ e for θ + α, then subtract α to get θ, taking care to remain within the required interval.
解 3 sin θ + 4 cos θ = 2 时，将左边写成 R sin(θ + α)。这里 R = √3² + 4² = 5，tan α = 4/3 ⇒ α ≈ 0.927 rad。方程变为 sin(θ + 0.927) = 2/5。先求解 θ + α，再减去 α 得到 θ，同时注意保持在规定的区间内。
10. General Strategies and Common Pitfalls | 通解策略与常见易错点
π, and double it if the argument is 2θ. Carefully identify all possible quadrants from the sign of the ratio, and never cancel a trigonometric term unless you know it is non-zero – factorise instead.
处理任何三角方程时，首先要明确界定区间。写出不等式，例如 0 ≤ θ ≤ 2π，如果参数是 2θ 则把区间相应加倍。根据三角比的符号仔细找出所有可能的象限，并且除非确定某项不为零，否则不要约去三角项——应当移项并提取公因子。
Also watch out for extraneous solutions when squaring both sides, and always verify solutions in the original equation. In proof questions, start from one side and transform it into the other using known identities, showing every algebraic step. Legibility and logical flow matter as much as the final result.
同时要注意两边平方时可能引入增根，并始终将解代回原方程验证。在证明题中，从一边出发，利用已知恒等式逐步化为另一边，展示每一步代数变形。整洁的书写与清晰的逻辑与最终结果同样重要。
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