📚 pdfjoiner_(4)-205: Trigonometric Equations and Identities | 三角方程与恒等式：Edexcel A-Level数学
fjoiner_(4)-205 resource, provides a comprehensive revision of trigonometric equations and identities for Edexcel students, covering radian measure, key identities, solution strategies, and common pitfalls.
解三角方程是A-Level数学的核心技能，对纯数学和应用模块都至关重要。本文基于pdfjoiner_(4)-205资源，为Edexcel学生系统复习三角方程与恒等式，涵盖弧度制、核心恒等式、解题策略和常见错误。
1. Introduction to Trigonometric Equations | 三角方程简介
n a given interval, often 0 ≤ θ < 2π or 0° ≤ θ < 360°. Edexcel exam questions frequently require solutions in radians.
三角方程涉及通常用θ或x表示的未知角。解方程意味着找出在给定区间内满足方程的所有角度，通常为0 ≤ θ < 2π或0° ≤ θ < 360°。Edexcel考试题常要求用弧度制给出解。
Unlike algebraic equations, trigonometric equations have infinitely many solutions due to periodicity. Therefore, we restrict solutions to a specific domain, and we use the symmetry and periodicity of trig functions to generate all answers.
与代数方程不同，由于周期性，三角方程有无限多个解。因此，我们将解限制在特定域内，并利用三角函数的对称性和周期性来生成所有答案。
2. Radian Measure and the Unit Circle | 弧度制与单位圆
adius. The conversion is π rad = 180°, so to convert degrees to radians multiply by π/180.
弧度制对于A-Level三角学至关重要。一弧度是弧长等于半径的圆心角。转换关系为π rad = 180°，因此角度转弧度乘以π/180。
180° = π rad, so 1° = π/180 rad, 1 rad ≈ 57.3°
Common radian values you must memorise are π/6 (30°), π/4 (45°), π/3 (60°), π/2 (90°), π (180°), and 2π (360°). Solutions to equations are almost always required in exact multiples of π.
你必须熟记的常见弧度值包括π/6 (30°)、π/4 (45°)、π/3 (60°)、π/2 (90°)、π (180°) 和 2π (360°)。方程的解几乎总是要求用π的精确倍数形式表示。
| Degrees | Radians |
|---|---|
| 0° | 0 |
| 30° | π/6 |
| 45° | π/4 |
| 60° | π/3 |
| 90° | π/2 |
| 180° | π |
| 270° | 3π/2 |
| 360° | 2π |
The table above shows standard degree-to-radian conversions you must be fluent with when giving final answers.
上表列出了标准角度与弧度转换，你在给出最终答案时必须熟练运用。
3. Basic Trigonometric Equations and Principal Values | 基本三角方程与主值
e first step is to find the principal value using the inverse trigonometric functions on your calculator, ensuring it is in the correct mode (radians).
最简单的三角方程是sin θ = k、cos θ = k和tan θ = k。第一步是使用计算器上的反三角函数求主值，确保处于正确的模式（弧度）。
= 5π/6. You can use the CAST diagram or graph to find other solutions within the interval.
对于sin θ = 0.5，主值为π/6。但sin θ在第一和第二象限为正，因此第二个解是π - π/6 = 5π/6。你可以使用CAST图或图像找出区间内的其他解。
If sin θ = k, then θ = sin⁻¹(k), π - sin⁻¹(k) ± 2nπ
¹(k)) plus multiples of 2π. For tan θ = k, solutions repeat every π, so θ = tan⁻¹(k) + nπ.
类似地，对于cos θ = k，解为θ = cos⁻¹(k)和-cos⁻¹(k)（或2π - cos⁻¹(k)）再加2π的整数倍。tan θ = k的解每π重复一次，因此θ = tan⁻¹(k) + nπ。
4. Using Trigonometric Identities | 使用三角恒等式
Many equations require the use of trigonometric identities to simplify. The two most fundamental identities you must know are sin²θ + cos²θ ≡ 1 and tan θ ≡ sin θ / cos θ.
许多方程需要用三角恒等式化简。你必须掌握的两个最基本恒等式是sin²θ + cos²θ ≡ 1和tan θ ≡ sin θ / cos θ。
sin²θ + cos²θ ≡ 1, tan θ ≡ sin θ / cos θ (cos θ ≠ 0)
3, replace sin²θ with 1 - cos²θ to form a quadratic in cos θ. Then solve the quadratic and find all θ values in the given interval.
例如，解方程2 sin²θ + 3 cos θ = 3，用1 - cos²θ替换sin²θ，形成关于cos θ的二次方程。然后解二次方程，求出给定区间内的所有θ值。
Other identities like 1 + cot²θ ≡ cosec²θ and 1 + tan²θ ≡ sec²θ are also useful. When the equation involves multiple functions, always try to express everything in terms of one trig function using identities.
像1 + cot²θ ≡ cosec²θ和1 + tan²θ ≡ sec²θ等恒等式也很有用。当方程包含多个函数时，始终尝试用恒等式将所有函数表示成同一个三角函数。
5. Quadratic Trigonometric Equations | 二次三角方程
Factorising or using the quadratic formula gives possible values for sin θ, which must lie between -1 and 1. Reject any values outside this range.
二次三角方程可写成a sin²θ + b sin θ + c = 0形式。因式分解或使用求根公式得出sin θ的可能值，这些值必须在-1到1之间。舍去超出此范围的值。
2 sin θ + 1)(sin θ - 1) = 0, so sin θ = -1/2 or sin θ = 1. Then solve each basic equation.
考虑2 sin²θ - sin θ - 1 = 0。因式分解：(2 sin θ + 1)(sin θ - 1) = 0，所以sin θ = -1/2或sin θ = 1。然后分别解基本方程。
6. Equations with Multiple Angles | 多倍角方程
Equations like sin 2θ = √3 u = 2θ, solve for u within the expanded interval, then divide by the coefficient to get θ.
像sin 2θ = √3/2或cos 3θ = 0.5这样的方程需要仔细处理。首先求多倍角的主值，例如令u = 2θ，在扩大后的区间内解出u，然后
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