Edexcel A-Level Pure Mathematics: Trigonometric Equations and Identities | Edexcel A-Level 纯数学：三角方程与恒等式

📚 Edexcel A-Level Pure Mathematics: Trigonometric Equations and Identities | Edexcel A-Level 纯数学：三角方程与恒等式

Trigonometric equations and identities are central to the Edexcel A-Level Pure Mathematics specification. They appear in Paper 1 and Paper 2, often worth 6-12 marks per question, and they underpin techniques in differentiation, integration, vectors and mechanics.

三角方程与恒等式是 Edexcel A-Level 纯数学考试的核心内容，常见于 Paper 1 和 Paper 2，每题通常占 6-12 分，并为微分、积分、向量和力学等模块提供基础工具。

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1. Radian Measure and Arc Length | 弧度制与弧长

Edexcel A-Level trig questions often switch between degrees and radians. One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius, so π radians = 180°.

Edexcel A-Level 三角题经常需要在角度制和弧度制之间切换。1 弧度是指弧长等于半径的圆弧所对的圆心角，因此 π 弧度 = 180°。

arc length = rθ, sector area = ½r²θ

When using calculus with trigonometric functions, angles must always be in radians unless a question explicitly states otherwise.

在涉及三角函数的微积分中，除非题目明确说明，否则角度必须使用弧度制。

• To convert degrees to radians: multiply by π/180.
• To convert radians to degrees: multiply by 180/π.

度数转弧度：乘以 π/180；弧度转度数：乘以 180/π。

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2. Exact Values of Trigonometric Functions | 特殊角的三角函数精确值

Exact values for 30°, 45° and 60° are tested regularly, especially in surd form. You must be able to recall them quickly.

30°、45° 和 60° 的精确值经常考查，尤其是根式形式，必须能够快速写出。

θ	sin θ	cos θ	tan θ
0° = 0	0	1	0
30° = π/6	1/2	√3/2	1/√3
45° = π/4	1/√2	1/√2	1
60° = π/3	√3/2	1/2	√3
90° = π/2	1	0	undefined

These exact values allow you to give answers such as sin⁻¹(1/2) = π/6 exactly, rather than a decimal approximation.

利用这些精确值，你可以精确地写出例如 sin⁻¹(1/2) = π/6，而不必给出小数近似值。

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3. Pythagorean Identities | 毕达哥拉斯恒等式

The fundamental identity sin²θ + cos²θ = 1 is used to convert between sin and cos, to prove other identities, and to solve quadratic-like equations.

基本恒等式 sin²θ + cos²θ = 1 用于 sin 和 cos 之间的转换、证明其他恒等式以及求解二次型方程。

sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
1 + cot²θ = cosec²θ

These are derived by dividing the first identity by cos²θ or sin²θ respectively. They are especially useful when an equation contains both tan θ and sec θ, or both cot θ and cosec θ.

这些恒等式可以通过将第一个恒等式分别除以 cos²θ 或 sin²θ 得到。当方程同时包含 tan θ 和 sec θ，或 cot θ 和 cosec θ 时，它们特别有用。

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4. Compound Angle and Double Angle Identities | 两角和差与二倍角公式

The compound angle formulae are often printed in the Edexcel formula booklet, but you should still practise using them accurately with signs and quadrants.

两角和差公式通常印在 Edexcel 公式手册中，但你仍然需要准确练习它们在符号和象限上的使用。

sin(A ± B) = sinA cosB ± cosA sinB
cos(A ± B) = cosA cosB ∓ sinA sinB
tan(A ± B) = (tanA ± tanB) ÷ (1 ∓ tanA tanB)

The double angle formulae can be obtained by setting A = B = θ. For cos 2θ, there are three equally important forms.

二倍角公式可以通过令 A = B = θ 得到。对于 cos 2θ，有三种等价且都很重要的形式。

sin 2θ = 2sinθ cosθ
cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ
tan 2θ = 2tanθ ÷ (1 − tan²θ)

Choosing the correct form of cos 2θ is a key exam skill. For example, if an equation contains sin²θ, using cos 2θ = 1 − 2sin²θ often simplifies the working.

选择 cos 2θ 的正确形式是一项关键考试技能。例如，如果方程含有 sin²θ，使用 cos 2θ = 1 − 2sin²θ 通常可以简化计算。

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5. Solving Basic Trigonometric Equations | 基本三角方程求解

To solve a basic trig equation, first isolate the trig function, identify the principal value, then apply symmetry and periodicity to find all solutions in the given domain.

求解基本三角方程时，首先分离三角函数，确定主值，然后利用对称性和周期性求出给定区间内的所有解。

If sin θ = k: θ = α + 2nπ or θ = π − α + 2nπ

If cos θ = k: θ = ±α + 2nπ.

如果 cos θ = k，则 θ = ±α + 2nπ。

If tan θ = k: θ = α + nπ

Here α is the principal value and n is an integer. Always check which solutions lie inside the requested interval.

其中 α 是主值，n 是整数。始终要检查哪些解落在题目要求的区间内。

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6. Equations with Multiple Angles and Extended Domains | 多倍角与区间扩展

When the equation involves a multiple angle, such as sin(2x − 30°) = 0.5, the angle inside the bracket must be solved first over its own expanded domain.

当方程含有倍角时，例如 sin(2x − 30°) = 0.5，必须先对括号内的角在其扩展后的区间上进行求解。

Example: Solve sin(2x − 30°) = 0.5 for 0° ≤ x ≤ 360°.

示例：在 0° ≤ x ≤ 360° 内求解 sin(2x − 30°) = 0.5。

First write the expanded domain: if 0° ≤ x ≤ 360°, then −30° ≤ 2x − 30° ≤ 690°. The basic solutions for sin u = 0.5 in this expanded interval are u = 30°, 150°, 390°, 510°.

首先写出扩展区间：若 0° ≤ x ≤ 360°，则 −30° ≤ 2x − 30° ≤ 690°。在该扩展区间内，sin u = 0.5 的基本解为 u = 30°、150°、390°、510°。

Then solve for x: 2x − 30° = 30° gives x = 30°; 2x − 30° = 150° gives x = 90°; 2x − 30° = 390° gives x = 210°; 2x − 30° = 510° gives x = 270°.

然后解出 x：2x − 30° = 30° 得 x = 30°；2x − 30° = 150° 得 x = 90°；2x − 30° = 390° 得 x = 210°；2x − 30° = 510° 得 x = 270°。

The final solution set is x = 30°, 90°, 210°, 270°.

最终解集为 x = 30°、90°、210°、270°。

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7. Quadratic Forms in sin, cos and tan | 含 sin、cos、tan 的二次型方程

Many harder Edexcel questions reduce to a quadratic equation in sin θ, cos θ or tan θ. Factorising is usually the fastest method.

许多较难的 Edexcel 题目最终会转化为关于 sin θ、cos θ 或 tan θ 的二次方程。因式分解通常是最快的方法。

Example: Solve 2cos²θ − cosθ − 1 = 0 for 0 ≤ θ < 2π.

示例：在 0 ≤ θ < 2π 内求解 2cos²θ − cosθ − 1 = 0。

Factorise: (2cosθ + 1)(cosθ − 1) = 0, so cosθ = −1/2 or cosθ = 1.

因式分解得 (2cosθ + 1)(cosθ − 1) = 0，因此 cosθ = −1/2 或 cosθ = 1。

For cosθ = −1/2, the solutions in [0, 2π) are θ = 2π/3 and θ = 4π/3. For cosθ = 1, the solution is θ = 0.

对于 cosθ = −1/2，在 [0, 2π) 内的解为 θ = 2π/3 和 θ = 4π/3。对于 cosθ = 1，解为 θ = 0。

If an equation contains both sin²θ and cosθ, use sin²θ = 1 − cos²θ to rewrite it as a quadratic in cosθ only.

如果方程同时含有 sin²θ 和 cosθ，可使用 sin²θ = 1 − cos²θ 将其化为只含 cosθ 的二次方程。

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8. Harmonic Form R sin(θ ± α) and R cos(θ ± α) | 辅助角形式

Expressing a sinθ + b cosθ in the form R sin(θ + α) is a standard Edexcel technique for finding maxima, minima and solving equations.

将 a sinθ + b cosθ 表示为 R sin(θ + α) 是 Edexcel 的标准技巧，用于求最大值、最小值以及解方程。

a sinθ + b cosθ = R sin(θ + α), where R = √(a² + b²) and tanα = b/a

Example: Express 3sinθ + 4cosθ in the form R sin(θ + α).

示例：将 3sinθ + 4cosθ 表示为 R sin(θ + α) 的形式。

R = √(3² + 4²) = 5, and tanα = 4/3, so α ≈ 53.13° or 0.927 radians. Therefore 3sinθ + 4cosθ = 5sin(θ + 0.927).

R = √(3² + 4²) = 5，且 tanα = 4/3，所以 α ≈ 53.13° 或 0.927 弧度。因此 3sinθ + 4cosθ = 5sin(θ + 0.927)。

The maximum value of this expression is 5 and the minimum value is −5, which is much quicker than differentiating or completing the square.

该表达式的最大值为 5，最小值为 −5，这比求导或配方法要快得多。

For equations such as R sin(θ + α) = c, solve for θ + α first, then subtract α from each valid solution.

对于 R sin(θ + α) = c 这类方程，先解出 θ + α，再从每个有效解中减去 α。

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9. Proving Trigonometric Identities | 证明三角恒等式

Proof questions often require you to start on one side of the identity and manipulate it until it matches the other side. Common tools are the Pythagorean identity, double angle formulae and common denominators.

证明题通常要求从恒等式的一侧开始变形，直到与另一侧相同。常用工具包括毕达哥拉斯恒等式、二倍角公式和通分。

Example: Prove that (sinθ + cosθ)² = 1 + sin 2θ.

示例：证明 (sinθ

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