Trigonometric Identities and Equations for Edexcel A-Level Pure Maths | 爱德思 A-Level 纯数学：三角恒等式与方程

📚 Trigonometric Identities and Equations for Edexcel A-Level Pure Maths | 爱德思 A-Level 纯数学：三角恒等式与方程

Trigonometric identities and equations form a central part of the Edexcel A-Level Pure Mathematics specification. They appear in both AS and A2 units, often combined with calculus, vectors and modelling. This revision guide covers the core identities, key solution techniques and common exam pitfalls.

三角恒等式与方程是爱德思 A-Level 纯数学大纲的核心内容。它们出现在 AS 和 A2 单元中，常与微积分、向量和建模结合考查。本复习指南涵盖核心恒等式、关键求解技巧和常见考试陷阱。

1. The Pythagorean Identities | 勾股恒等式

The three Pythagorean identities are derived directly from the unit circle definition of sine and cosine. They allow you to rewrite one trigonometric function in terms of another, which is essential for solving equations and proving more complex identities.

三个勾股恒等式直接来源于单位圆上正弦和余弦的定义。它们允许你将一个三角函数用另一个表示，这对于解方程和证明更复杂的恒等式至关重要。

sin²θ + cos²θ = 1

1 + tan²θ = sec²θ

1 + cot²θ = cosec²θ

You should memorise all three forms. The second and third are obtained by dividing the first identity by cos²θ and sin²θ respectively. In exam questions, these identities are often used to change a quadratic in sine into a quadratic in cosine, or vice versa.

你应当熟记这三种形式。第二和第三个恒等式分别由第一个恒等式除以 cos²θ 和 sin²θ 得到。在考试题中，这些恒等式常用于将正弦的二次式转化为余弦的二次式，或反过来。

• Use sin²θ = 1 − cos²θ when a question contains only sin²θ and cosθ.
• Use cos²θ = 1 − sin²θ when a question contains only cos²θ and sinθ.
• Use tan²θ = sec²θ − 1 when a question contains tan²θ and secθ.

• 当题目只含 sin²θ 和 cosθ 时，使用 sin²θ = 1 − cos²θ。
• 当题目只含 cos²θ 和 sinθ 时，使用 cos²θ = 1 − sin²θ。
• 当题目含 tan²θ 和 secθ 时，使用 tan²θ = sec²θ − 1。

---

2. Compound Angle Formulae | 复合角公式

terms of the individual angles. They are given in the Edexcel formula booklet, but recognising when to apply them is a key skill.

复合角公式将 A ± B 的正弦、余弦和正切用单个角表示。这些公式在爱德思公式手册中给出，但识别何时应用它们是一项关键技能。

sin(A ± B) = sinA cosB ± cosA sinB

cos(A ± B) = cosA cosB ∓ sinA sinB

tan(A ± B) = (tanA ± tanB) ÷ (1 ∓ tanA tanB)

Notice the sign change in the cosine formula: cos(A + B) has a minus sign, while cos(A − B) has a plus sign. For tangent, the sign in the numerator matches the sign on the left, but the sign in the denominator is opposite.

注意余弦公式中的符号变化：cos(A + B) 带负号，而 cos(A − B) 带正号。对于正切，分子中的符号与左边一致，但分母中的符号相反。

° = sin(45° + 30°), or to simplify expressions like sinθ cosφ + cosθ sinφ into sin(θ + φ).

这些公式经常用于求精确值，例如 sin75° = sin(45° + 30°)，或将 sinθ cosφ + cosθ sinφ 简化为 sin(θ + φ)。

---

3. Double Angle Formulae | 二倍角公式

are listed separately because they are tested very heavily in Edexcel A-Level Pure Maths.

二倍角公式是复合角公式在 A = B = θ 时的特殊情况。由于它们在爱德思 A-Level 纯数学中考查非常频繁，因此被单独列出。

sin2θ = 2sinθ cosθ

cos2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ

tan2θ = 2tanθ ÷ (1 − tan²θ)

The three versions of cos2θ are all equivalent. You choose the most useful form depending on whether you want the expression entirely in terms of cosθ, entirely in terms of sinθ, or as a difference of squares.

cos2θ 的三种形式完全等价。根据你是想将表达式完全用 cosθ 表示、完全用 sinθ 表示，还是表示为平方差，来选择最有用的形式。

• Use cos2θ = 2cos²θ − 1 to convert cos²θ into cos2θ.
• Use cos2θ = 1 − 2sin²θ to convert sin²θ into cos2θ.

• 使用 cos2θ = 2cos²θ − 1 将 cos²θ 转化为 cos2θ。
• 使用 cos2θ = 1 − 2sin²θ 将 sin²θ 转化为 cos2θ。

or example when integrating sin²x or cos²x, and in solving equations such as sin2θ = sinθ.

这些恒等式在积分中必不可少，例如对 sin²x 或 cos²x 积分时，以及在解 sin2θ = sinθ 这类方程时。

---

4. The R Addition Formula | R 加法公式

The R addition formula, also called the harmonic form, is used to rewrite expressions of the type a sinθ + b cosθ as a single sine or cosine function. This is particularly useful for solving equations and finding maximum and minimum values.

R 加法公式，也称为简谐形式，用于将 a sinθ + b cosθ 型表达式改写为单个正弦或余弦函数。这在解方程以及求最大值和最小值时特别有用。

a sinθ + b cosθ = R sin(θ + α)

a sinθ − b cosθ = R sin(θ − α)

a cosθ + b sinθ = R cos(θ − α)

a cosθ − b sinθ = R cos(θ + α)

Here R = √a² + b² k the quadrant carefully. The value of R is always positive, and α is usually given in radians or degrees to 1 decimal place unless the question asks for an exact value.

这里 R = √a² + b²，α = arctan(b ÷ a)，但你必须仔细检查象限。R 的值始终为正，α 通常以弧度或度给出到 1 位小数，除非题目要求精确值。

For example, 3 sinθ + 4 cosθ can be written as 5 sin(θ + 53.1°), since R = √3² + 4² = 5 and α = arctan(4 ÷ 3) ≈ 53.1°.

例如，3 sinθ + 4 cosθ 可以写成 5 sin(θ + 53.1°)，因为 R = √3² + 4² = 5 且 α = arctan(4 ÷ 3) ≈ 53.1°。

---

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

se the symmetry of the sine, cosine and tangent graphs to generate all possible solutions.

解三角方程需要你求出给定区间内的所有 θ 值，通常是 0 ≤ θ < 360° 或 0 ≤ θ < 2π。你必须利用正弦、余弦和正切图像的对称性来生成所有可能的解。

• For sinθ = k, solutions are θ and 180° − θ in the range 0° to 360°.
• For cosθ = k, solutions are θ and 360° − θ in the range 0° to 360°.
• For tanθ = k, solutions repeat every 180°, so add 180° to the principal value.

• 对于 sinθ = k，在 0° 到 360° 范围内解为 θ 和 180° − θ。
• 对于 cosθ = k，在 0° 到 360° 范围内解为 θ 和 360° − θ。
• 对于 tanθ = k，解每 180° 重复一次，因此将主值加上 180°。

Always begin by finding the principal value using your calculator. Then sketch the relevant graph or use a CAST diagram to identify all solutions in the required interval.

始终先用计算器求出主值。然后画出相关图像或使用 CAST 图来确定所需区间内的所有解。

ve in the first and second quadrants, the solutions are θ = 30° and θ = 180° − 30° = 150°.

例如，解 sinθ = 0.5，其中 0° ≤ θ < 360°。主值为 30°。由于正弦在第一和第二象限为正，解为 θ = 30° 和 θ = 180° − 30° = 150°。

---

广告 / Sponsored

6. Solving Equations Using Identities | 利用恒等式解方程

te the equation in terms of a single trigonometric function using one of the Pythagorean identities.

许多考试题要求你解含有多个三角函数的方程，例如 3 cos²θ + sinθ = 1。第一步是利用勾股恒等式将方程改写为只含一个三角函数的形式。

Replace cos²θ with 1 − sin²θ to obtain a quadratic in sinθ. Then solve the quadratic by factorising or using the quadratic formula, and finally solve the basic trigonometric equations.

将 cos²θ 替换为 1 − sin²θ，得到一个关于 sinθ 的二次方程。然后通过因式分解或求根公式解二次方程，最后解基本三角方程。

2)(sinθ − 1) = 0, giving sinθ = −2/3 or sinθ = 1. Then find all solutions in the interval.

例如，解 3 cos²θ + sinθ = 1，其中 0° ≤ θ < 360°。代入 cos²θ = 1 − sin²θ 得到 3(1 − sin²θ) + sinθ = 1，化简为 3 sin²θ − sinθ − 2 = 0。因式分解为 (3 sinθ + 2)(sinθ − 1) = 0，得到 sinθ = −2/3 或 sinθ = 1。然后求出区间内的所有解。

Be careful to check that any solutions found lie within the interval given. Also watch for extraneous solutions caused by squaring both sides of an equation.

注意检查求得的解是否在给定区间内。同时注意由于方程两边平方而产生的增根。

---

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

Proof questions require you to show that one side of an equation can be transformed into the other using known identities. You should work on one side only, usually the more complicated side, and simplify it step by step.

证明题要求你使用已知恒等式将等式的一边转化为另一边。你应当只处理一边，通常是较复杂的一边，并逐步化简。

Common strategies include converting everything to sine and cosine, using the Pythagorean identities to replace squares, and using double angle or compound angle formulae to combine terms.

常用策略包括将所有函数转换为正弦和余弦，使用勾股恒等式替换平方项，以及使用二倍角或复合角公式合并项。

use the identity sin²θ + cos²θ = 1 to replace 1 − sin²θ with cos²θ. The proof is complete.

例如，证明 (1 + sinθ)(1 − sinθ) = cos²θ。展开左边得到 1 − sin²θ，然后利用恒等式 sin²θ + cos²θ = 1 将 1 − sin²θ 替换为 cos²θ。证明完成。

Always state the identity you are using at each step, such as 'using sin²θ + cos²θ = 1' elps the examiner follow your reasoning and can earn method marks even if you make a slip.

每一步都要说明你使用的恒等式，例如“使用 sin²θ + cos²θ = 1”。这有助于阅卷人理解你的推理，即使你出现小错误也能获得方法分。

---

8. Modelling with Trigonometric Functions | 三角函数的建模

Trigonometric functions are often used to model periodic phenomena such as tides, temperature changes, or the motion of a pendulum. The R addition formula is particularly useful for writing a model in a form that reveals its amplitude and phase shift.

三角函数常用于模拟周期性现象，如潮汐、温度变化或摆的运动。R 加法公式特别适用于将模型写成能显示振幅和相位移动的形式。

(ct − d), where a is the vertical shift, b is the amplitude, c affects the period, and d affects the horizontal shift. The period is 2π ÷ c for sine and cosine models.

一个典型的模型可能是 h(t) = a + b sin(ct − d)，其中 a 是垂直位移，b 是振幅，c 影响周期，d 影响水平位移。对于正弦和余弦模型，周期为 2π ÷ c。

When solving modelling questions, always link the mathematical solutions back to the real-world context. For example, if the model gives the height of a tide, you must state the times when the tide reaches a certain height.

在解建模题时，始终将数学解联系回实际情境。例如，如果模型给出潮汐高度，你必须说明潮汐达到某一高度的时刻。

Exam questions often ask for the maximum and minimum values of a function like 5 sinθ + 12 cosθ. Using the R addition formula, you can write it as 13 sin(θ + 67.4°), so the maximum is 13 and the minimum is −13.

考试题常要求求 5 sinθ + 12 cosθ 这类函数的最大值和最小值。使用 R 加法公式，你可以将其写成 13 sin(θ + 67.4°)，因此最大值为 13，最小值为 −13。

---

9. Common Mistakes and Exam Tips | 常见错误与应试技巧

Students often lose marks on trigonometric equations by forgetting to find all solutions in the given interval, or by using the wrong sign in the compound angle formulae.

学生在三角方程上常因忘记求出给定区间内的所有解，或在复合角公式中使用错误符号而失分。

• Always check the interval. Solutions outside the interval must be discarded.
• Use a CAST diagram or graph to generate all solutions, not just the calculator value.
• Remember that sin⁻¹, cos⁻¹ and tan⁻¹ on a calculator return only the principal value.
• Do not divide both sides of an equation by a trigonometric expression, as this can lose solutions. Instead, factorise.

• 始终检查区间。超出区间的解必须舍去。
• 使用 CAST 图或图像生成所有解，而不仅仅是计算器给出的值。
• 记住计算器上的 sin⁻¹、cos⁻¹ 和 tan⁻¹ 只返回主值。
• 不要将方程两边除以一个三角函数表达式，因为这会丢解。应当因式分解。

When proving identities, avoid working on both sides simultaneously. Transform one side into the other and state the identities used.

在证明恒等式时，避免同时处理两边。将一边转化为另一边，并说明所使用的恒等式。

---

10. Practice Questions and Worked Examples | 练习题与例题解析

Work through the following examples to consolidate your understanding. Cover the solutions before attempting each question, then check your method.

通过以下例题巩固理解。在尝试每道题之前先遮盖答案，然后检查你的方法。

Question 1: Solve 2 sin²θ − cosθ − 1 = 0 for 0° ≤ θ < 360°.

问题 1： 解 2 sin²θ − cosθ − 1 = 0，其中 0° ≤ θ < 360°。

r cosθ = −1. The solutions are θ = 60°, 300° from cosθ = 0.5, and θ = 180° from cosθ = −1.

解：将 sin²θ 替换为 1 − cos²θ 得到 2(1 − cos²θ) − cosθ − 1 = 0，化简为 2 cos²θ + cosθ − 1 = 0。因式分解为 (2 cosθ − 1)(cosθ + 1) = 0，得到 cosθ = 0.5 或 cosθ = −1。解为 θ = 60°, 300° 来自 cosθ = 0.5，以及 θ = 180° 来自 cosθ = −1。

Question 2: n(θ + α), where R > 0 and 0° < α < 90°. Hence solve 3 sinθ + 4 cosθ = 2 for 0° ≤ θ < 360°.

问题 2： 将 3 sinθ + 4 cosθ 写成 R sin(θ + α) 的形式，其中 R > 0 且 0° < α < 90°。由此解 3 sinθ + 4 cosθ = 2，其中 0° ≤ θ < 360°。

Solution: R = √3² + 4² e by adding 360°: θ ≈ 103.3° + 360° is outside the interval. The only solution is θ ≈ 103.3°.

解：R = √3² + 4² = 5，α = arctan(4 ÷ 3) ≈ 53.1°。因此 3 sinθ + 4 cosθ = 5 sin(θ + 53.1°)。方程变为 5 sin(θ + 53.1°) = 2，所以 sin(θ + 53.1°) = 0.4。设 x = θ + 53.1°。则 sinx = 0.4，在第一个周期内得到 x ≈ 23.6° 或 x ≈ 156.4°。因此 θ = x − 53.1°，得到 θ ≈ −29.5°（舍去）和 θ ≈ 103.3°。考虑下一个周期加 360°：θ ≈ 103.3° + 360° 超出区间。唯一解是 θ ≈ 103.3°。

Practising a wide range of past paper questions will help you recognise the patterns quickly under exam conditions.

练习大量的历年真题将帮助你在考试条件下快速识别题型。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）