Edexcel A-Level Maths: Trigonometric Identities & Equations | Edexcel A-Level 数学：三角恒等式与方程

📚 Edexcel A-Level Maths: Trigonometric Identities & Equations | Edexcel A-Level 数学：三角恒等式与方程

Trigonometric identities and equations are central to Edexcel A-Level Mathematics, appearing in Pure Paper 1 and Paper 2. This guide covers the key identities, the standard solution methods, the calculus connections and the common exam pitfalls that cost marks.

三角恒等式与方程是 Edexcel A-Level 数学的核心内容，在 Pure Paper 1 和 Paper 2 中频繁出现。本指南涵盖关键恒等式、标准求解方法、与微积分的联系，以及最易丢分的常见考试陷阱。

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1. Identity vs Equation | 恒等式与方程的区别

an identity because it holds for every θ, and so is tan θ = sin θ/cos θ wherever cos θ ≠ 0.

三角恒等式对变量所有允许取值都成立。例如 sin²θ + cos²θ = 1 是恒等式，因为它对任意 θ 都成立；只要 cos θ ≠ 0，tan θ = sin θ/cos θ 也是恒等式。

re θ = 30° and θ = 150°. In the exam you must decide whether you are proving an identity or solving an equation, because the two tasks are marked completely differently.

而方程只在某些特定取值下成立。例如 sin θ = 0.5 是方程，在区间 0 ≤ θ < 360° 内其解为 θ = 30° 与 θ = 150°。考试中你必须区分自己是在证明恒等式还是在求解方程，因为二者的评分方式截然不同。

When proving an identity, never move terms across the equals sign. Start from the more complicated side and manipulate it, using the standard identities, until it matches the other side. Write each step on a new line and keep one side unchanged as your target.

证明恒等式时，绝不可把项在等号两侧移来移去。应从较复杂的一侧出发，利用标准恒等式进行变形，直至与另一侧完全相同。每一步另起一行书写，并把另一侧固定为你的目标。

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2. Reciprocal and Quotient Identities | 倒数与商数恒等式

= 1/sin θ and cot θ = 1/tan θ = cos θ/sin θ, valid wherever the denominators are non-zero.

倒数三角函数包括正割、余割和余切。它们定义为 sec θ = 1/cos θ、cosec θ = 1/sin θ、cot θ = 1/tan θ = cos θ/sin θ，在分母不为零处成立。

These definitions are especially useful when simplifying rational expressions or changing a problem into sine and cosine. Most exam questions that look unfamiliar with sec, cosec and cot become routine once everything is rewritten in terms of sin θ and cos θ.

这些定义在化简分式或将问题化为正弦和余弦时特别有用。多数涉及 sec、cosec、cot 的题目看似陌生，一旦全部改写成 sin θ 与 cos θ，就变成了常规题。

English	中文	Formula
Secant	正割	sec θ = 1/cos θ
Cosecant	余割	cosec θ = 1/sin θ
Cotangent	余切	cot θ = cos θ/sin θ = 1/tan θ

s θ and sin θ lie between −1 and 1. Any solution that contradicts these ranges must be rejected.

注意取值范围：sec θ ≥ 1 或 sec θ ≤ −1，cosec θ ≥ 1 或 cosec θ ≤ −1，因为 cos θ 与 sin θ 都介于 −1 与 1 之间。任何与此矛盾的解都必须舍去。

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3. Pythagorean Identities | 毕达哥拉斯恒等式

ding it by cos²θ gives 1 + tan²θ = sec²θ, and dividing by sin²θ gives 1 + cot²θ = cosec²θ.

主要的毕达哥拉斯恒等式是 sin²θ + cos²θ = 1。将其除以 cos²θ 得 1 + tan²θ = sec²θ，除以 sin²θ 得 1 + cot²θ = cosec²θ。

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

= 1 − sin²θ. These are the versions you actually substitute into equations and integrals.

变形后的形式同样重要：sin²θ = 1 − cos²θ 与 cos²θ = 1 − sin²θ。真正代入方程和积分中的往往正是这两种形式。

qually, 2cos²θ − 1 = cos 2θ, which is essential for integration and for solving equations.

这些恒等式允许你将 2cos²θ − 1 等表达式用 sin θ 或 cos θ 表示：由 cos²θ = 1 − sin²θ 得 2cos²θ − 1 = 1 − 2sin²θ。同时 2cos²θ − 1 = cos 2θ，这对积分和方程求解至关重要。

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4. Compound Angle Formulae | 复合角公式

Compound angle formulae express the sine, cosine and tangent of a sum or difference of two angles. They are given in the formula booklet, but you are expected to use them fluently in both directions.

复合角公式用两个角的正弦、余弦和正切表示它们和与差的正弦、余弦和正切。公式手册中有给出，但考试要求你能熟练双向使用。

sin(A + B) = sin A cos B + cos A sin B
sin(A − B) = sin A cos B − cos A sin B
cos(A + B) = cos A cos B − sin A sin B
cos(A − B) = cos A cos B + sin A sin B

/(1 + tan A tan B). Note the sign reversal in the denominator compared with the numerator.

正切的两个公式为 tan(A + B) = (tan A + tan B)/(1 − tan A tan B)，tan(A − B) = (tan A − tan B)/(1 + tan A tan B)。注意分母的符号与分子相反。

x, and showing that a given expression is constant. Always check whether the question says 'hence' - if it does, you must use the previous part rather than starting again.

典型用法包括求精确值，如 sin 75° = sin(45° + 30°)；化简表达式，如 sin 3x cos x − cos 3x sin x = sin 2x；以及证明某表达式为常数。务必留意题目是否写了 'hence'（由此）——若有，就必须使用上一问的结果，而不能另起炉灶。

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5. Double Angle and Half Angle Formulae | 倍角与半角公式

mulae produces the double angle formulae, which are among the most heavily examined results in the whole specification.

在复合角公式中令 B = A，就得到倍角公式，这是整个考纲中考查最频繁的结果之一。

sin 2θ = 2 sin θ cos θ
cos 2θ = cos²θ − sin²θ = 1 − 2sin²θ = 2cos²θ − 1
tan 2θ = 2 tan θ/(1 − tan²θ)

on involves sine, use cos 2θ = 1 − 2sin²θ; if it involves cosine, use cos 2θ = 2cos²θ − 1.

cos 2θ 的三种形式并非可有可无的补充：选对其中一种通常就是证明题的关键一步。若式中其余部分都是正弦，则用 cos 2θ = 1 − 2sin²θ；若其余部分是余弦，则用 cos 2θ = 2cos²θ − 1。

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They turn squared trigonometric terms into linear ones, which is exactly what you need before integrating.

将其变形，即得半角（降幂）形式 sin²θ = (1 − cos 2θ)/2 与 cos²θ = (1 + cos 2θ)/2。它们把三角函数的平方项化为一次项，而这正是积分前所必需的。

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6. The R-Formula (Harmonic Form) | 辅助角公式（和差化积）

sults are a sin θ + b cos θ = R sin(θ + α) and a sin θ + b cos θ = R cos(θ − α), where R = √a² + b² and tan α = b/a.

形如 a sin θ + b cos θ 的表达式都可以写成单个正弦或余弦。标准结果为 a sin θ + b cos θ = R sin(θ + α) 或 a sin θ + b cos θ = R cos(θ − α)，其中 R = √a² + b²，tan α = b/a。

a sin θ + b cos θ = R sin(θ + α), R = √a² + b², tan α = b/a

Always expand your answer back out to check it. Because the maximum of sin(θ + α) is 1 and the minimum is −1, the expression has maximum value R and minimum value −R; this is how questions about greatest and least values, and about the first time a maximum occurs, are answered.

务必把结果展开回去检验。由于 sin(θ + α) 的最大值为 1、最小值为 −1，该表达式的最大值为 R、最小值为 −R；关于最大值、最小值以及首次取得最大值的时刻等问题，正是这样作答的。

0.4, and you solve it exactly as you would solve any basic sine equation, using θ + α as the new variable.

例题：3 sin θ + 4 cos θ = 5 sin(θ + α)，其中 tan α = 4/3，故 α ≈ 53.13°。于是 3 sin θ + 4 cos θ = 2 变为 sin(θ + α) = 0.4，此时把 θ + α 当作新变量，按解基本正弦方程的步骤求解即可。

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7. Solving Trigonometric Equations: Intervals and Graphs | 解三角方程：区间与图像

Every trigonometric equation must be solved over the interval stated in the question. Find the principal value with your calculator, then use the symmetry of the sine, cosine and tangent graphs - or a CAST diagram - to find all remaining solutions inside the interval.

每个三角方程都必须在题目给定的区间内求解。先用计算器求出主值，再利用正弦、余弦、正切图像的对称性（或 CAST 图）找出区间内其余所有解。

0°. Then 2x = 30°, 150°, 390°, 510°, giving x = 15°, 75°, 195°, 255°. Forgetting to double the interval is the single most common source of lost marks.

当角度是 θ 的倍数时，区间也要相应变换。例如在 0 ≤ x ≤ 360° 上解 sin 2x = 0.5，首先注意 2x 落在 0 ≤ 2x ≤ 720° 内，于是 2x = 30°、150°、390°、510°，得 x = 15°、75°、195°、255°。忘记把区间翻倍，是丢分最常见的原因。

270°, 450°, 630° and θ = 45°, 135°, 225°, 315°. Alternatively, cos²θ = 1/2 gives cos θ = ±√2/2 and leads to the same four solutions.

第二道例题：在 0 ≤ θ < 360° 上解 2cos²θ − 1 = 0。利用 cos 2θ = 2cos²θ − 1 得 cos 2θ = 0，故 2θ = 90°、270°、450°、630°，θ = 45°、135°、225°、315°。或者由 cos²θ = 1/2 得 cos θ = ±√2/2，同样得到这四个解。

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8. Quadratic and Factorable Trigonometric Equations | 二次型与可因式分解的三角方程

solve the resulting basic equations. Always reject any root outside the range −1 ≤ u ≤ 1.

许多方程可化为关于 sin θ、cos θ 或 tan θ 的二次方程。令 u = sin θ，因式分解，解出 u，再解所得的基本方程。务必舍去所有落在 −1 ≤ u ≤ 1 之外根。

2sin θ − 1)(sin θ + 1) = 0, so sin θ = 1/2 or sin θ = −1. Hence θ = 30°, 150° or θ = 270°.

例题：在 0 ≤ θ < 360° 上解 2sin²θ + sin θ − 1 = 0。因式分解得 (2sin θ − 1)(sin θ + 1) = 0，故 sin θ = 1/2 或 sin θ = −1。于是 θ = 30°、150° 或 θ = 270°。

0 is a quadratic in sec θ, giving sec θ = 1 or sec θ = 2, hence cos θ = 1 or cos θ = 1/2.

当方程中同时出现正弦与 sec、cosec、cot 时，先把所有量都化为 sin 与 cos。例如 sec²θ − 3sec θ + 2 = 0 是关于 sec θ 的二次方程，得 sec θ = 1 或 sec θ = 2，即 cos θ = 1 或 cos θ = 1/2。

nfirming that cos θ = 0 does not satisfy the equation - otherwise you will lose solutions.

另一类标准题通过两边同除以 cos θ 并利用 tan θ = sin θ/cos θ 来解，但前提是先确认 cos θ = 0 不满足原方程，否则会丢失解。

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9. Using Identities in Calculus | 恒等式在微积分中的应用

Identities are not only a Pure algebra topic; they are the standard tool for differentiating and integrating products and powers of trigonometric functions.

恒等式不只属于纯代数题型，它还是对三角函数乘积与幂进行求导和积分的标准工具。

x = sin 2x. This double-angle simplification is usually the form the mark scheme expects.

对 y = sin²x 求导时，写成 y = sin x · sin x 并用乘积法则，得 dy/dx = 2 sin x cos x = sin 2x。评分标准通常期望的正是这个倍角化简后的形式。

2 + sin 2x/4 + c. Any integral of an even power of sine or cosine is handled the same way.

积分时使用降幂形式：∫ sin²x dx = ∫ (1 − cos 2x)/2 dx = x/2 − sin 2x/4 + c；同理 ∫ cos²x dx = x/2 + sin 2x/4 + c。任何偶数次幂的正弦或余弦积分都可用同样方法处理。

With radians, remember that the derivative of sin x is cos x and the derivative of cos x is −sin x only when the angle is measured in radians, which is why calculus questions in Paper 2 are always set in radians.

在弧度制下，sin x 的导数为 cos x，cos x 的导数为 −sin x，这仅在角度以弧度计量时成立；因此 Paper 2 中的微积分题总是使用弧度。

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10. General Solutions and Exact Values | 通解与特殊角精确值

θ = k, θ = ±arccos k + 360°n; for tan θ = k, θ = arctan k + 180°n, where n is any integer.

通解把解集推广到全体实数。对 sin θ = k，θ = arcsin k + 360°n 或 θ = 180° − arcsin k + 360°n；对 cos θ = k，θ = ±arccos k + 360°n；对 tan θ = k，θ = arctan k + 180°n，其中 n 为任意整数。

In radian work,

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