Trigonometric Identities | 三角恒等式

📚 Trigonometric Identities | 三角恒等式

Trigonometric identities are the backbone of A-Level problem-solving, enabling you to simplify expressions, prove relationships, and crack trigonometric equations. In this Edexcel Mathematics revision guide, we walk through the principal identities you must master — from the Pythagorean and compound angle formulas to double angle and harmonic forms — with worked examples and exam-focused advice.

三角恒等式是 A-Level 数学解题的支柱，能够帮你化简表达式、证明关系式并求解三角方程。在这篇 Edexcel 数学复习指南中，我们将梳理必须掌握的核心恒等式 —— 从勾股恒等式、复合角公式到二倍角与辅助角形式 —— 并配有例题和应试建议。

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1. The Pythagorean Identities | 勾股恒等式

ds for every angle θ. This single relationship, derived from the unit circle, gives rise to two other powerful identities by division.

所有三角恒等式的基石是 sin²θ + cos²θ = 1，它对任意角 θ 都成立。这个源自单位圆的关系式，通过除法可以衍生出另外两个强有力的恒等式。

sin²θ + cos²θ = 1

1 + tan²θ = sec²θ

1 + cot²θ = csc²θ

These identities are essential when you need to replace squared functions to simplify an expression or solve an equation.

将 sin²θ + cos²θ = 1 除以 cos²θ 得到 1 + tan²θ = sec²θ，在 cosθ ≠ 0 时成立。类似地，除以 sin²θ 得到 1 + cot²θ = csc²θ，前提是 sinθ ≠ 0。当你需要用平方函数的替换来化简表达式或解方程时，这些恒等式至关重要。

= sin²θ. This trick of turning a sum into a tidy single term is typical of identity work.

例如，化简 sin⁴θ + sin²θ cos²θ，提取 sin²θ：sin²θ(sin²θ + cos²θ) = sin²θ × 1 = sin²θ。这种把和式变成整洁单项的技巧正是恒等式运用的典型模式。

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2. Compound Angle Formulas | 复合角公式

rically and are used to find exact values of trigonometric ratios for non-standard angles.

加法（复合角）公式使你能够处理 sin(A ± B)、cos(A ± B) 和 tan(A ± B)。它们由几何推导而来，可用于求非特殊角的三角比精确值。

sin(A ± B) = sin A cos B ± cos A sin B

cos(A ± B) = cos A cos B ∓ sin A sin B

tan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B)

Notice the sign reversal in the cosine identity: the 'plus' on the left gives a minus inside, and the 'minus' gives a plus. The tangent formula has a denominator of 1 ∓ tan A tan B — always the opposite sign to the numerator's ±. These patterns often catch students out, so commit them to memory.

注意余弦恒等式中的符号反转：左边的「加」在括号内产生减号，「减」则产生加号。正切公式的分母是 1 ∓ tan A tan B —— 总是与分子的 ± 符号相反。这些模式常让学生出错，务必牢记。

° as 45° − 30° and apply sin(A − B) = sin A cos B − cos A sin B. Using sin 45° = cos 45° = √2/2, sin 30° = 1/2, cos 30° = √3/2 gives sin 15° = (√2/2)(√3/2) − (√2/2)(1/2) = √6/4 − √2/4 = (√6 − √2)/4.

一个经典应用：不用计算器求 sin 15° 的精确值。将 15° 写成 45° − 30°，然后应用 sin(A − B) = sin A cos B − cos A sin B。代入 sin 45° = cos 45° = √2/2，sin 30° = 1/2，cos 30° = √3/2 得到 sin 15° = (√2/2)(√3/2) − (√2/2)(1/2) = √6/4 − √2/4 = (√6 − √2)/4。

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3. Double Angle Formulas | 二倍角公式

ties immediately delivers the double angle formulas. These are among the most-used identities in integration, differentiation, and equation solving.

将 B = A 代入复合角公式，立即得到二倍角公式。这些是积分、微分和方程求解中最常用的恒等式之一。

sin 2A = 2 sin A cos A

cos 2A = cos²A − sin²A = 2 cos²A − 1 = 1 − 2 sin²A

tan 2A = 2 tan A / (1 − tan²A)

ot of algebra. For instance, to express sin²θ in terms of cos 2θ, use cos 2θ = 1 − 2 sin²θ ⇒ sin²θ = (1 − cos 2θ)/2.

cos 2A 的三种形式可互相转换；选择与题目其余部分匹配的形式可以省去大量的代数运算。例如，要用 cos 2θ 表示 sin²θ，可用 cos 2θ = 1 − 2 sin²θ ⇒ sin²θ = (1 − cos 2θ)/2。

Example: Simplify (2 sin x cos x) / (cos²x − sin²x). The numerator is sin 2x, the denominator is cos 2x, so the fraction simplifies to tan 2x. Recognising these patterns quickly is a key exam skill.

例题：化简 (2 sin x cos x) / (cos²x − sin²x)。分子是 sin 2x，分母是 cos 2x，因此分式化简为 tan 2x。快速识别这些模式是一项关键的考试技能。

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4. Half-Angle Identities from Double Angles | 由二倍角公式导出的半角恒等式

Rearranging the double angle formulas for cos 2θ gives the so-called half-angle identities, which are extremely useful for integrating powers of sine and cosine.

重新整理 cos 2θ 的二倍角公式可得到所谓的半角恒等式，这些在积分正弦和余弦的幂次时极其有用。

sin²θ = (1 − cos 2θ) / 2

cos²θ = (1 + cos 2θ) / 2

ey also appear when solving equations like 2 cos²x − 1 = 0, which can be rewritten as cos 2x = 0.

这些恒等式将平方降次为线性三角项，非常适用于进一步数学中处理 ∫ sin²x dx 或 ∫ cos²x dx。它们在解诸如 2 cos²x − 1 = 0 的方程时也会出现，该方程可重写为 cos 2x = 0。

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5. Simplifying Expressions Using Identities | 使用恒等式化简表达式

Exam questions often ask you to 'simplify fully' a trigonometric expression. The strategy is to spot substitutions that reduce the number of distinct functions or lower the degree.

考题常要求你「完全化简」某个三角表达式。策略是找出能减少不同函数种类或降低次数的替换。

n becomes cos θ + sin θ cos θ = cos θ(1 + sin θ). Every step relies on definitions and the Pythagorean identity.

例 1：化简 (1 / sec θ) + (sin²θ / tan θ)。写出 sec θ = 1/cos θ，则 1/sec θ = cos θ。而 sin²θ/tan θ = sin²θ/(sin θ/cos θ) = sin θ cos θ。表达式变为 cos θ + sin θ cos θ = cos θ(1 + sin θ)。每一步都依赖定义和勾股恒等式。

²θ − tan²θ, which by the Pythagorean identity equals 1. These short, elegant chains of reasoning test your fluency.

例 2：证明 (sec θ − tan θ)(sec θ + tan θ) = 1。相乘得 sec²θ − tan²θ，由勾股恒等式等于 1。这种简短而优雅的推理链条考查你对恒等式的熟练程度。

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

When proving an identity, you usually start from the more complicated side and transform it step by step until you reach the other side. Always work on one side only and cite the identities you use.

证明恒等式时，通常从较复杂的一边入手，逐步变形直至到达另一边。始终只操作一边，并注明你所用的恒等式。

Example: Prove (1 − cos 2θ) / sin 2θ ≡ et sin θ / cos θ = tan θ. Hence the identity holds for all θ where both sides are defined.

例题：证明 (1 − cos 2θ) / sin 2θ ≡ tan θ。从左边开始。利用 cos 2θ = 1 − 2 sin²θ，分子变为 2 sin²θ。分母 sin 2θ = 2 sin θ cos θ。约去 2 sin θ 得到 sin θ / cos θ = tan θ。因此该恒等式在所有使两边有定义的 θ 下成立。

Another common style: Prove (sin θ + cos θ)² ≡ u obtain 1 + sin 2θ. Simple expansion, backed by correct identity recall, is all you need.

另一种常见题型：证明 (sin θ + cos θ)² ≡ 1 + sin 2θ。展开左边：sin²θ + 2 sin θ cos θ + cos²θ。利用 sin²θ + cos²θ = 1 以及 2 sin θ cos θ = sin 2θ，得到 1 + sin 2θ。只需简单的展开和正确的恒等式记忆即可完成。

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7. Solving Trigonometric Equations | 解三角方程

Identities are indispensable for solving equations where more than one trigonometric function appears. The goal is to rewrite the equation in terms of a single function — sin θ, cos θ, tan θ, or a double angle — and then factor or solve the resulting polynomial.

在求解含多种三角函数的方程时，恒等式不可或缺。目标是将方程改写为只含一种函数 —— sin θ、cos θ、tan θ 或一个二倍角 —— 然后因式分解或求解所得的多项式。

x = 2 cos²x for 0° ≤ x ≤ 360°. Replace cos²x with 1 − sin²x to get 3 sin x = 2(1 − sin²x) ⇒ 2 = 0. Factor as (2 sin x − 1)(sin x + 2) = 0. Since sin x = −2 has no solution, sin x = ½ ⇒ x = 30°, 150°.

例题：解 3 sin x = 2 cos²x，其中 0° ≤ x ≤ 360°。用 1 − sin²x 替换 cos²x，得到 3 sin x = 2(1 − sin²x) ⇒ 2 sin²x + 3 sin x − 2 = 0。因式分解为 (2 sin x − 1)(sin x + 2) = 0。由于 sin x = −2 无解，故 sin x = ½ ⇒ x = 30°, 150°。

Example with double angle: Solve sin 2x = cos x for 0 ≤ x ≤ 2π. Use sin 2x = 2 sin x cos x ⇒ 2 sin x cos x − cos x = 0 ⇒ ives x = π/6, 5π/6. Always check the domain and list all solutions in the required format.

含有二倍角的例题：解 sin 2x = cos x，0 ≤ x ≤ 2π。利用 sin 2x = 2 sin x cos x ⇒ 2 sin x cos x − cos x = 0 ⇒ cos x (2 sin x − 1) = 0。于是 cos x = 0

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