A-Level Mathematics: Trigonometric Identities | A-Level 数学：三角恒等式

📚 A-Level Mathematics: Trigonometric Identities | A-Level 数学：三角恒等式

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variable where both sides are defined. They form the backbone of A-Level trigonometry, enabling you to simplify expressions, prove equations, and solve complex problems across pure mathematics, calculus, and coordinate geometry.

三角恒等式是涉及三角函数的等式，它们在两边都有定义的所有变量取值下都成立。它们构成 A-Level 三角学的核心支柱，使你能够化简表达式、证明等式，并解决纯数学、微积分和坐标几何中的复杂问题。

1. The Fundamental Identities | 基本恒等式

The two fundamental identities that all others build upon are the tangent identity and the Pythagorean identity. You must be able to recall these instantly and apply them fluently in both directions.

所有其他恒等式都建立在两个基本恒等式之上：正切恒等式和毕达哥拉斯恒等式。你必须能够立即回忆起它们，并在两个方向上流畅运用。

tan θ = sin θ ÷ cos θ cot θ = cos θ ÷ sin θ

sin² θ + cos² θ = 1

These definitions imply a direct algebraic relationship: sin θ = cos θ × tan θ. This is frequently used when rewriting trigonometric expressions in terms of a single function.

这些定义蕴含着一个直接的代数关系：sin θ = cos θ × tan θ。这在将三角表达式改写为单一函数时经常使用。

From the Pythagorean identity, we can rearrange to isolate any single function: sin² θ = 1 − cos² θ, or cos² θ = 1 − sin² θ. This is useful when substituting into expressions involving squares of sine or cosine.

由毕达哥拉斯恒等式，我们可以变形以隔离任意单一函数：sin² θ = 1 − cos² θ，或 cos² θ = 1 − sin² θ。这在代入涉及正弦或余弦平方的表达式中非常有用。

2. Pythagorean Identities in Secant and Cosecant | 正割余割形式的毕达哥拉斯恒等式

Dividing the identity sin² θ + cos² θ = 1 by cos² θ yields a second form; dividing by sin² θ yields a third. These are equally important in A-Level examinations.

将 sin² θ + cos² θ = 1 两边除以 cos² θ 得到第二种形式；两边除以 sin² θ 得到第三种形式。这些在 A-Level 考试中同等重要。

tan² θ + 1 = sec² θ

1 + cot² θ = cosec² θ

A common examination question asks you to simplify an expression such as (sec θ − tan θ)(sec θ + tan θ). Recognising the difference of two squares, this becomes sec² θ − tan² θ, which equals 1 by the identity above.

一个常见的考试问题是化简如 (sec θ − tan θ)(sec θ + tan θ) 的表达式。识别出平方差后，它变为 sec² θ − tan² θ，根据上述恒等式等于 1。

When you see a fraction with sec² θ or cosec² θ in the denominator, rewriting it using 1 + tan² θ or 1 + cot² θ often converts the expression into a more manageable algebraic form.

当你看到分母中含有 sec² θ 或 cosec² θ 的分数时，使用 1 + tan² θ 或 1 + cot² θ 改写，通常能将表达式转化为更易处理的代数形式。

3. Compound Angle Formulas | 复合角公式

These formulas express sine, cosine, and tangent of sums or differences of two angles. They are derived from geometric considerations and are given in the Edexcel formula booklet, but you must know how to apply them correctly.

这些公式表达了两个角的和或差的正弦、余弦和正切。它们由几何推导而来，在 Edexcel 公式册中有给出，但你必须知道如何正确应用。

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

The mnemonic “C-C-S-S” for cosine helps: cos(A + B) = cos A cos B − sin A sin B, with signs reversed in the minus case. For sine, the sign in the expansion matches the sign in the bracket.

记忆口诀“余余正正”有助于余弦：cos(A + B) = cos A cos B − sin A sin B，在减号情形下符号相反。对于正弦，展开式中的符号与括号中的符号一致。

A classic A-Level example: express cos(x + 60°) in terms of sin x and cos x. Expanding gives cos x cos 60° − sin x sin 60°, then substituting cos 60° = ½ and sin 60° = √3⁄2 yields ½ cos x − (√3⁄2) sin x.

一个经典 A-Level 例子：用 sin x 和 cos x 表示 cos(x + 60°)。展开得 cos x cos 60° − sin x sin 60°，再代入 cos 60° = ½ 和 sin 60° = √3⁄2，得到 ½ cos x − (√3⁄2) sin x。

The tangent compound formula is less commonly tested but appears in proof questions. Note the sign pattern: tan(A + B) has + in the numerator and − in the denominator.

正切的复合角公式在证明题中偶有出现。注意符号模式：tan(A + B) 的分子为 +，分母为 −。

tan(A + B) = (tan A + tan B) ÷ (1 − tan A tan B)

tan(A − B) = (tan A − tan B) ÷ (1 + tan A tan B)

4. Double Angle Formulas | 二倍角公式

Setting B = A in the compound angle formulas gives the double angle formulas. These are among the most heavily tested identities in A-Level examinations, appearing in algebra, calculus, and numerical methods questions.

在复合角公式中令 B = A 即得二倍角公式。它们是 A-Level 考试中考查最频繁的恒等式之一，出现在代数、微积分和数值方法题目中。

sin 2A = 2 sin A cos A

cos 2A = cos² A − sin² A = 2 cos² A − 1 = 1 − 2 sin² A

tan 2A = 2 tan A ÷ (1 − tan² A)

The three equivalent forms of cos 2A are crucial. The form cos 2A = 2 cos² A − 1 rearranges to cos² A = ½(1 + cos 2A), which is routinely used to integrate cos² x.

cos 2A 的三种等价形式至关重要。形式 cos 2A = 2 cos² A − 1 变形为 cos² A = ½(1 + cos 2A)，常用于对 cos² x 积分。

Similarly, cos 2A = 1 − 2 sin² A rearranges to sin² A = ½(1 − cos 2A). These half-angle squared formulas convert even powers of sine and cosine into first powers of cos 2A, making integration straightforward.

类似地，cos 2A = 1 − 2 sin² A 变形为 sin² A = ½(1 − cos 2A)。这些半角平方公式将正弦和余弦的偶次幂转化为 cos 2A 的一次幂，使积分变得直接简单。

5. Half Angle Identities | 半角恒等式

Closely related to the double angle formulas are the half-angle identities. While the formulas themselves appear in the booklet, recognising when to apply them is a skill examiners actively test.

与二倍角公式密切相关的是半角恒等式。虽然公式本身出现在公式册中，但识别何时应用它们是考官积极考查的技能。

sin² (θ⁄2) = (1 − cos θ) ÷ 2

cos² (θ⁄2) = (1 + cos θ) ÷ 2

tan (θ⁄2) = (1 − cos θ) ÷ sin θ = sin θ ÷ (1 + cos θ)

A typical question asks you to evaluate an integral such as ∫ sin² x dx. Replacing sin² x with ½(1 − cos 2x) gives ½x − ¼ sin 2x + C — a result obtained in two lines that would otherwise require integration by parts.

一个典型问题要求计算积分如 ∫ sin² x dx。将 sin² x 替换为 ½(1 − cos 2x) 得到 ½x − ¼ sin 2x + C —— 两行即可得出结果，而用其他方法则需分部积分。

In proving the half-angle identity tan(θ⁄2) = sin θ ÷ (1 + cos θ), start from the right-hand side: substitute sin θ = 2 sin(θ⁄2) cos(θ⁄2) and 1 + cos θ = 2 cos²(θ⁄2), then cancel the common factor.

在证明半角恒等式 tan(θ⁄2) = sin θ ÷ (1 + cos θ) 时，从右边开始：代入 sin θ = 2 sin(θ⁄2) cos(θ⁄2) 和 1 + cos θ = 2 cos²(θ⁄2)，然后约去公因式。

6. Product-to-Sum and Sum-to-Product | 积化和差与和差化积

These identities convert products of trigonometric functions into sums or differences, and vice versa. They are essential when solving equations containing both sin A cos B terms or when integrating products like sin 3x cos 2x.

这些恒等式将三角函数之积转化为和或差，反之亦然。它们在求解含 sin A cos B 项的方程或积分 sin 3x cos 2x 这样的乘积时必不可少。

sin A cos B = ½ [sin(A + B) + sin(A − B)]

cos A sin B = ½ [sin(A + B) − sin(A − B)]

cos A cos B = ½ [cos(A + B) + cos(A − B)]

sin A sin B = ½ [cos(A − B) − cos(A + B)]

To integrate ∫ sin 3x cos 5x dx, apply the first product-to-sum formula with A = 3x and B = 5x. The integrand becomes ½[sin 8x + sin(−2x)], which simplifies to ½(sin 8x − sin 2x), and each term integrates by inspection.

要积分 ∫ sin 3x cos 5x dx，令 A = 3x 和 B = 5x，应用第一个积化和差公式。被积函数变为 ½[sin 8x + sin(−2x)]，化简为 ½(sin 8x − sin 2x)，每一项都可以直接积分。

For the sum-to-product direction, sin P + sin Q = 2 sin((P+Q)⁄2) cos((P−Q)⁄2) can be derived by writing P = A + B and Q = A − B and solving simultaneously. This conversion helps when factoring expressions like sin x + sin 3x = 0.

对于和差化积方向，sin P + sin Q = 2 sin((P+Q)⁄2) cos((P−Q)⁄2) 可通过设 P = A + B、Q = A − B 并联立求解推导出来。这种转换有助于因式分解如 sin x + sin 3x = 0 的表达式。

7. Solving Trigonometric Equations Using Identities | 利用恒等式求解三角方程

Many A-Level equations cannot be solved directly and require strategic use of identities to reduce the equation to a single trigonometric function with one argument.

许多 A-Level 方程无法直接求解，需要策略性地使用恒等式将方程化为单一三角函数且单一角度。

Consider the equation 2 sin² θ = cos θ + 1. Replacing sin² θ with 1 − cos² θ gives 2(1 − cos² θ) = cos θ + 1. Rearranging produces 2 cos² θ + cos θ − 1 = 0, a quadratic in cos θ, which factors to (2 cos θ − 1)(cos θ + 1) = 0.

考虑方程 2 sin² θ = cos θ + 1。将 sin² θ 替换为 1 − cos² θ 得 2(1 − cos² θ) = cos θ + 1。整理得 2 cos² θ + cos θ − 1 = 0，这是关于 cos θ 的二次方程，因式分解为 (2 cos θ − 1)(cos θ + 1) = 0。

Each linear factor gives solutions: cos θ = ½ gives θ = 60°, 300°, and cos θ = −1 gives θ = 180° in the interval 0 ≤ θ < 360°. This "reduce to quadratic" strategy is the single most common trigonometric equation technique at A-Level.

每个线性因式给出解：cos θ = ½ 得 θ = 60°、300°，cos θ = −1 得 θ = 180°（在 0 ≤ θ < 360° 范围内）。这种“化为二次方程”策略是 A-Level 中最常见的三角方程技巧。

When an equation mixes sin 2x and sin x, such as sin 2x = sin x, expand sin 2x to obtain 2 sin x cos x = sin x. Subtract and factor: sin x (2 cos x − 1) = 0. The solutions arise from sin x = 0 and cos x = ½.

当方程同时包含 sin 2x 和 sin x 时，例如 sin 2x = sin x，展开 sin 2x 得 2 sin x cos x = sin x。移项并因式分解：sin x (2 cos x − 1) = 0。解来自 sin x = 0 和 cos x = ½。

8. Proving Trigonometric Identities | 三角恒等式的证明

Proof questions require you to start from one side of the equation and transform it step by step until it becomes identical to the other side. The key is systematic strategy, not inspiration.

证明题要求你从等式的一边出发，逐步变形直到它变得与另一边完全相同。关键是系统性的策略，而非灵光一现。

Here is a reliable strategy sequence:

以下是一个可靠的策略序列：

Start with the more complicated side, since it offers more room for transformation.
从较复杂的一边开始，因为它提供更多变换空间。
Convert all functions to sine and cosine using the fundamental identities.
用基本恒等式将所有函数转化为正弦和余弦。
Look for common denominators to combine fractions.
寻找公分母以合并分数。
Apply Pythagorean identities to eliminate 1 ± sin² θ-type expressions.
应用毕达哥拉斯恒等式消除 1 ± sin² θ 型表达式。
Factorise where possible to reveal cancellations.
在可能处因式分解以揭示约分。

Worked example: prove that (1 − cos θ) ÷ sin θ = sin θ ÷ (1 + cos θ). Begin with the left side and multiply numerator and denominator by (1 + cos θ): (1 − cos θ)(1 + cos θ) ÷ [sin θ (1 + cos θ)] = (1 − cos² θ) ÷ [sin θ (1 + cos θ)] = sin² θ ÷ [sin θ (1 + cos θ)] = sin θ ÷ (1 + cos θ).

示例：证明 (1 − cos θ) ÷ sin θ = sin θ ÷ (1 + cos θ)。从左边开始，分子分母同乘 (1 + cos θ)：(1 − cos θ)(1 + cos θ) ÷ [sin θ (1 + cos θ)] = (1 − cos² θ) ÷ [sin θ (1 + cos θ)] = sin² θ ÷ [sin θ (1 + cos θ)] = sin θ ÷ (1 + cos θ)。

Proof questions reward clear, annotated working. Every line must state both the algebraic manipulation and the identity used, since partial credit is awarded for correct transformations even if you cannot finish the proof.

证明题重视清晰、有注释的步骤。每一步都必须陈述所用的代数操作和恒等式，因为即使无法完成证明，正确的变换也能获得部分分数。

9. R sin(x + α) Form | R sin(x + α) 形式

The expression a sin x + b cos x can always be rewritten as a single sine (or cosine) function with an amplitude R and phase shift α. This is a variant of the compound angle formulas used in reverse.

表达式 a sin x + b cos x 总是可以改写为具有振幅 R 和相移 α 的单一正弦（或余弦）函数。这是复合角公式的逆向应用变体。

a sin x + b cos x = R sin(x + α)

R = √(a² + b²), tan α = b ÷ a

To derive this, expand R sin(x + α) = R sin x cos α + R cos x sin α. Matching coefficients requires R cos α = a and R sin α = b. Squaring and adding gives R² = a² + b², and dividing gives tan α = b ÷ a.

为推导此形式，展开 R sin(x + α) = R sin x cos α + R cos x sin α。对比系数需要 R cos α = a 和 R sin α = b。平方相加得 R² = a² + b²，两式相除得 tan α = b ÷ a。

This technique is essential for finding the maximum and minimum of a sin x + b cos x: since sine ranges from −1 to 1, the maximum is R and the minimum is −R. It also converts equations like 3 sin x + 4 cos x = 2 into the simple form 5 sin(x + α) = 2.

此技巧对于求 a sin x + b cos x 的最大值和最小值至关重要：由于正弦取值范围为 −1 到 1，最大值为 R，最小值为 −R。它还将 3 sin x + 4 cos x = 2 这样的方程化为简单形式 5 sin(x + α) = 2。

Note that sin x − 2 cos x = R sin(x − α) uses a minus sign, with tan α = 2 ÷ 1. You must draw the right triangle to determine which quadrant α lies in, based on the signs of a and b.

注意 sin x − 2 cos x = R sin(x − α) 使用减号，其中 tan α = 2 ÷ 1。你必须画出直角三角形，根据 a 和 b 的符号确定 α 所在的象限。

10. Exam Tips and Common Mistakes | 考试技巧与常见错误

Examiner reports show that students lose more marks on trigonometric identities through sign errors and incorrect argument manipulation than through lack of formula knowledge.

考官报告显示，学生在三角恒等式上失分更多是由于符号错误和角度变换不当，而非缺乏公式知识。

Here is a checklist of the most frequent pitfalls and how to avoid them:

以下是最常见陷阱及其规避方法的清单：

Sign errors in compound angles: sin(x − y) = sin x cos y − cos x sin y, not plus. Always test with a numerical value such as x = 90°.
复合角符号错误：sin(x − y) = sin x cos y − cos x sin y，而不是加号。始终用数值如 x = 90° 验证。
Confusing the three forms of cos 2A: check whether you need the forms with cos², sin², or both by looking at what the rest of the expression contains.
混淆 cos 2A 的三种形式：通过观察表达式其余部分包含 cos²、sin² 还是两者来确定需要哪种形式。
Forgetting factorisation: equations like sin x (2 cos x − 1) = 0 must be solved by setting each factor to zero, but never divide both sides by sin x, as this discards solutions.
忘记因式分解：像 sin x (2 cos x − 1) = 0 这样的方程必须将每个因式置零求解，但绝不能两边同除以 sin x，因为这会丢失解。
Overlooking the range: for equations in degrees, always list solutions in the interval 0 ≤ θ < 360°, adding 360°n or 180°n as required before matching to the range.
忽略范围：对于度数方程，务必列出 0 ≤ θ < 360° 区间内的解，先在通解中加上 360°n 或 180°n 再匹配到指定范围。
Unverified proofs: after completing a proof, substitute a test angle such as 30° into both sides to confirm equality. This catches algebraic slips immediately.
未验证的证明：完成证明后，将测试角如 30° 代入两边确认相等，这能立即发现代数错误。

11. Practice Problems | 练习题

Work through these problems in order of increasing difficulty, timing yourself as you would in an examination. For each problem, state which identities you used.

按难度递增的顺序完成以下练习，像考试一样给自己计时。对每个问题，说明你使用了哪些恒等式。

Simplify: (1 − sin² θ) tan² θ.
化简：(1 − sin² θ) tan² θ。
Prove: cos(90° − x) = sin x and sin(90° − x) = cos x, using the compound angle formulas.
证明：使用复合角公式证明 cos(90° − x) = sin x 和 sin(90° − x) = cos x。
Solve, for 0 ≤ x < 360°, the equation 2 cos² x + 3 sin x − 3 = 0.
求解方程 2 cos² x + 3 sin x − 3 = 0，其中 0 ≤ x < 360°。
Express 12 sin x + 5 cos x in the form R sin(x + α), stating R and α to 2 decimal places. Hence find its maximum value and the smallest positive x at which it occurs.
将 12 sin x + 5 cos x 表示为 R sin(x + α) 的形式，R 和 α 精确到两位小数。由此求其最大值及取得最大值的最小正 x。
Find ∫ sin² 3x dx.
求 ∫ sin² 3x dx。
Given that tan A = 3⁄4 and tan B = 5⁄12, both A and B acute, compute tan(A + B) and deduce A + B.
已知 tan A = 3⁄4、tan B = 5⁄12，且 A、B 均为锐角，计算 tan(A + B) 并由此推断 A + B。

Check your answers carefully against the original question each time. For the proof questions, verify numerically at a convenient angle before moving on.

每次都要仔细对照原题检查答案。对于证明题，先在一个方便的角度进行数值验证，再继续下一题。

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A-Level Mathematics: Trigonometric Identities | A-Level 数学：三角恒等式