📚 Trigonometric Identities: Complete A-Level Guide | 三角函数恒等式：A-Level 完整指南
Trigonometric identities are equations involving sine, cosine, tangent and their reciprocal functions that hold for every permitted value of the variable, not merely for a few special solutions. In the Edexcel A-Level Mathematics specification these identities are the working tools used to simplify complicated expressions, to prove new results, to solve trigonometric equations over a stated interval, and to model periodic behaviour such as tides, temperatures and alternating currents.
三角恒等式是包含正弦、余弦、正切及其倒数函数、并且在变量所有允许取值下都成立的等式，而并非只对少数特殊解成立。在 Edexcel A-Level 数学考纲中，这些恒等式是核心工具：用于化简复杂表达式、证明新结论、在给定区间内求解三角方程，以及对潮汐、气温、交流电等周期性现象建模。
e formulae, half-angle (power-reducing) forms, the harmonic form R sin(θ ± α), equation solving, proof technique and modelling. Every result below is examinable and appears repeatedly across Pure Mathematics 1, Pure Mathematics 2 and the Year 2 papers.
本文涵盖毕达哥拉斯恒等式、倒数与商数恒等式、三个毕达哥拉斯形式、和角与二倍角公式、半角（降幂）形式、辅助角形式 R sin(θ ± α)、方程求解、证明技巧以及建模应用。下面每一个结论都属于可考内容，并在 Pure Mathematics 1、Pure Mathematics 2 以及 Year 2 的试卷中反复出现。
1. What Makes an Identity Different | 恒等式与方程的区别
An identity is written with the triple bar symbol ≡ rather than the equals sign =. The statement sin² θ + cos² θ ≡ n equation such as sin θ = 0.5 is only true for particular values of θ, and solving it is a different type of task from proving an identity.
恒等式用三横线符号 ≡ 书写，而不是等号 =。sin² θ + cos² θ ≡ 1 对任何可以代入的角 θ 都成立，包括负角、大于 360° 的角以及用弧度表示的角。相反，像 sin θ = 0.5 这样的方程只对某些特定的 θ 值成立，求解它与证明恒等式是两类完全不同的任务。
Examiners test identities in four distinct ways. First, simplification: replacing 1 − sin² θ by cos² θ to shorten an expression. Second, proof: showing that a complicated left-hand side equals a simple right-hand side. Third, equation solving: using an identity to convert an equation involving several trigonometric functions into one involving a single function. Fourth, modelling: writing a real periodic quantity as a single sine or cosine so that its amplitude, period and maximum value can be read off directly.
考官从四个不同角度考查恒等式。第一，化简：把 1 − sin² θ 换成 cos² θ 以缩短表达式。第二，证明：证明复杂的左边等于简单的右边。第三，解方程：用恒等式把含有多个三角函数的方程转化为只含一个三角函数的方程。第四，建模：把一个真实的周期量写成单个正弦或余弦，从而直接读出振幅、周期和最大值。
One piece of notation causes more lost marks than any other. In sin² θ the superscript belongs to the function value, so sin² θ means (sin θ)². It never means sin(θ²). The same rule applies to cos² θ, tan² θ, sec² θ, cosec² θ and cot² θ.
有一个记号造成的失分比其他任何细节都多。在 sin² θ 中，上标属于函数值，所以 sin² θ 表示 (sin θ)²，绝不表示 sin(θ²)。同样的规则适用于 cos² θ、tan² θ、sec² θ、cosec² θ 和 cot² θ。
2. The Pythagorean Identity from the Unit Circle | 由单位圆得到毕达哥拉斯恒等式
ircle is defined by the equation x² + y² = 1, so every point on it satisfies that relation. Substituting x = cos θ and y = sin θ into the circle equation gives the single most important identity in the whole course.
对于从 x 轴正方向逆时针量出的任意角 θ，单位圆上对应点的坐标为 (cos θ, sin θ)。单位圆由方程 x² + y² = 1 定义，因此圆上每一点都满足该关系。把 x = cos θ 和 y = sin θ 代入圆的方程，就得到整个课程中最重要的一个恒等式。
sin² θ + cos² θ ≡ 1
This identity is valid in degrees and in radians, for positive and negative angles, and for angles of any size whatsoever. It is the reason the sine and cosine graphs never leave the interval [−1, 1]: if either function exceeded 1 in magnitude, the sum of squares would exceed 1.
这个恒等式在角度制和弧度制下都成立，对正角、负角以及任意大小的角都成立。它也正是正弦与余弦图像始终不离开区间 [−1, 1] 的原因：如果其中任一函数的绝对值超过 1，平方和就会大于 1。
The identity also gives an immediate bound that is frequently needed in modelling and in inequality questions, where a quantity such as 3 sin θ + 4 cos θ must be shown to lie between two limits.
该恒等式还立刻给出一个上下界，在建模和不等式问题中经常需要用到，例如要证明像 3 sin θ + 4 cos θ 这样的量介于两个极限之间。
3. Rearranging the Pythagorean Identity | 毕达哥拉斯恒等式的变形
The identity sin² θ + cos² θ ≡ 1 is far more useful when rearranged. Subtracting sin² θ from both sides and subtracting cos² θ from both sides produces two equivalent forms that convert between sine and cosine at will.
把 sin² θ + cos² θ ≡ 1 变形后用途大得多。从两边分别减去 sin² θ 和 cos² θ，就得到两个等价形式，可以在正弦与余弦之间自由转换。
sin² θ ≡ 1 − cos² θ and cos² θ ≡ 1 − sin² θ
pses to 5. In an equation such as 2 sin² θ + cos θ = 2, replacing sin² θ by 1 − cos² θ leaves a quadratic in cos θ alone, which is then solvable by factorising.
这些变形能把混合表达式变成只含一个函数的表达式。例如 3 − 3 sin² θ 变为 3 cos² θ，而 5 cos² θ + 5 sin² θ 直接化简为 5。在 2 sin² θ + cos θ = 2 这样的方程中，用 1 − cos² θ 替换 sin² θ，就只剩下关于 cos θ 的二次式，可通过因式分解求解。
When the target is a proof, look for the number 1 appearing on its own. Any isolated 1 in a trigonometric expression can be replaced by sin² θ + cos² θ, which often creates the common factor or the cancellation that finishes the proof.
在证明题中，要留意单独出现的数字 1。表达式中任何一个孤立的 1 都可以换成 sin² θ + cos² θ，这往往能制造出公因式或消去项，从而完成证明。
4. Reciprocal and Quotient Identities | 倒数与商数恒等式
Edexcel A-Level also requires confident use of the reciprocal trigonometric functions. They are defined from sin θ, cos θ and tan θ as follows, and each one is undefined wherever its denominator is zero.
Edexcel A-Level 还要求熟练使用倒数三角函数。它们由 sin θ、cos θ 和 tan θ 定义如下，并且当分母为零时均无定义。
tan θ ≡ sin θ / cos θ, cot θ ≡ cos θ / sin θ ≡ 1 / tan θ
sec θ ≡ 1 / cos θ, cosec θ ≡ 1 / sin θ
These definitions are not optional extras: they appear in differentiation, in integration, in trigonometric equations and in modelling. Note that Edexcel uses the notation cosec θ rather than csc θ, and that cot θ is the reciprocal of tan θ, not its inverse function.
这些定义不是可有可无的补充：它们会出现在微分、积分、三角方程以及建模中。注意 Edexcel 使用 cosec θ 而不是 csc θ 的写法，并且 cot θ 是 tan θ 的倒数，而不是它的反函数。
- θ and sec θ are undefined when cos θ = 0, that is at θ = 90°, 270°, ... or π/2, 3π/2, ...
- cosec θ are undefined when sin θ = 0, that is at θ = 0°, 180°, 360°, ... or 0, π, 2π, ...
- Since |sin θ| ≤ 1, we have |cosec θ| ≥ 1; since |cos θ| ≤ 1, we have |sec θ| ≥ 1.
- cot θ can take any real value, and so can tan θ.
- 当 cos θ = 0 时，tan θ 与 sec θ 无定义，即 θ = 90°、270°……或 π/2、3π/2……
- 当 sin θ = 0 时，cot θ 与 cosec θ 无定义，即 θ = 0°、180°、360°……或 0、π、2π……
- 由于 |sin θ| ≤ 1，所以 |cosec θ| ≥ 1；由于 |cos θ| ≤ 1，所以 |sec θ| ≥ 1。
- cot θ 可以取任意实数值，tan θ 也是如此。
A useful check when simplifying is to convert everything into sine and cosine first. Almost every reciprocal or quotient expression in an exam simplifies within two lines once it is written purely in terms of sin θ and cos θ.
化简时一个有用的检查方法是先把一切都化为正弦和余弦。考试中几乎每一个倒数或商数表达式，只要改写成完全由 sin θ 和 cos θ 表示，两三行之内就能化简完成。
5. The Three Pythagorean Identities | 三个毕达哥拉斯恒等式
Dividing sin² θ + cos² θ ≡ 1 through by cos² θ produces the second Pythagorean identity, and dividing by sin² θ produces the third. Together the three form a family that examiners expect you to recognise instantly.
把 sin² θ + cos² θ ≡ 1 两边同时除以 cos² θ，得到第二个毕达哥拉斯恒等式；两边除以 sin² θ，得到第三个。三者构成一个整体，考官期望你能一眼识别。
sin² θ + cos² θ ≡ 1
1 + tan² θ ≡ sec² θ
1 + cot² θ ≡ cosec² θ
The second identity is rearranged as sec² θ − tan² θ ≡ 1, which is the standard way to simplify expressions mixing sec² θ and tan² θ before integration. The third is rearranged as cosec² θ − cot² θ ≡ 1, used in the same way for cosec and cot.
第二个恒等式可变形为 sec² θ − tan² θ ≡ 1，这是积分前化简同时含 sec² θ 与 tan² θ 的表达式时的标准做法。第三个可变形为 cosec² θ − cot² θ ≡ 1，处理 cosec 与 cot 时用法相同。
| Identity | 恒等式 | Rearranged forms | 变形形式 |
|---|---|
| sin² θ + cos² θ ≡ 1 | sin² θ ≡ 1 − cos² θ; cos² θ ≡ 1 − sin² θ |
| 1 + tan² θ ≡ sec² θ | tan² θ ≡ sec² θ − 1; sec² θ − tan² θ ≡ 1 |
| 1 + cot² θ ≡ cosec² θ | cot² θ ≡ cosec² θ − 1; cosec² θ − cot² θ ≡ 1 |
ypical exam question asks you to solve 2 tan² θ + sec θ = 1 for 0 ≤ θ ≤ 360°. Using tan² θ ≡ + 3)(sec θ − 1) = 0, giving sec θ = 1 or sec θ = −3/2 and hence a finite set of solutions.
一道典型考题是求 2 tan² θ + sec θ = 1 在 0 ≤ θ ≤ 360° 内的解。利用 tan² θ ≡ sec² θ − 1，方程化为 2 sec² θ + sec θ − 3 = 0，因式分解为 (2 sec θ + 3)(sec θ − 1) = 0，得 sec θ = 1 或 sec θ = −3/2，从而得到有限个解。
6. Compound Angle Formulae | 和角公式
The compound angle formulae express the sine, cosine and tangent of a sum or difference in terms of the individual angles. They are given in the Edexcel formula booklet, but you must be able to use them fluently and in both directions, because they are also the derivation route to the double angle formulae.
和角公式用两个单独的角来表示它们和或差的正弦、余弦与正切。这些公式在 Edexcel 公式册中给出，但你必须能熟练双向使用，因为它们也是推导二倍角公式的途径。
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 formula and in the denominator of the tangent formula: where the left-hand side has a plus sign, the right-hand side has a minus sign on that particular term. Writing the wrong sign is one of the most common errors in the whole topic.
注意余弦公式以及正切公式分母中的符号反转：左边是加号的位置，右边对应项就是减号。写错符号是本专题最常见的错误之一。
standard. For example, 15° = 45° − 30°, so sin 15° = sin 45° cos 30° − cos 45° sin 30° = (√6 − √2) / 4 ≈ 0.259. Similarly 75° = 45° + 30° gives sin 75° = (√6 + √2) / 4.
这些公式可以给出非特殊角的精确值。例如 15° = 45° − 30°，所以 sin 15° = sin 45° cos 30° − cos 45° sin 30° = (√6 − √2) / 4 ≈ 0.259。同理 75° = 45° + 30°，得 sin 75° = (√6 + √2) / 4。
mplicated expression becomes a single sine term and can be integrated or solved almost immediately.
在 Year 2 中，同样的公式会反向使用来合并表达式。组合 sin 3x cos x + cos 3x sin x 恰好等于 sin(3x + x) = sin 4x，于是一个复杂表达式变成单个正弦项，几乎可以立即积分或求解。
7. Double Angle Formulae | 二倍角公式
mulae produces the double angle formulae. Cosine has three equally valid versions, and choosing the right one is the key skill in this section.
在合角公式中令 B = A，就得到二倍角公式。余弦有三个同样有效的版本，选对版本是本节的关键技能。
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 version of cos 2A you choose depends on what else is in the expression. If the rest of the question involves only sine, use cos 2A ≡ 1 − 2 sin² A. If it involves only cosine, use cos 2A ≡ 2 cos² A − 1. If it involves both, use cos 2A ≡ cos² A − sin² A.
选择哪个 cos 2A 版本取决于表达式中还有什么。如果题目其余部分只涉及正弦，就用 cos 2A ≡ 1 − 2 sin² A；如果只涉及余弦，就用 cos 2A ≡ 2 cos² A − 1；如果两者都有，就用 cos 2A ≡ cos² A − sin² A。
ving sin 2θ = sin θ, on the other hand, uses 2 sin θ cos θ = sin θ, which rearranges to sin θ (2 cos θ − 1) = 0. Never divide both sides by sin θ, because that destroys the solutions sin θ = 0.
例如，求 3 cos 2θ + 5 cos θ + 1 = 0 在 0 ≤ θ ≤ 360° 内的解时，应选 2 cos² θ − 1 这个版本，得到 6 cos² θ + 5 cos θ − 2 = 0，进而 (3 cos θ − 1)(2 cos θ + 2) = 0。而解 sin 2θ = sin θ 时则用 2 sin θ cos θ = sin θ，整理为 sin θ (2 cos θ − 1) = 0。切记不要两边同除 sin θ，那会丢掉 sin θ = 0 的解。
8. Half-Angle and Power-Reducing Forms | 半角与降幂形式
Rearranging the second and third versions of the cosine double angle formula gives the half-angle forms, also called power-reducing forms because they replace a squared trigonometric function by a first-power expression.
把余弦二倍角公式的第二、第三个版本变形，就得到半角形式，也叫降幂形式，因为它们把三角函数的平方换成一次表达式。
sin² A ≡ (1 − cos 2A) / 2
cos² A ≡ (1 + cos 2A) / 2
tan² A ≡ (1 − cos 2A) / (1 + cos 2A)
These forms matter mainly for integration. A term such as sin² x cannot be integrated directly with the standard results, but replacing it with (1 − cos 2x) / 2 turns it into two elementary terms. The same trick handles cos² x and combinations such as sin² x cos² x.
这些形式主要用于积分。像 sin² x 这样的项无法直接用标准结果积分，但把它换成 (1 − cos 2x) / 2 后就变成两个基本项。同样的技巧适用于 cos² x 以及 sin² x cos² x 这类组合。
In an equation such as 4 sin² θ = 3, either take square roots immediately to get sin θ = ±√3 solve for 2θ. Both routes are accepted; the first is usually faster in degrees questions.
在需要消去平方项的方程中它们同样有用。在 4 sin² θ = 3 这类方程中，既可以直接开方得 sin θ = ±√3/2，也可以用半角形式得到 2(1 − cos 2θ) = 3 再解 2θ。两种途径都被接受，在角度制问题中前者通常更快。
9. Solving Trigonometric Equations with Identities | 用恒等式求解三角方程
Most A-Level trigonometric equations of this type follow the same three-step pattern: replace every term with a single trigonometric function using the identities, form a quadratic or linear equation in that function, then solve within the given interval.
这类 A-Level 三角方程大多遵循同一个三步模式：用恒等式把每一项都换成同一个三角函数，得到关于该函数的二次或一次方程，然后在给定区间内求解。
s θ + 1)(cos θ + 1) = 0. Hence cos θ = −1/2, giving θ = 120° and 240°, or cos θ = −1, giving θ = 180°. The full solution set is θ = 120°, 180°, 240°.
例题：求 2 sin² θ = 3 cos θ + 3 在 0 ≤ θ ≤ 360° 内的解。用 1 − cos² θ 替换
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图