Compound Angle Formulas: The Complete Guide to A-Level Trigonometry — A-Level数学：和角公式在解题中的应用

📚 Compound Angle Formulas: The Complete Guide to A-Level Trigonometry — A-Level数学：和角公式在解题中的应用

和角公式（Compound Angle Formulas）是A-Level数学三角函数章节的核心内容，也是考试中出现频率最高的考点之一。无论是求解三角方程、证明恒等式，还是处理函数的最值与积分问题，和角公式都扮演着不可替代的角色。本篇文章将系统梳理和角公式的全部要点，从公式本身、推导方法、记忆技巧，到各类题型的解题框架，帮助你在考试中熟练运用。

Compound angle formulas are the heart of the trigonometry chapter in A-Level Mathematics and one of the most frequently tested topics in examinations. Whether you are solving trigonometric equations, proving identities, or tackling maximum and minimum problems and integration, compound angle formulas play an indispensable role. This article systematically reviews every key point: the formulas themselves, their derivations, memory techniques, and step-by-step frameworks for each question type, so you can use them with confidence in the exam.

一、和角公式是什么？三大基本公式全览 | What Are Compound Angle Formulas? The Three Core Identities

所谓和角公式，就是描述两个角之和或之差的正弦、余弦、正切如何用这两个角各自的正弦、余弦、正切来表示的一组恒等式。在A-Level考试中，你需要掌握以下三个最基本的形式，它们构成了整个三角函数公式体系的基石。

Compound angle formulas are a set of identities that express the sine, cosine and tangent of the sum or difference of two angles in terms of the trigonometric functions of the individual angles. In the A-Level exam, you need to master the three most basic forms below, which form the foundation of the entire trigonometric formula system.

公式 Formula	展开形式 Expanded Form	中文名称 Chinese Name
sin(A + B)	sinA cosB + cosA sinB	正弦和角公式
sin(A – B)	sinA cosB – cosA sinB	正弦差角公式
cos(A + B)	cosA cosB – sinA sinB	余弦和角公式
cos(A – B)	cosA cosB + sinA sinB	余弦差角公式
tan(A + B)	(tanA + tanB) / (1 – tanA tanB)	正切和角公式
tan(A – B)	(tanA – tanB) / (1 + tanA tanB)	正切差角公式

注意观察正弦与余弦公式中符号的规律：正弦公式中，sin(A + B)展开后中间是加号，sin(A – B)展开后中间是减号，符号与括号内的符号保持一致；而余弦公式恰好相反，cos(A + B)展开后中间是减号，cos(A – B)展开后中间是加号。这个”正弦同号、余弦异号”的规律是记忆公式的关键。

Pay close attention to the sign patterns in the sine and cosine formulas: for sine, sin(A + B) expands with a plus sign in the middle and sin(A – B) with a minus sign, matching the sign inside the brackets; for cosine, the pattern is reversed, with cos(A + B) expanding to a minus sign and cos(A – B) to a plus sign. This rule of “sine keeps the sign, cosine flips it” is the key to memorising the formulas.

二、公式从哪来？从余弦差角公式出发的完整推导 | Where Do the Formulas Come From? Deriving Everything from cos(A—B)

很多同学觉得和角公式是一堆需要死记硬背的结论，其实整个公式族可以从一个最基本的公式出发逐步推导出来。在Edexcel的A-Level教材中，余弦差角公式 cos(A – B) = cosA cosB + sinA sinB 是通过单位圆上的坐标几何方法证明的，一旦你理解了它的来历，其余所有公式都可以顺理成章地推出。

Many students treat compound angle formulas as a pile of results to memorise by rote, but the whole family can actually be derived step by step from a single basic formula. In the Edexcel A-Level textbooks, cos(A – B) = cosA cosB + sinA sinB is proved using coordinate geometry on the unit circle. Once you understand where it comes from, all the other formulas follow naturally.

推导思路如下：在单位圆上取两点P和Q，点P对应的角度为A，点Q对应的角度为B，那么两点之间的夹角为(A – B)。利用两点间距离公式计算线段PQ的长度，同时利用余弦定理计算同一个长度，将两个表达式相等，经过整理即可得到余弦差角公式。这个证明方法体现了坐标几何与三角学的完美结合，也是考试中常见的推导证明题。

Here is the idea: on the unit circle, take two points P and Q, where P corresponds to angle A and Q to angle B, so the angle between them is (A – B). Compute the length of segment PQ using the distance formula, then compute the same length using the cosine rule; equating the two expressions and simplifying yields the cosine difference formula. This proof beautifully combines coordinate geometry with trigonometry and is a common derivation question in exams.

得到cos(A – B)之后，其余公式的推导路径为：第一步，用(-B)替换B得到cos(A + B) = cosA cosB – sinA sinB；第二步，利用余角关系sinθ = cos(90° – θ)将cos(A + B)变形，得到sin(A – B)和sin(A + B)的公式；第三步，利用tanθ = sinθ / cosθ，将正弦公式除以余弦公式，得到正切的和差公式。整个推导链条清晰完整，建议你亲手推导一遍，这比单纯背诵有效得多。

Once cos(A – B) is established, the derivation path for the rest is: first, replace B with (-B) to get cos(A + B) = cosA cosB – sinA sinB; second, use the complementary-angle relation sinθ = cos(90° – θ) to transform cos(A + B), yielding the formulas for sin(A – B) and sin(A + B); third, use tanθ = sinθ / cosθ, dividing the sine formula by the cosine formula, to obtain the tangent sum and difference formulas. The whole chain is clear and complete; I strongly recommend deriving it by hand once, which is far more effective than rote memorisation.

三、倍角公式：和角公式的直接推论 | Double Angle Formulas: Direct Consequences of Compound Angles

在A-Level考试中，倍角公式的出现频率甚至高于和角公式本身。所谓倍角公式，就是在和角公式中令B = A，得到的关于2A的表达式。倍角公式不是一套需要单独记忆的新公式，而是和角公式的特例，理解这一点能大大减轻你的记忆负担。

In the A-Level exam, double angle formulas appear even more frequently than the compound angle formulas themselves. The double angle formulas are obtained by setting B = A in the compound angle formulas, giving expressions involving 2A. They are not a separate set of formulas to memorise but special cases of the compound angle formulas; understanding this greatly reduces your memory load.

令B = A，由sin(A + B)得到sin2A = 2sinA cosA；由cos(A + B)得到cos2A = cos²A – sin²A；由tan(A + B)得到tan2A = 2tanA / (1 – tan²A)。其中cos2A的公式特别重要，因为它有三种等价形式：cos2A = cos²A – sin²A = 2cos²A – 1 = 1 – 2sin²A。后两种形式是通过sin²A + cos²A = 1这个基本恒等式变形得到的。

Setting B = A, we get sin2A = 2sinA cosA from sin(A + B), cos2A = cos²A – sin²A from cos(A + B), and tan2A = 2tanA / (1 – tan²A) from tan(A + B). The cos2A formula is especially important because it has three equivalent forms: cos2A = cos²A – sin²A = 2cos²A – 1 = 1 – 2sin²A. The latter two forms are obtained by rearranging the fundamental identity sin²A + cos²A = 1.

考试中最常见的考法之一，是给出cos2A的值（例如cos2A = 1/3），要求求出sinA或cosA的值。此时你需要根据2cos²A – 1 = cos2A或1 – 2sin²A = cos2A这两个形式直接解出cos²A或sin²A，再根据A所在的象限确定正负号。这类题目考查的是公式的逆向运用能力，需要多加练习。

One of the most common exam questions gives the value of cos2A (for example cos2A = 1/3) and asks you to find sinA or cosA. Here you use the forms 2cos²A – 1 = cos2A or 1 – 2sin²A = cos2A to solve directly for cos²A or sin²A, then determine the sign from the quadrant in which A lies. These questions test your ability to use the formulas in reverse and require plenty of practice.

四、降幂公式：处理sin²x与cos²x的利器 | Power-Reduction Identities: Handling sin²x and cos²x

把倍角公式中的cos2A = 2cos²A – 1和cos2A = 1 – 2sin²A稍作移项，就得到了降幂公式（Power-Reduction Identities）：cos²A = (1 + cos2A) / 2，sin²A = (1 – cos2A) / 2。这两个公式的核心用途是把二次的三角函数降为一次，从而把难以直接积分的表达式转化为可以逐项积分的形式。

Rearranging the double angle formulas cos2A = 2cos²A – 1 and cos2A = 1 – 2sin²A gives the power-reduction identities: cos²A = (1 + cos2A) / 2 and sin²A = (1 – cos2A) / 2. Their core purpose is to reduce a squared trigonometric function to first power, converting expressions that are hard to integrate directly into forms that can be integrated term by term.

积分场景是降幂公式最典型的应用。例如计算 ∫ sin²x dx，直接积分没有现成的公式，但利用降幂公式改写为 ∫ (1 – cos2x)/2 dx = x/2 – sin2x/4 + C，就可以顺利求解。同理，∫ cos²x dx = x/2 + sin2x/4 + C。在Edexcel A-Level数学Paper 2和Paper 3中，这类积分题几乎每年都会出现。

Integration is the most typical application of the power-reduction identities. For example, to compute ∫ sin²x dx, there is no ready-made integration formula, but rewriting it via the power-reduction identity gives ∫ (1 – cos2x)/2 dx = x/2 – sin2x/4 + C, which can be solved smoothly. Similarly, ∫ cos²x dx = x/2 + sin2x/4 + C. In Edexcel A-Level Mathematics Paper 2 and Paper 3, such integration questions appear almost every year.

除了积分，降幂公式还常用于证明恒等式和化简表达式。当你看到一个式子里同时出现sin²x和cos²x，或者sin²x与sin2x混在一起时，第一反应就应该是尝试用降幂公式把所有二次项统一成cos2x的形式，这样往往能让表达式结构变得清晰，为后续的因式分解或合并同类项创造便利条件。

Besides integration, the power-reduction identities are also commonly used to prove identities and simplify expressions. When you see sin²x and cos²x together in one expression, or sin²x mixed with sin2x, your first instinct should be to use the power-reduction identities to unify all squared terms into cos2x form. This often clarifies the structure of the expression and paves the way for factorisation or collecting like terms.

五、辅助角公式（R公式）：把asinθ + bcosθ化成一个正弦 | The R-Formula: Rewriting asinθ + bcosθ as a Single Sine

辅助角公式，通常称为R公式，是和角公式最重要的应用之一。它的内容是：任意形如a sinθ + b cosθ的表达式都可以改写为R sin(θ + α)的形式，其中R = √(a² + b²)，α由tanα = b/a确定，具体取值取决于a、b的符号所在的象限。

The auxiliary angle formula, usually called the R-formula, is one of the most important applications of compound angle formulas. It states that any expression of the form a sinθ + b cosθ can be rewritten as R sin(θ + α), where R = √(a² + b²) and α is determined by tanα = b/a, with its exact value depending on the quadrant determined by the signs of a and b.

这个公式的推导其实就是在反用sin(A + B) = sinA cosB + cosA sinB。把R sin(θ + α)展开得到R sinθ cosα + R cosθ sinα，令它等于a sinθ + b cosθ，比较系数得到R cosα = a，R sinα = b，两式平方相加得到R² = a² + b²，两式相除得到tanα = b/a。整个推导一气呵成，也解释了为什么R公式又叫”合一变形”。

The derivation is simply the reverse use of sin(A + B) = sinA cosB + cosA sinB. Expanding R sin(θ + α) gives R sinθ cosα + R cosθ sinα; setting this equal to a sinθ + b cosθ and comparing coefficients gives R cosα = a and R sinα = b. Squaring and adding the two equations yields R² = a² + b², while dividing them gives tanα = b/a. The derivation is smooth and also explains why the R-formula is called “combining into one”.

R公式最大的价值在于：它把一个看似复杂的二元三角函数表达式压缩成单个正弦函数，从而可以直接讨论其最值、周期、零点，甚至画出它的图像。例如函数y = 3sinx + 4cosx可以改写为y = 5sin(x + α)，其中tanα = 4/3，于是最大值显然是5，最小值是-5，周期仍为360°。这类题目在考试中属于高频考点。

The greatest value of the R-formula is that it compresses a seemingly complicated two-term trigonometric expression into a single sine function, allowing you to discuss its maximum, minimum, period, zeros, and even sketch its graph directly. For example, y = 3sinx + 4cosx can be rewritten as y = 5sin(x + α) with tanα = 4/3, so the maximum is clearly 5, the minimum is -5, and the period remains 360°. Such questions are high-frequency exam items.

六、解三角方程：用和角公式把方程化到最简 | Solving Trigonometric Equations: Simplifying with Compound Angles

解三角方程是A-Level考试的基础题型，而与和角公式结合的方程题则是中高难度的区分题。典型的情况有两种：一种是方程中含有sin(2x + 30°)这样的复合角表达式，另一种是方程中含有sin2x、cos2x这样的倍角形式，需要先化简再求解。

Solving trigonometric equations is a basic question type in A-Level exams, and equations combined with compound angle formulas are the medium-to-high difficulty differentiators. Two typical cases exist: equations containing compound angle expressions like sin(2x + 30°), and equations containing double angle forms like sin2x or cos2x that must be simplified before solving.

处理第一类方程时，把(2x + 30°)看成一个整体变量t，先求出t在给定区间内的所有取值，再解出x。以方程sin(2x + 30°) = 1/2在0° ≤ x ≤ 180°内为例：令t = 2x + 30°，则30° ≤ t ≤ 390°，在这个区间内sin t = 1/2的解为t = 30°, 150°, 390°，于是2x + 30°分别等于这三个值，解得x = 0°, 60°, 180°。注意一定要先扩大变量的范围，否则会漏解。

For the first type, treat (2x + 30°) as a single variable t, first find all values of t within the given interval, then solve for x. Take sin(2x + 30°) = 1/2 for 0° ≤ x ≤ 180° as an example: let t = 2x + 30°, so 30° ≤ t ≤ 390°. Within this interval the solutions of sin t = 1/2 are t = 30°, 150°, 390°, so 2x + 30° equals each of these in turn, giving x = 0°, 60°, 180°. Always expand the range of the new variable first, otherwise you will miss solutions.

处理第二类方程时，关键是识别出可以统一的角度形式。例如方程sin2x = cosx，左边是倍角，右边是一次角，直接比较无从下手。正确的做法是把sin2x展开为2sinx cosx，得到2sinx cosx = cosx，移项并因式分解为cosx(2sinx – 1) = 0，于是cosx = 0或sinx = 1/2，分别求解后合并即可。因式分解是这类题的核心技巧。

For the second type, the key is to recognise which angle form can be unified. Take sin2x = cosx: the left side is a double angle while the right side is a single angle, so direct comparison gets nowhere. The correct approach is to expand sin2x as 2sinx cosx, giving 2sinx cosx = cosx; then rearrange and factorise as cosx(2sinx – 1) = 0, so cosx = 0 or sinx = 1/2, solved separately and combined. Factorisation is the core technique for this type.

七、恒等式证明：从左边到右边的系统方法 | Proving Identities: A Systematic Left-to-Right Approach

恒等式证明题要求学生证明等式两边对定义域内所有取值都成立。这类题没有固定的套路，但有非常有效的通用策略：通常从结构更复杂的一边出发，利用和角公式、倍角公式将其化简，逐步向结构更简单的一边靠拢，最终两边完全一致。

Identity proof questions require you to show that the two sides of an equation hold for all values in the domain. There is no fixed formula for these problems, but there is a highly effective general strategy: start from the more complicated side, use compound and double angle formulas to simplify it step by step, gradually moving toward the simpler side until the two sides match exactly.

一个经典例子是证明cos(60° – x) = (√3 cosx + sinx) / 2。左边用余弦差角公式展开：cos(60° – x) = cos60° cosx + sin60° sinx = (1/2)cosx + (√3/2)sinx。把两项通分合并，就得到(√3 sinx + cosx) / 2，即(√3 cosx + sinx) / 2，与右边完全一致，证明完成。整个过程中关键是熟记特殊角的三角函数值。

A classic example is proving cos(60° – x) = (√3 cosx + sinx) / 2. Expand the left side using the cosine difference formula: cos(60° – x) = cos60° cosx + sin60° sinx = (1/2)cosx + (√3/2)sinx. Combining the two terms over a common denominator gives (√3 sinx + cosx) / 2, which is (√3 cosx + sinx) / 2, exactly matching the right side, completing the proof. The key throughout is remembering the exact trigonometric values of special angles.

另一个常见技巧是”1的妙用”，即把常数1替换为sin²x + cos²x。例如证明恒等式sin2x / (1 + cos2x) = tanx时，分子sin2x展开为2sinx cosx，分母1 + cos2x用2cos²x替换，整个分式变为2sinx cosx / (2cos²x) = sinx / cosx = tanx，一步到位。这类题目考查的是你对公式各种等价形式的熟悉程度。

Another common trick is the “clever use of 1”, replacing the constant 1 with sin²x + cos²x. For example, to prove sin2x / (1 + cos2x) = tanx, expand the numerator as 2sinx cosx and replace the denominator 1 + cos2x with 2cos²x; the whole fraction becomes 2sinx cosx / (2cos²x) = sinx / cosx = tanx in one step. These questions test how familiar you are with the various equivalent forms of the formulas.

八、最值与值域：R公式在函数分析中的应用 | Maxima and Minima: Applying the R-Formula to Function Analysis

求三角函数的最值和值域是A-Level考试的另一类高频题目。当函数形如y = a sinθ + b cosθ时，直接讨论最值比较困难，但利用R公式改写为y = R sin(θ + α)之后，由于正弦函数的值域是[-1, 1]，函数的最值一目了然：最大值R，最小值-R。

Finding the maximum, minimum and range of trigonometric functions is another high-frequency question type in A-Level exams. When a function has the form y = a sinθ + b cosθ, discussing its extrema directly is difficult, but after rewriting it as y = R sin(θ + α) via the R-formula, the range of sine is [-1, 1], so the extrema are immediately clear: maximum R, minimum -R.

如果题目进一步要求”求取得最大值时θ的取值”，那么令sin(θ + α) = 1，即θ + α = 90° + 360°k，解出θ即可。例如y = 5sin(x + α)的最大值为5，当x + α = 90°，即x = 90° – α时取得。需要注意的是，若题目给定的区间限制x的范围，则要检查解出的x是否落在区间内。

If the question further asks “find the value of θ at which the maximum occurs”, set sin(θ + α) = 1, i.e. θ + α = 90° + 360°k, and solve for θ. For example, y = 5sin(x + α) has maximum 5, attained when x + α = 90°, i.e. x = 90° – α. Note that if the question restricts x to a given interval, you must check whether the solutions fall inside that interval.

还有一类衍生题型：把R公式与恒等式结合，求形如y = sinx + √3 cosx + 2的函数最值。先把前两项合并为2sin(x + 60°)，整个函数变为y = 2sin(x + 60°) + 2，于是最大值4，最小值0。这类题考查的是”先合一、后平移”的两步思路，步骤清晰、不易出错。

There is also a derived question type that combines the R-formula with identity work, such as finding the extrema of y = sinx + √3 cosx + 2. First combine the first two terms into 2sin(x + 60°), so the whole function becomes y = 2sin(x + 60°) + 2, giving maximum 4 and minimum 0. These questions test the two-step idea of “combine first, then translate”, which is clear and hard to get wrong.

九、微积分中的和角公式：求导与积分 | Compound Angles in Calculus: Differentiation and Integration

和角公式在微积分中的应用主要体现在三个方面：对复合角三角函数的求导、对二次三角函数表达式的积分、以及对有理分式形式的三角函数的处理。掌握这些应用，能让你的Paper 2和Paper 3得分能力大幅提升。

The applications of compound angle formulas in calculus mainly appear in three areas: differentiating compound-angle trigonometric functions, integrating squared trigonometric expressions, and handling trigonometric expressions in rational form. Mastering these applications significantly boosts your score potential in Paper 2 and Paper 3.

求导方面，链式法则与和角公式经常一起出现。例如y = sin(2x + 1)的导数为2cos(2x + 1)，这里的”2″来自对(2x + 1)求导的内层导数。更复杂的例子如y = sin²x，可以先用倍角公式改写为y = (1 – cos2x)/2，再求导得到dy/dx = sin2x；也可以直接用链式法则：dy/dx = 2sinx cosx = sin2x，两种方法结果一致，可以互相验证。

For differentiation, the chain rule and compound angle formulas often appear together. For example, the derivative of y = sin(2x + 1) is 2cos(2x + 1), where the “2” comes from differentiating the inner function (2x + 1). For a more complex example like y = sin²x, you can either rewrite it as y = (1 – cos2x)/2 using the double angle formula and differentiate to get dy/dx = sin2x, or use the chain rule directly: dy/dx = 2sinx cosx = sin2x. Both methods agree, so they can be used to check each other.

积分方面，除了前面提到的降幂公式处理sin²x和cos²x之外，还有一种常见题型是积分∫ sinx cosx dx。此时把被积函数改写为(1/2)sin2x，积分结果为-(1/4)cos2x + C。另外，遇到∫ sin(ax + b) dx这类复合角积分，直接利用换元法或公式∫ sin(ax + b) dx = -(1/a)cos(ax + b) + C即可，注意不要漏掉系数1/a。

For integration, besides using the power-reduction identities on sin²x and cos²x mentioned earlier, a common question type is ∫ sinx cosx dx. Rewrite the integrand as (1/2)sin2x, giving the result -(1/4)cos2x + C. Also, for compound-angle integrals like ∫ sin(ax + b) dx, use substitution or the standard formula ∫ sin(ax + b) dx = -(1/a)cos(ax + b) + C directly, taking care not to forget the factor 1/a.

十、常见易错点：符号、象限与定义域 | Common Pitfalls: Signs, Quadrants and Domains

和角公式相关的错误通常集中在几个固定的地方。第一个易错点是符号问题：很多同学在展开cos(A + B)时误写成cosA cosB + sinA sinB，或者在展开sin(A – B)时把中间的减号写错。记住”正弦同号、余弦异号”的口诀可以有效避免这类错误。

Errors related to compound angle formulas tend to cluster in a few fixed places. The first pitfall is signs: many students mistakenly expand cos(A + B) as cosA cosB + sinA sinB, or get the minus sign wrong when expanding sin(A – B). Remembering the mantra “sine keeps the sign, cosine flips it” effectively prevents this type of error.

第二个易错点是象限判断。在R公式中确定α时，仅仅知道tanα = b/a是不够的，因为正切函数在第二象限和第四象限都是负的，在第三象限和第一象限都是正的。你必须结合a和b的具体符号来判断α所在的象限。例如a < 0、b > 0时，α应在第二象限。这是R公式题失分的重灾区。

The second pitfall is quadrant determination. When determining α in the R-formula, knowing only tanα = b/a is not enough, because tangent is negative in the second and fourth quadrants and positive in the first and third. You must combine the actual signs of a and b to determine the quadrant of α. For example, when a < 0 and b > 0, α lies in the second quadrant. This is a major source of lost marks in R-formula questions.

第三个易错点是解方程时的漏解。当方程中含有2x、3x这样的倍角变量时，变量范围会相应扩大为原来的2倍、3倍，很多同学忘记扩大范围，导致丢解。正确的做法是先写出新变量t = 2x + 30°的完整取值范围，在扩大后的范围内求所有解，再逐一解出x。宁可多写几步，也不要漏掉任何解。

The third pitfall is missing solutions when solving equations. When an equation contains a multiple angle like 2x or 3x, the variable range expands correspondingly to 2 or 3 times the original; many students forget to expand the range and therefore lose solutions. The correct procedure is to write out the full range of the new variable t = 2x + 30°, find all solutions within the expanded range, then solve for x one by one. Better to write a few extra steps than to miss any solution.

十一、典型真题演练：三步解题框架 | Worked Examples: A Three-Step Framework

下面用一个完整的真题风格题目演示和角公式的综合运用。例题：已知函数f(x) = 4sinx – 3cosx。(a) 将f(x)改写为R sin(x – α)的形式，其中R > 0且0° < α < 90°；(b) 求f(x)的最大值和最小值；(c) 解方程f(x) = 2，其中0° ≤ x ≤ 360°。

Here is a complete exam-style question demonstrating the integrated use of compound angle formulas. Example: Given f(x) = 4sinx – 3cosx. (a) Rewrite f(x) in the form R sin(x – α) where R > 0 and 0° < α < 90°; (b) Find the maximum and minimum values of f(x); (c) Solve f(x) = 2 for 0° ≤ x ≤ 360°.

第一步，确定R和α。由R² = 4² + (-3)² = 25得R = 5。因为我们要写成R sin(x – α) = R sinx cosα – R cosx sinα，比较系数得R cosα = 4，R sinα = 3，所以tanα = 3/4，α = 36.87°（用计算器求得，保留到小数点后两位）。于是f(x) = 5sin(x – 36.87°)。

Step one, determine R and α. From R² = 4² + (-3)² = 25 we get R = 5. Since we want R sin(x – α) = R sinx cosα – R cosx sinα, comparing coefficients gives R cosα = 4 and R sinα = 3, so tanα = 3/4 and α = 36.87° (from the calculator, correct to two decimal places). Hence f(x) = 5sin(x – 36.87°).

第二步，求最值。因为-1 ≤ sin(x – 36.87°) ≤ 1，所以f(x)的最大值为5，最小值为-5。这一步几乎不需要计算，R公式的价值就在于此。

Step two, find the extrema. Since -1 ≤ sin(x – 36.87°) ≤ 1, the maximum of f(x) is 5 and the minimum is -5. This step requires almost no computation; that is exactly the value of the R-formula.

第三步，解方程。由f(x) = 2得sin(x – 36.87°) = 2/5。令t = x – 36.87°，则-36.87° ≤ t ≤ 323.13°。在计算器上求得sin t = 2/5的主解为t = 23.58°，另一个解为t = 180° – 23.58° = 156.42°，均在范围内。于是x = t + 36.87°，得到x = 60.45°和x = 193.29°。注意所有角度均保留两位小数，并检查每个解都在给定区间内。

Step three, solve the equation. From f(x) = 2 we get sin(x – 36.87°) = 2/5. Let t = x – 36.87°, so -36.87° ≤ t ≤ 323.13°. The calculator gives the principal solution of sin t = 2/5 as t = 23.58°, and the other solution is t = 180° – 23.58° = 156.42°, both within range. Hence x = t + 36.87°, giving x = 60.45° and x = 193.29°. Keep all angles to two decimal places and check that every solution lies in the given interval.

十二、复习清单与考试策略 | Revision Checklist and Exam Strategy

临近考试时，建议按以下清单进行系统复习。第一，能默写六个和角公式并能从cos(A – B)推导全部公式；第二，熟练运用三个倍角公式，特别是cos2A的三种等价形式及其与降幂公式的相互转化；第三，掌握R公式的完整流程，包括象限判断；第四，熟悉解三角方程的换元与因式分解两种核心方法。

As the exam approaches, revise systematically using this checklist. First, be able to write out all six compound angle formulas from memory and derive them all from cos(A – B); second, use the three double angle formulas fluently, especially the three equivalent forms of cos2A and their conversion to and from the power-reduction identities; third, master the full R-formula procedure including quadrant determination; fourth, be familiar with the two core methods for solving trigonometric equations: substitution and factorisation.

考试策略方面，注意以下三点。第一，和角公式的题目往往分值较高且步骤较多，务必写出完整的中间步骤，即使最后结果算错，步骤分也能挽回大部分分数；第二，使用计算器求解α或方程解时，注意角度模式（度数或弧度）要与题目一致，Edexcel题目中带°符号的用度数模式，否则用弧度模式；第三，解完方程后一定要把答案代回原方程验证，这能拦截绝大多数粗心错误。

For exam strategy, note the following three points. First, compound angle questions tend to carry high marks and require many steps; always write out the complete intermediate steps, because even if the final answer is wrong, method marks will recover most of the credit. Second, when using the calculator to find α or solve equations, make sure the angle mode (degrees or radians) matches the question: use degree mode when the question contains the ° symbol, otherwise radian mode. Third, always substitute your answers back into the original equation to verify; this catches the vast majority of careless errors.

最后，把和角公式放进你的”公式网络”中理解：它与倍角公式、降幂公式、R公式、积化和差公式（Further Maths内容）共同构成一个有机整体。理解了公式之间的推导关系，你就再也不需要死记硬背，考试中遇到任何变形都能从容应对。

Finally, understand compound angle formulas as part of your “formula network”: they form an organic whole together with the double angle formulas, power-reduction identities, the R-formula, and the product-to-sum formulas (Further Maths content). Once you understand the derivation relationships between the formulas, you will never need to memorise by rote again, and you will handle any variation calmly in the exam.

Summary | 总结

本文系统讲解了A-Level数学和角公式的全部核心内容：六大和角公式及其符号规律、从cos(A – B)出发的完整推导、三个倍角公式与cos2A的三种形式、降幂公式在积分中的应用、R公式的推导与最值应用、解三角方程的换元与因式分解方法、恒等式证明的通用策略，以及求导积分、常见易错点和完整的三步解题框架。掌握这些内容，你就拿到了三角函数高分题的关键钥匙。

This article has systematically explained everything about compound angle formulas in A-Level Mathematics: the six compound angle formulas and their sign patterns, the complete derivation starting from cos(A – B), the three double angle formulas and the three forms of cos2A, the power-reduction identities in integration, the derivation of the R-formula and its application to extrema, the substitution and factorisation methods for solving trigonometric equations, general strategies for proving identities, differentiation and integration, common pitfalls, and a complete three-step solving framework. Master these and you hold the key to high-scoring trigonometry questions.

建议你把本文中的表格和例题整理到自己的笔记中，并额外练习至少十道相关真题，特别是近五年的Edexcel Paper 1和Paper 2真题。三角函数部分的题目规律性强，练得越多，考场上就越有把握。祝你在A-Level数学考试中取得理想成绩。

We recommend copying the tables and worked examples in this article into your own notes and practising at least ten more related past-paper questions, especially Edexcel Paper 1 and Paper 2 questions from the last five years. Trigonometry questions follow strong patterns; the more you practise, the more confident you will be in the exam hall. Wishing you excellent results in your A-Level Mathematics examination.

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