📚 Mastering Trigonometric Differentiation for Edexcel A-Level Pure Maths | 掌握Edexcel A-Level纯数学三角函数求导
Trigonometric differentiation runs through the whole of the Edexcel A-Level Pure Mathematics specification, and it appears most heavily in Paper 1 and Paper 2. You need to differentiate sin x, cos x and tan x fluently, then combine those results with the chain rule, the product rule and the quotient rule. This guide sets out the standard results, the derivation of tan x, a full set of worked examples, the integration links you are expected to recognise, and the specific traps that cost marks in the exam hall.
三角函数求导贯穿 Edexcel A-Level 纯数学大纲，在 Paper 1 与 Paper 2 中出现频率最高。你需要熟练求出 sin x、cos x、tan x 的导数，并把这些结果与链式法则、乘法法则、除法法则结合使用。本文系统整理基本结论、tan x 的推导过程、成套例题、必须能识别的积分联系，以及考场上最容易丢分的具体陷阱。
1. Why Trigonometric Differentiation Matters | 为什么三角函数求导重要
In the Edexcel specification, differentiating trigonometric functions is a core Pure Mathematics skill rather than a stand-alone topic. It feeds directly into questions on stationary points, equations of tangents and normals, rates of change, curve sketching and integration. A single weak link here can therefore pull down your performance across several unrelated-looking questions in the same paper.
在 Edexcel 大纲中，三角函数求导不是孤立知识点，而是纯数学的核心技能。它直接服务于驻点、切线与法线方程、变化率、曲线作图以及积分等题型。因此这一环节只要薄弱，就会拖累同一份试卷中好几个看似无关的题目。
Most exam questions are not simply "differentiate sin x". They present composite arguments such as sin(3x + 1), products such as x² cos x, or quotients such as sin x / x, and they expect you to select the right rule instantly. The examiner is testing structure recognition, not memory of isolated results.
大多数考题并不是简单地要求"对 sin x 求导"。它们给出复合角如 sin(3x + 1)、乘积如 x² cos x、商如 sin x / x，并期望你立刻选对法则。考官考查的是结构识别能力，而不是孤立结论的记忆。
A useful way to think about it: the three standard derivatives are the vocabulary, and the chain, product and quotient rules are the grammar. You need both, and neither alone will carry you through a full question.
一个有用的思路是：三个基本导数是词汇，链式、乘法、除法法则是语法。两者缺一不可，单独掌握任何一项都无法完整做完一道大题。
2. Radians First: The Condition That Makes the Formulas True | 弧度制前提：公式成立的条件
ly when the angle x is measured in radians. If x were measured in degrees, differentiating sin x° would give (π/180) cos x°, which is a completely different answer. In A-Level calculus, always work in radians unless a question explicitly hands you an angle in degrees and asks you to convert first.
基本结论 d/dx (sin x) = cos x 与 d/dx (cos x) = −sin x 仅在角 x 以弧度为单位时成立。若 x 以度为单位，对 sin x° 求导会得到 (π/180) cos x°，答案完全不同。在 A-Level 微积分中，除非题目明确给出角度并要求先换算，否则一律使用弧度。
This has a practical calculator consequence. If your calculator is still in degree mode, then sin(π/6) will not return 0.5, and every subsequent numerical check will mislead you. Check the mode indicator at the start of the paper, not after question 3.
这有一个实际的计算器后果。如果你的计算器仍处于角度模式，sin(π/6) 不会返回 0.5，之后所有数值检验都会误导你。请在开考时确认模式指示，而不是做到第 3 题才发现。
ral counterpart of the small-angle approximation sin θ ≈ θ. That neatness disappears entirely in degrees.
弧度制要求也解释了这些函数在零点附近的平滑性：y = sin x 在 x = 0 处的梯度恰好为 1，这正是小角近似 sin θ ≈ θ 的自然对应。若使用角度，这种简洁性会完全消失。
3. Standard Derivatives of sin x and cos x | sin x 与 cos x 的基本导数
Everything starts from two results, and you should be able to write them down without hesitation:
一切都从两个结论开始，你应当能够毫不犹豫地写出它们：
d/dx (sin x) = cos x
d/dx (cos x) = −sin x
Notice the negative sign in the derivative of cos x. This is the single most common place to lose a mark in the whole topic, and it is completely avoidable. Write the negative sign down before you begin any problem involving cos x, even if you intend to simplify later.
注意 cos x 导数中的负号。这是整个专题中最常见的失分点，而且完全可以避免。在开始任何涉及 cos x 的题目之前，先把负号写下来，即使你打算稍后化简。
It helps to remember the cyclic pattern. Differentiating sin x moves you to cos x; differentiating cos x moves you to −sin x; differentiating again gives −cos x; and once more returns you to sin x. So the fourth derivative of sin x is sin x itself, a fact that occasionally appears in questions about repeated differentiation.
记住这个循环规律会很有帮助。对 sin x 求导得到 cos x；再对 cos x 求导得到 −sin x；再求导得到 −cos x；再求导又回到 sin x。所以 sin x 的四阶导数就是 sin x 本身，这一点偶尔会出现在关于重复求导的题目中。
0 and cos x has gradient −1. These values are a fast sanity check when you sketch a curve.
你还应当能说出关键点处的梯度。在 x = 0 处，sin x 的梯度为 1，cos x 的梯度为 0。在 x = π/2 处，sin x 的梯度为 0，cos x 的梯度为 −1。作图时，这些数值是快速的自检手段。
4. The Derivative of tan x | tan x 的导数
The third standard result is:
第三个基本结论是：
d/dx (tan x) = sec²x = 1 / cos²x
uestion is entirely plausible and it also proves you understand where the result comes from.
不要指望公式手册给出这一条。请准备好用除法法则对 tan x = sin x / cos x 在两行内完成推导，因为考查推导过程的题目完全可能出现，而且推导本身也能证明你理解结论的来源。
Setting u = sin x and v = cos x, the quotient rule gives:
令 u = sin x，v = cos x，由除法法则可得：
dy/dx = (v·du/dx − u·dv/dx) / v²
= (cos x · cos x − sin x · (−sin x)) / cos²x
= (cos²x + sin²x) / cos²x = 1 / cos²x = sec²x
The step where cos²x + sin²x collapses to 1 is the Pythagorean identity, and examiners like to see it written explicitly. It is a standard mark on derivation questions, so do not skip it.
cos²x + sin²x 化为 1 的那一步用到了毕达哥拉斯恒等式，考官希望看到它被明确写出。在推导题中这是一个标准得分点，不要跳过。
Because sec²x is always positive, tan x is strictly increasing on every interval where it is defined. Its gradient is never zero, so tan x has no stationary points at all - a fact that surprises some students who expect a turning point between consecutive asymptotes.
由于 sec²x 恒为正，tan x 在其定义的每个区间上严格递增。它的梯度永不为零，因此 tan x 完全没有驻点——这一点会让一些期待在相邻渐近线之间找到拐点的学生感到意外。
5. The Chain Rule with Trigonometric Functions | 链式法则与三角函数
Almost every exam question uses a composite argument, so the chain rule is the workhorse of this topic. The pattern is simple: differentiate the outer function, then multiply by the derivative of the inside.
几乎所有考题都使用复合角，因此链式法则是本专题的主力工具。规律很简单：先对外层函数求导，再乘以内部函数的导数。
d/dx sin(ax + b) = a cos(ax + b)
d/dx cos(ax + b) = −a sin(ax + b)
d/dx tan(ax + b) = a sec²(ax + b)
Here are four representative cases, all of which have appeared in past papers in some form:
以下四个典型案例，都曾以某种形式出现在历年真题中：
y = sin(3x + 1) gives dy/dx = 3 cos(3x + 1). The 3 is essential.
y = sin(3x + 1)，则 dy/dx = 3 cos(3x + 1)。系数 3 必不可少。
y = cos(x²) gives dy/dx = −2x sin(x²). The minus sign and the 2x both matter.
y = cos(x²)，则 dy/dx = −2x sin(x²)。负号与 2x 都不能漏。
dy/dx = 2 sin x cos x = sin 2x. This double-angle simplification is often the final mark.
y = sin²x，写成 (sin x)²，则 dy/dx = 2 sin x cos x = sin 2x。这个二倍角化简常常就是最后一分。
y = √sin x gives dy/dx = cos x / (2√sin x), valid where sin x > 0.
y = √sin x，则 dy/dx = cos x / (2√sin x)，在 sin x > 0 处成立。
s away a mark immediately, and it is the kind of slip that also ruins any later stationary-point work built on that derivative.
最常见的错误是漏掉内部导数。把 d/dx sin 3x 写成 cos 3x 而不是 3 cos 3x，会立刻丢掉一分，而且这类失误还会连带毁掉后续建立在该导数之上的驻点计算。
6. Combining with the Product Rule | 与乘法法则的结合
When a trigonometric function is multiplied by a polynomial, an exponential or another trigonometric function, use the product rule:
当三角函数与多项式、指数函数或另一个三角函数相乘时，使用乘法法则：
d/dx (uv) = u·dv/dx + v·du/dx
Worked cases worth memorising as patterns:
以下例题值得作为模式记住：
y = x sin x gives dy/dx = sin x + x cos x.
y = x sin x，则 dy/dx = sin x + x cos x。
y = x² cos x gives dy/dx = 2x cos x − x² sin x.
y = x² cos x，则 dy/dx = 2x cos x − x² sin x。
y = eˣ sin x gives dy/dx = eˣ sin x + eˣ cos x = eˣ(sin x + cos x).
y = eˣ sin x，则 dy/dx = eˣ sin x + eˣ cos x = eˣ(sin x + cos x)。
ives dy/dx = cos²x − sin²x = cos 2x, which agrees perfectly with differentiating ½ sin 2x.
y = sin x cos x，则 dy/dx = cos²x − sin²x = cos 2x，与对 ½ sin 2x 求导的结果完全一致。
-rule question into a single chain-rule step. Examiners accept either route, but the identity route is faster and less error-prone under time pressure.
最后一个例子很受青睐，因为它有两种解法。先识别二倍角恒等式 sin x cos x = ½ sin 2x，可以把乘法法则题变成一步链式法则。考官接受任一解法，但在时间压力下，恒等式路线更快、更不易出错。
Keep the product rule order consistent. Since addition is commutative the order of the two terms does not affect correctness, but writing u first every time reduces the chance of mixing up which derivative belongs to which factor.
乘法法则的书写顺序保持一致。由于加法可交换，两项顺序不影响正确性，但每次都把 u 写在前面，可以降低把导数与因式配错的概率。
7. Quotients and Reciprocal Trigonometric Functions | 除法法则与倒数三角函数
For a quotient, use:
对于商式，使用：
d/dx (u/v) = (v·du/dx − u·dv/dx) / v²
Two standard illustrations:
两个标准示例：
y = sin x / x gives dy/dx = (x cos x − sin x) / x².
y = sin x / x，则 dy/dx = (x cos x − sin x) / x²。
+ sin x). The algebra simplifies more than students expect, which is why it makes a good exam question.
y = cos x / (1 + sin x)，则 dy/dx = −1 / (1 + sin x)。化简结果比学生预想的更简洁，所以它是很好的考题素材。
x = −(1 + sin x), using sin²x + cos²x = 1. Dividing by (1 + sin x)² leaves −1/(1 + sin x).
简单验证第二个结果。分子为 (−sin x)(1 + sin x) − cos x(cos x) = −sin x − sin²x − cos²x = −(1 + sin x)，其中用到 sin²x + cos²x = 1。除以 (1 + sin x)² 后得到 −1/(1 + sin x)。
The reciprocal functions cosec x, sec x and cot x are not required for standard Edexcel A-Level Mathematics, but they do appear in Further Mathematics. If you meet them, the derivatives are:
倒数函数 cosec x、sec x、cot x 不属于标准 Edexcel A-Level 数学的必考内容，但会出现在进阶数学中。如果你遇到它们，导数为：
d/dx (cosec x) = −cosec x cot x
d/dx (sec x) = sec x tan x
d/dx (cot x) = −cosec²x
Notice the pattern: every derivative of a "co-" function carries a minus sign. It is a small memory hook that saves a lot of trouble.
注意这个规律：每个"co-"函数的导数都带负号。这个小记忆钩子能省下不少麻烦。
8. Worked Example: Tangents and Normals | 例题：切线与法线
the tangent at the point where x = π/6, and the equation of the normal at the same point.
考虑曲线 y = sin 2x。求 x = π/6 处的切线方程，以及同一点处的法线方程。
Step one: differentiate. Using the chain rule, dy/dx = 2 cos 2x.
第一步：求导。由链式法则，dy/dx = 2 cos 2x。
tep two: evaluate the gradient. At x = π/6, 2x = π/3, so cos 2x = ½ and dy/dx = 2 × ½ = 1.
第二步：计算梯度。当 x = π/6 时，2x = π/3，故 cos 2x = ½，dy/dx = 2 × ½ = 1。
Step three: find the y-coordinate. y = sin(π/3) = √3 / 2.
第三步：求纵坐标。y = sin(π/3) = √3 / 2。
Step four: write the tangent. Using y − y₁ = m(x − x₁):
第四步：写出切线。由 y − y₁ = m(x − x₁)：
y − √3/2 = 1 · (x − π/6)
normal gradient is the negative reciprocal of the tangent gradient, so m = −1. Therefore:
第五步：法线梯度是切线梯度的负倒数，故 m = −1。因此：
y − √3/2 = −1 · (x − π/6)
Two marks here are for method rather than arithmetic: selecting the chain rule, and correctly taking the negative reciprocal. Keep answers in exact form with surds and π. Converting to decimals loses accuracy marks in Edexcel papers, even when the decimal is correct to three places.
这里有兩分属于方法分而不是算术分：正确选择链式法则，以及正确取负倒数。答案请保留根号与 π 的精确形式。在 Edexcel 试卷中，即使小数精确到三位，把结果化成小数也会丢掉精度分。
9. Stationary Points and the Second Derivative | 驻点与二阶导数
nchanged from polynomial work: set dy/dx = 0, solve within the interval, then use the second derivative or a gradient table to classify.
常见的综合题会要求你求出三角函数曲线在给定区间上的驻点并判断其性质。方法与其他函数完全一致：令 dy/dx = 0，在区间内求解，再用二阶导数或梯度表判断类型。
Take y = sin x + cos x on the interval 0 ≤ x ≤ 2π.
取 y = sin x + cos x，区间为 0 ≤ x ≤ 2π。
dy/dx = cos x − sin x = 0 ⟹ tan x = 1
Within the interval, tan x = 1 gives x = π/4 and x = 5π/4. The corresponding y values are √2 and −√2.
在区间内，tan x = 1 给出 x = π/4 与 x = 5π/4，对应的 y 值分别为 √2 与 −√2。
For classification, differentiate again:
判断类型需要再求一次导：
d²y/dx² = −sin x − cos x
At x = π/4 this is −√2 < 0, so (π/4, √2) is a maximum. At x = 5π/4 it is +√2 > 0, so (5π/4, −√2 sing the second solution is a classic error when students solve tan x = 1 and stop at π/4.
在 x = π/4 处该值为 −√2 < 0，故 (π/4, √2) 为极大值点。在 x = 5π/4 处该值为 +√2 > 0，故 (5π/4, −√2) 为极小值点。注意必须检查区间内所有解；学生解 tan x = 1 后停在 π/4 而漏掉第二个解，是经典错误。
When the interval is stated in radians, give your answer in radians. A solution of 45° where π/4 was expected will be penalised even though the geometry is identical.
当区间以弧度给出时，答案也应以弧度表示。当题目期望 π/4 而你写出 45° 时，尽管几何意义相同，仍会被扣分。
10. The Link to Integration | 与积分的联系
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