📚 Binomial Expansion, Trigonometric Ratios and Identities | 二项展开、三角比与三角恒等式
This revision guide covers the core pure mathematics topics usually found on pages 134-169 of the AS/A Level Pure Mathematics course: binomial expansion, trigonometric ratios, exact values, and trigonometric identities. These ideas sit at the heart of the Edexcel, AQA, OCR and Cambridge specifications, and they appear in both calculator and non-calculator papers year after year.
本复习指南涵盖 A-Level 纯数学课程第 134-169 页常见的核心主题：二项展开、三角比、特殊角的精确值以及三角恒等式。这些内容是 Edexcel、AQA、OCR 与剑桥考试局大纲的核心，年复一年地出现在允许使用计算器和不允许使用计算器的试卷中。
The two halves of the topic look unrelated at first, but both reward the same skills: fluency with algebra, careful bookkeeping of indices, and the confidence to work from a standard formula rather than memorising a hundred special cases. Work through the sections in order, then return to the worked examples when you revise.
这两个部分乍看互不相关，但都考察同样的能力：熟练的代数运算、对指数的细致处理，以及敢于从标准公式出发而不是死记上百个特例的信心。请按顺序通读各小节，复习时再回到例题部分重做一遍。
1. Pascal's Triangle and nCr | 帕斯卡三角形与组合数
Pascal's triangle is built from the coefficients that appear when (a + b)ⁿ is expanded. The first rows are 1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1, and every entry after the first row is the sum of the two entries directly above it.
帕斯卡三角形由展开 (a + b)ⁿ 时出现的系数构成。前几行为 1；1 1；1 2 1；1 3 3 1；1 4 6 4 1，除第一行外，每一项都等于其正上方两项之和。
The entry in row n at position r is written ⁿCᵣ and is read as "n choose r". It counts the number of ways of choosing r objects from n distinct objects when the order does not matter, and its value is given by the factorial formula below.
第 n 行第 r 个系数写作 ⁿCᵣ，读作 "n 选 r"。它表示从 n 个不同物体中不计次序地选出 r 个的方法数，其数值由下面的阶乘公式给出。
ⁿCᵣ = n! / [r!(n - r)!]
For example ⁵C₂ = 5! / (2! × 3! ₙ₋ᵣ, which is exactly why every row of the triangle reads the same forwards and backwards.
例如 ⁵C₂ = 5! / (2! × 3!) = 120 / 12 = 10。由公式可立即得到两个有用结论：ⁿC₀ = 1、ⁿCₙ = 1，以及 ⁿCᵣ = ⁿCₙ₋ᵣ，这正是三角形每一行正读倒读都一样的原因。
Most calculators have an nCr key, which is safer for large values of n. For small n the triangle is usually quicker and it also shows you the whole pattern at a glance, which is helpful when a question asks you to state a full expansion.
多数计算器都有 nCr 键，当 n 较大时用它更保险。当 n 较小时，用三角形通常更快，而且能一眼看清整体规律，这在题目要求写出完整展开式时特别有用。
- Row 0: 1 Row 1: 1 1 Row 2: 1 2 1 Row 3: 1 3 3 1
- ⁿCᵣ is only defined when 0 ≤ r ≤ n, and r must be a non-negative integer here.
- 考试提示 / Exam tip: always write the general term before substituting numbers; it keeps your working method-mark friendly.
2. The Binomial Theorem for Positive Integers | 正整数指数的二项式定理
When n is a positive integer, (a + b)ⁿ expands into exactly n + 1 terms. The powers of a decrease from n down to 0 while the powers of b increase from 0 up to n, so every term has total degree n.
当 n 为正整数时，(a + b)ⁿ 展开后恰好有 n + 1 项。a 的次数从 n 递减到 0，同时 b 的次数从 0 递增到 n，因此每一项的总次数都是 n。
(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² + ... + ⁿCᵣ aⁿ⁻ʳ bʳ + ... + ⁿCₙ bⁿ
is written below. Being able to quote it instantly saves a great deal of time in the exam, because most questions ask for one particular term rather than the whole expansion.
从 r = 0 数起的通项如下。能够立刻写出通项在考试中能节省大量时间，因为多数题目只要求某一个特定项，而不是整个展开式。
Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ , r = 0, 1, 2, ..., n
x² + 10x³ + 5x⁴ + x⁵. Notice how the coefficients 1, 5, 10, 10, 5, 1 match row 5 of Pascal's triangle, and how the indices always add up to 5.
例如 (1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵。注意系数 1、5、10、10、5、1 正是帕斯卡三角形第 5 行，而各项指数之和始终为 5。
A common error is to forget the coefficient of x when b is something like 3x. Treat the whole of b, including its sign and its numerical factor, as one bracket and raise the entire bracket to the power r.
常见错误是当 b 为 3x 之类的式子时忘记 x 的系数。要把整个 b（连同符号与数字系数）当作一个整体括号，并把整个括号升到 r 次幂。
3. Worked Examples: Finding a Particular Term | 例题：求特定项
Example 1. the coefficient of x³ in the expansion of (2 + 3x)⁵. The general term is Tᵣ₊₁ = ⁵Cᵣ (2)⁵⁻ʳ (3x)ʳ, so the power of x is r. We need r = 3, giving ⁵C₃ × 2² × 3³ = 10 × 4 × 27 = 1080.
例 1. 求 (2 + 3x)⁵ 展开式中 x³ 的系数。通项为 Tᵣ₊₁ = ⁵Cᵣ (2)⁵⁻ʳ (3x)ʳ，故 x 的次数是 r。令 r = 3，得 ⁵C₃ × 2² × 3³ = 10 × 4 × 27 = 1080。
Example 2. Find the coefficient of x² in the expansion of (1 - 3x)⁶. Here Tᵣ₊₁ = ⁶Cᵣ (1)⁶⁻ʳ (-3x)ʳ h r = 2 we get ⁶C₂ × (-3)² = 15 × 9 = 135. The negative sign disappears because r is even.
例 2. 求 (1 - 3x)⁶ 展开式中 x² 的系数。此处 Tᵣ₊₁ = ⁶Cᵣ (1)⁶⁻ʳ (-3x)ʳ。取 r = 2 得 ⁶C₂ × (-3)² = 15 × 9 = 135。由于 r 为偶数，负号消失。
Example 3. Find the term independent of x in (x + 2/x)⁶. Write the general term as ⁶Cᵣ x⁶⁻ʳ (2x⁻¹)ʳ = ⁶Cᵣ 2ʳ x⁶⁻²ʳ term to be constant we need 6 - 2r = 0, so r = 3 and the term is ⁶C₃ × 2³ = 20 × 8 = 160.
例 3. 求 (x + 2/x)⁶ 中的常数项。写出通项 ⁶Cᵣ x⁶⁻ʳ (2x⁻¹)ʳ = ⁶Cᵣ 2ʳ x⁶⁻²ʳ。要使该项与 x 无关，需 6 - 2r = 0，故 r = 3，该项为 ⁶C₃ × 2³ = 20 × 8 = 160。
- Always state the value of r before you compute the coefficient.
- Check the sign carefully: only odd powers of a negative bracket produce a minus sign.
- “Independent of x” means the total index of x is zero; “coefficient of xᵏ” means the total index equals k.
4. Expanding (a + bx)ⁿ and Estimating | 展开 (a + bx)ⁿ 与近似估计
(1 + bx/a)ⁿ. This makes it obvious where the powers of a go and reduces arithmetic slips.
当括号为 (a + bx)ⁿ 时，先提出 a，使第二项成为真分数：(a + bx)ⁿ = aⁿ (1 + bx/a)ⁿ。这样能清楚看出 a 的幂次分布，并减少算术失误。
The same structure extends beyond positive integers. For any rational n and |x| < 1, the expansion continues forever and takes the form shown here, where the coefficients use falling factorials.
同样的结构可以推广到正整数以外。对任意有理数 n 且 |x| < 1，展开式可以无限延续，形式如下，其系数使用降阶乘。
(1 + x)ⁿ = 1 + nx + [n(n - 1)/2!] x² + [n(n - 1)(n - 2)/3!] x³ + ... (|x| < 1)
¹ = 1 - x + x² - x³ + ... and (1 + x)½ = 1 + x/2 - x²/8 + x³/16 - ..., both valid only for |x| < 1. The condition of validity is a mark in its own right.
值得记住的两种情形是 (1 + x)⁻¹ = 1 - x + x² - x³ + ... 与 (1 + x)½ = 1 + x/2 - x²/8 + x³/16 - ...，二者都只在 |x| < 1 时成立。适用条件本身就是得分点。
21896, which is very close to the true value 1.21899. Including more terms always improves the estimate when |x| is small.
二项展开还可用于近似计算。例如 (1 + 0.02)¹⁰ ≈ 1 + 10(0.02) + 45(0.02)² + 120(0.02)³ = 1 + 0.2 + 0.018 + 0.00096 ≈ 1.21896，与真值 1.21899 非常接近。当 |x| 较小时，多取几项总能提高精度。
5. Trigonometric Ratios in Right-Angled Triangles | 直角三角形中的三角比
In a right-angled triangle the three basic ratios are defined with respect to an acute angle θ: sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. The mnemonic SOHCAHTOA is the fastest way to recall them.
在直角三角形中，三个基本三角比针对锐角 θ 定义：正弦为对边比斜边，余弦为邻边比斜边，正切为对边比邻边。口诀 SOHCAHTOA 是记忆它们最快的方法。
sin θ = opp / hyp cos θ = adj / hyp tan θ = opp / adj = sin θ / cos θ
Because both sine and cosine are ratios of lengths divided by the hypotenuse, they must always lie between -1 and 1. The tangent ratio has no such restriction and can take any real value.
由于正弦与余弦都是长度与斜边之比，它们的取值必然在 -1 与 1 之间。正切则没有这个限制，可以取任意实数值。
In an exam, if a diagram is not given, sketch one and label the known side and angle before choosing a ratio. That single habit prevents the most common mistake of all, which is using the wrong pair of sides.
考试中若题目没有配图，先自己画一个直角三角形，标出已知的边和角，再选择三角比。这个习惯可以避免最常见的错误：用错了一对边。
- Given two sides, use Pythagoras to find the third before using any ratio.
- Given one side and one angle, choose the ratio that links the known side to the required side.
- Angles of elevation and depression are always measured from the horizontal.
6. The Unit Circle, CAST and Radians | 单位圆、CAST 图与弧度制
Extending the ratios beyond 90° requires the unit circle. For a point P on the circle of radius 1 at angle θ measured anticlockwise from the positive x-axis, the coordinates of P are (cos θ, sin θ), and tan θ is the gradient of OP.
要把三角比推广到 90° 以外，需要借助单位圆。设 P 是半径为 1 的圆上一点，θ 为从 x 轴正方向逆时针量得的角度，则 P 的坐标为 (cos θ, sin θ)，而 tan θ 是直线 OP 的斜率。
This picture gives the CAST rule: in the first quadrant all three ratios are positive; in the second only sine is positive; in the third only tangent is positive; in the fourth only cosine is positive. When several angles share the same sine, cosine or tangent value, the circle shows you exactly which ones they are.
这一图形给出 CAST 法则：第一象限三个三角比均为正；第二象限只有正弦为正；第三象限只有正切为正；第四象限只有余弦为正。当多个角具有相同的正弦、余弦或正切值时，单位圆能直接告诉你它们分别是哪些角。
180°. Radians are the default in calculus, and the standard results you must know are shown below.
角度可用度或弧度度量，换算关系为 π 弧度 = 180°。弧度是微积分中的默认单位，必须掌握的标准结果如下。
arc length s = rθ sector area A = ½ r²θ (θ in radians)
Note the warning: these two formulae only work when θ is in radians. If a question gives the angle in degrees, convert it first, otherwise the answer will be wrong by a factor of about 57.
注意：这两个公式只在 θ 用弧度表示时成立。若题目给的是度数，必须先换算，否则答案会相差约 57 倍。
7. Exact Values and Special Angles | 特殊角与精确值
Certain angles appear so often that their trigonometric values must be known exactly, without a calculator. They come from the two standard triangles: the isosceles right triangle with sides 1, 1, √2, and the half-equilateral triangle with sides 1, √3, 2.
有些角出现得极其频繁，其三角值必须不依赖计算器而精确掌握。它们来自两个标准三角形：边长为 1、1、√2 的等腰直角三角形，以及边长为 1、√3、2 的等边三角形之半。
| θ | 0 | 30° | 45° | 60° | 90° |
| radians | 0 | π/6 | π/4 | π/3 | π/2 |
| sin θ | 0 | 1/2 | √2/2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | √2/2 | 1/2 | 0 |
| tan θ | 0 | √3/3 | 1 | √3 | undefined |
drants follow from symmetry. For example sin 150° = sin 30° = 1/2, cos 135° = -cos 45° = -√2/2, and tan 240° = tan 60° = √3. Always identify the quadrant first, then decide the sign.
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