📚 Using Trigonometric Identities | 利用三角恒等式
Trigonometric identities are the backbone of A-level mathematics, enabling you to simplify expressions, solve equations, and tackle calculus problems with confidence. Mastering the art of choosing the right identity and manipulating expressions flexibly is essential for success in the Edexcel course. This article explores the key identities you must know and demonstrates how to use them effectively across a range of exam-style contexts.
三角恒等式是 A-level 数学的支柱，能让你自信地化简表达式、解方程并处理微积分问题。掌握选择合适的恒等式并灵活变形的技巧，对于 Edexcel 课程的成功至关重要。本文探讨你必须掌握的核心恒等式，并展示如何在各种考试题型中有效运用它们。
1. Fundamental Pythagorean Identities | 基本的毕达哥拉斯恒等式
You will constantly rely on the identity sin²θ + cos²θ ≡ 1 and its rearranged forms. Dividing through by cos²θ gives 1 + tan²θ ≡ sec²θ, while dividing by sin²θ yields 1 + cot²θ ≡ csc²θ. The quotient identity tanθ ≡ sinθ/cosθ is also fundamental.
你会不断用到恒等式 sin²θ + cos²θ ≡ 1 及其变形。两边除以 cos²θ 得到 1 + tan²θ ≡ sec²θ，而除以 sin²θ 则得到 1 + cot²θ ≡ csc²θ。商数恒等式 tanθ ≡ sinθ/cosθ 同样基本。
When proving identities, look for chances to replace expressions like 3cos²θ with 3(1 – sin²θ) or to handle sec²θ – 1 by converting it to tan²θ. These substitutions are especially useful in trigonometric equations.
证明恒等式时，要寻找机会将类似 3cos²θ 替换为 3(1 – sin²θ)，或者将 sec²θ – 1 转化为 tan²θ。这些代换在三角方程中尤为有用。
2. Strategy for Proving Identities | 证明恒等式的策略
Begin with one side of the equation – usually the more complicated one – and transform it step by step until it matches the other side. Write each step clearly and state the identity used. Avoid moving terms across the ≡ sign unless you are careful to maintain equivalence.
从等式的一边开始——通常是较复杂的一边——逐步变换，直到与另一边相同。每一步写清楚，并注明所用的恒等式。除非能确保等价性，否则避免将项移到 ≡ 号另一边。
sinθ and cosθ, then combine into a single fraction. For example, to prove (1 + tan²θ)/secθ ≡ secθ, replace tan²θ with sec²θ – 1, then simplify using secθ = 1/cosθ.
常见技巧是把所有内容写成 sinθ 和 cosθ 的形式，然后合并成一个分数。例如，证明 (1 + tan²θ)/secθ ≡ secθ，可将 tan²θ 替换为 sec²θ – 1，再利用 secθ = 1/cosθ 化简。
3. Solving Simple Equations with Identities | 利用恒等式解简单方程
Equations like 2sin²x – cosx – 1 = 0 become manageable once you replace sin²x with 1 – lly 0 ≤ x < 2π, and remember to use the CAST diagram or graph to find all possible angles.
像 2sin²x – cosx – 1 = 0 这样的方程，一旦用 1 – cos²x 替换 sin²x，就变得易于处理，得到关于 cosx 的二次方程。始终检查解是否在给定区间（通常 0 ≤ x < 2π）内，并记得用 CAST 图或图像找出所有可能的角。
Factorising induces equations like cosx(2cosx – ½. Present angles in radians as required by Edexcel exams: x = π/2, 3π/2, π/3, 5π/3, etc.
因式分解产生 cosx(2cosx – 1) = 0 这样的方程，由此解出 cosx = 0 或 cosx = ½。按照 Edexcel 考试要求用弧度表示角：x = π/2、3π/2、π/3、5π/3 等。
4. Compound Angle Formulae: sin(A±B) and cos(A±B) | 和角公式：sin(A±B) 与 cos(A±B)
The compound angle identities are vital for expanding expressions like sin(θ + 30°) or cos(2θ – π/3). You must memorise:
和角恒等式对于展开 sin(θ + 30°) 或 cos(2θ – π/3) 至关重要。你必须熟记：
sin(A ± B) ≡ sinA cosB ± cosA sinB
cos(A ± B) ≡ cosA cosB ∓ sinA sinB
sin(A ± B) ≡ sinA cosB ± cosA sinB
cos(A ± B) ≡ cosA cosB ∓ sinA sinB
45°+30°). In pure examination questions, you often need to apply these formulae backwards to simplify an expression into a single trigonometric function.
仔细应用这些公式，可以推导出精确值，例如 cos 75° = cos(45°+30°)。在纯数考题中，常需反向应用这些公式，将表达式化简为单个三角函数。
5. Compound Angle Formula for tan(A±B) | 正切的和角公式
The tangent compound formula is:
正切的和角公式为：
tan(A ± B) ≡ (tanA ± tanB) / (1 ∓ tanA tanB)
tan(A ± B) ≡ (tanA ± tanB) / (1 ∓ tanA tanB)
°, which can be expressed as 60°+45°. Always pay attention to the sign in the denominator: '±' on top corresponds to '∓' on the bottom, which is a frequent source of errors.
在处理如 105°（可表示为 60°+45°）这样角的精确值时特别有用。务必注意分母中的符号：上面的“±”对应下面的“∓”，这是常见的错误来源。
6. Double Angle Identities | 倍角恒等式
The double angle formulae follow directly from the compound angle identities and must be at your fingertips:
倍角公式可直接从和角恒等式导出，你必须熟练运用：
| Identity | Forms |
| sin 2A | 2 sinA cosA |
| cos 2A | cos²A – sin²A = 2cos²A – 1 = 1 – 2sin²A |
| tan 2A | 2 tanA / (1 – tan²A) |
ates other functions or matches a given quadratic. In integration, the identity cos 2A = 1 – 2sin²A often rearranges to sin²A = (1 – cos 2A)/2.
cos 2A 的三种形式尤为强大，因为你能够选择合适的版本以消去其他函数或匹配给定的二次式。在积分中，恒等式 cos 2A = 1 – 2sin²A 常变形为 sin²A = (1 – cos 2A)/2。
7. Solving Equations Using Double Angles | 利用倍角公式解方程
both sin 2θ and cos θ, replace sin 2θ with 2 sinθ cosθ. Factorising then gives sinθ(2cosθ – ed-angle equation into a product of simpler functions, making it straightforward to find all solutions within the required domain.
当方程同时含有 sin 2θ 和 cos θ 时，将 sin 2θ 替换为 2 sinθ cosθ。因式分解得到 sinθ(2cosθ – 1) = 0。这种方法将混角方程转化为简单函数的乘积，便于找出指定区间内的所有解。
If the equation involves cos 2θ and cos θ only, use cos 2θ = 2cos²θ – θ. Always check for extraneous solutions and ensure your calculator is set to radian mode.
若方程仅包含 cos 2θ 和 cos θ，则利用 cos 2θ = 2cos²θ – 1 得到关于 cosθ 的二次方程。务必检查增根并确保计算器处于弧度模式。
8. Expressing a sinθ + b cosθ as R sin(θ ± α) | 将 a sinθ + b cosθ 化为 R sin(θ ± α)
ms require rewriting an expression like 3 sinθ + 4 cosθ in the form R sin(θ + α). Here R = √a² + b² ude, and α is found from tanα = b/a, with the quadrant determined by the signs of a and b.
许多试题要求将 3 sinθ + 4 cosθ 写成 R sin(θ + α) 的形式。这里 R = √a² + b² 给出振幅，而 α 通过 tanα = b/a 求得，象限由 a 和 b 的符号决定。
nd R sinα = b. This technique is also used to find maximum and minimum values: the maximum of R sin(θ + α) is simply R, occurring when sin(θ + α) = 1.
展开 R sin(θ + α) 需用和角公式：R sinθ cosα + R cosθ sinα。与原式对比系数得到 R cosα = a 和 R sinα = b。该技巧也用于求最值：R sin(θ + α) 的最大值就是 R，出现在 sin(θ + α) = 1 时。
9. Solving Equations with the R-Form | 利用 R 形公式解方程
usting for the interval. Remember to add or subtract multiples of 2π to get all solutions.
一旦表达式化为 R sin(θ + α) 的形式，像 R sin(θ + α) = c 这样的方程就变得可解：先写出 θ + α = arcsin(c/R)，再根据区间调整。记得通过加减 2π 的整数倍来获得所有解。
equation involves cos instead of sin, e.g., a sinθ + b cosθ can also be written as R cos(θ – β). The same principles apply: R = √a² + b² and tanβ = a/b with appropriate quadrant checks.
有时方程涉及 cos 而非 sin，例如 a sinθ + b cosθ 也可写成 R cos(θ – β)。同样原则适用：R = √a² + b²，tanβ = a/b，并进行相应的象限检查。
10. Half-Angle and Product-to-Sum Identities | 半角与积化和差恒等式
t always the headline act in Edexcel exams, half-angle identities such as sin(θ/2) = ±√[(1 – quently, you will use the product-to-sum formulae to integrate products like sin 3x cos x.
尽管在 Edexcel 考试中并非总是主角，半角恒等式如 sin(θ/2) = ±√[(1 – cosθ)/2] 可能出现在较难的证明或换元中。更常见的用武之地是，利用积化和差公式对 sin 3x cos x 这样的乘积进行积分。
The relevant identities are: sinA cosB = ½[sin(A+B) + sin(A–B)], cosA sinB = ½[sin(A+B) – sin(A–B)], cosA cosB = ½[cos(A+B) + cos(A–B)], and sinA sinB = ½[cos(A–B) – cos(A+B)]. Use these to turn products into sums that are easier to integrate.
相关的恒等式为：sinA cosB = ½[sin(A+B) + sin(A–B)]，cosA sinB = ½[sin(A+B) – sin(A–B)]，cosA cosB = ½[cos(A+B) + cos(A–B)]，以及 sinA sinB = ½[cos(A–B) – cos(A+B)]。利用它们将乘积转化为更易积分的和差形式。
11. Trigonometric Identities in Calculus | 微积分中的三角恒等式
vel unless you first rewrite them using cos 2x identities. For ∫ sin²x dx, use sin²x = ½(1 – cos 2x), leading to ½ x – ¼ sin 2x + C. Similar logic applies to ∫ cos²x dx, which becomes ½ x + ¼ sin 2x + C.
在 A-level 中，直接积分 sin²x 或 cos²x 是不可行的，除非先用 cos 2x 的恒等式改写。对于 ∫ sin²x dx，利用 sin²x = ½(1 – cos 2x) 得到 ½ x – ¼ sin 2x + C。同理，∫ cos²x dx 变为 ½ x + ¼ sin 2x + C。
x allows you to see connections quickly. In reverse, you can recognise integrals leading to ln|secx| or ln|sinx|.
对 tanx 求导得到 sec²x 是通过商法得出，但恒等式 1 + tan²x = sec²x 能让你迅速看出联系。反过来，你可以识别出积分为 ln|secx| 或 ln|sinx| 的情形。
12. Exam Tips and Common Pitfalls | 考试技巧与常见误区
Always write ≡ ptable when solving equations. In compound angle problems, double-check the sign for cos(A–B) – inus. When finding α in R-form, sketch the right-angled triangle to avoid quadrant errors.
证明恒等式时始终使用 ≡，但解方程时标准等号 = 也可以。在和角问题中，仔细核对 cos(A–B) 的符号——它是 cosA cosB + sinA sinB，而非减号。在 R 形公式中求 α 时，画出直角三角形以避免象限错误。
Don' ive all solutions in the specified interval using the general solution pattern: θ = nπ + (–1)ⁿα or the CAST method.
不要忘记平方一个方程可能会引入增根；务必通过代入来验证答案。对于 sinθ = k 这样的方程，利用通解模式 θ = nπ + (–1)ⁿα 或 CAST 方法给出指定区间内的所有解。
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