📚 Trigonometric Equations & Identities for Edexcel A-Level Maths | 三角方程与恒等式：Edexcel A-Level 数学核心突破
Trigonometric equations and identities form a central part of the Edexcel A-Level Pure Mathematics specification. Success in this topic depends on fluent use of the unit circle, exact values, radian measure and algebraic techniques applied to sin, cos and tan. This revision article explains the key ideas, common question types and exam strategies in a bilingual format.
三角方程与恒等式是 Edexcel A-Level 纯数学考试的核心内容。要在这个主题上拿高分，必须熟练掌握单位圆、特殊角的精确值、弧度制，以及对正弦、余弦和正切进行代数处理。本文以中英双语讲解核心概念、常见题型和考试策略。
1. The Unit Circle and Trigonometric Ratios | 单位圆与三角比
The unit circle gives a geometric definition of sine and cosine for any angle θ. A point P on the circle has coordinates (cos θ, sin θ), where θ is measured anticlockwise from the positive x-axis.
单位圆给出了任意角 θ 的正弦和余弦的几何定义。圆上一点 P 的坐标为 (cos θ, sin θ)，其中 θ 是从 x 轴正方向逆时针旋转的角度。
Because the radius is 1, the coordinates of P must satisfy x² + y² = 1, which immediately gives the fundamental identity:
因为半径是 1，点 P 的坐标必须满足 x² + y² = 1，这就直接得到了基本恒等式：
sin² θ + cos² θ = 1
For Edexcel questions, you should know the exact values of sin, cos and tan for 0°, 30°, 45°, 60° and 90°, and their radian equivalents 0, π/6, π/4, π/3 and π/2.
在 Edexcel 考试中，你必须牢记 0°、30°、45°、60° 和 90° 以及对应弧度 0、π/6、π/4、π/3、π/2 的正弦、余弦和正切精确值。
2. Key Pythagorean Identities | 核心毕达哥拉斯恒等式
From sin² θ + cos² θ = 1, dividing through by cos² θ gives the tangent form:
由 sin² θ + cos² θ = 1 两边同除以 cos² θ，可以得到正切形式的恒等式：
1 + tan² θ = sec² θ
Dividing the original identity by sin² θ instead gives the cotangent form:
如果将基本恒等式两边同除以 sin² θ，则得到余切形式的恒等式：
1 + cot² θ = cosec² θ
These identities are often used to rewrite expressions such as 2sin² x + 3cos x as a quadratic in cos x only, which can then be solved.
这些恒等式常用于把 2sin² x + 3cos x 一类表达式改写成只含 cos x 的二次式，从而进行求解。
3. Tangent, Cotangent, Secant and Cosecant | 正切、余切、正割与余割
Edexcel A-Level requires fluency with the reciprocal trigonometric functions. The definitions are:
Edexcel A-Level 要求熟练运用倒数三角函数。其定义如下：
- sec θ = 1 / cos θ
- cosec θ = 1 / sin θ
- cot θ = 1 / tan θ = cos θ / sin θ
You should also be comfortable using the relationships tan θ = sin θ / cos θ and cot θ = cos θ / sin θ when proving identities.
你还应能熟练使用 tan θ = sin θ / cos θ 以及 cot θ = cos θ / sin θ 这些关系式来证明恒等式。
A common exam step is to express everything in terms of sin and cos before simplifying. This often reveals hidden factors and cancellations.
考试中常用的一步是先把所有量都写成 sin 和 cos 的形式再化简。这样往往能暴露出隐藏的因式和可以约去的项。
4. Radian Measure and Arc Length | 弧度制与弧长
Angles in A-Level trigonometry are usually given in radians. One complete revolution is 2π radians, so π radians = 180°.
A-Level 三角学中的角通常用弧度表示。一整圈是 2π 弧度，因此 π 弧度 = 180°。
For a sector of a circle with radius r and angle θ in radians, the arc length s and area A are given by:
对于半径为 r、圆心角为 θ 弧度的扇形，弧长 s 和面积 A 的公式为：
s = rθ
A = ½ r² θ
Make sure your calculator is in radian mode when solving equations such as sin x = 0.4 in the range 0 ≤ x < 2π. A degree-mode answer will be wrong in radian questions.
在求解 0 ≤ x < 2π 范围内的 sin x = 0.4 这类方程时，务必让计算器处于弧度模式。弧度题型中如果使用角度模式，答案就会出错。
5. Graphs of sin x, cos x and tan x | 正弦、余弦与正切函数图像
The graph of y = sin x has period 2π and range [-1, 1]. It starts at 0, rises to 1 at π/2, returns to 0 at π, falls to -1 at 3π/2 and returns to 0 at 2π.
y = sin x 的图像周期为 2π，值域为 [-1, 1]。图像从 0 开始，在 π/2 处升至 1，在 π 处回到 0，在 3π/2 处降到 -1，在 2π 处回到 0。
The graph of y = cos x also has period 2π, but it starts at 1 when x = 0. The graph of y = tan x has period π and vertical asymptotes at x = π/2 + nπ.
y = cos x 的图像周期也是 2π，但在 x = 0 时从 1 开始。y = tan x 的周期为 π，并且在 x = π/2 + nπ 处有竖直渐近线。
| Function | Period | Range |
| sin x | 2π | [-1, 1] |
| cos x | 2π | [-1, 1] |
| tan x | π | ℝ |
Understanding these graphs helps you find all solutions in a given interval by using symmetry and periodicity rather than relying only on a calculator.
理解这些图像有助于利用对称性和周期性找出给定区间内的所有解，而不是只依赖计算器给出的一个解。
6. Double-Angle Formulae | 倍角公式
The double-angle formulae are essential in Edexcel Pure Mathematics. The three versions for cos 2θ are particularly important because they allow you to change the form of an expression to match a question.
倍角公式在 Edexcel 纯数学中至关重要。cos 2θ 的三种形式尤其重要，因为它们可以把表达式改写成适合题目的形式。
sin 2θ = 2 sin θ cos θ
cos 2θ = cos² θ – sin² θ
cos 2θ = 2 cos² θ – 1
cos 2θ = 1 – 2 sin² θ
tan 2θ = 2 tan θ / (1 – tan² θ)
For example, if an equation contains sin x cos x, replacing it with ½ sin 2x often makes the equation much easier to solve.
例如，如果方程中含有 sin x cos x，将它替换成 ½ sin 2x 通常会使方程更容易求解。
7. Compound-Angle Formulae | 和差角公式
The compound-angle formulae allow you to expand expressions such as sin(A ± B), cos(A ± B) and tan(A ± B). They are listed in the Edexcel formula booklet but must be used accurately under exam pressure.
和差角公式用于展开 sin(A ± B)、cos(A ± B) 和 tan(A ± B) 等表达式。它们虽然列在 Edexcel 公式表中，但考试时仍必须准确使用。
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)
These formulae are particularly useful when proving identities or when solving equations that involve angles like x + π/3. Expanding the compound angle often converts the expression into a standard linear combination of sin x and cos x.
这些公式在证明恒等式或求解含有 x + π/3 这类角的方程时特别有用。展开和差角后，表达式通常会转化为 sin x 和 cos x 的标准线性组合。
8. Solving Basic Trigonometric Equations | 基础三角方程求解
A typical Edexcel question gives an equation such as sin θ = 0.5 and asks for all values of θ in the interval 0 ≤ θ < 2π. The first step is to find the principal value using inverse trigonometric functions.
Edexcel 常见题型给出 sin θ = 0.5 这类方程，要求在 0 ≤ θ < 2π 内求出所有 θ 值。第一步是用反三角函数求出主值。
For sin θ = 0.5, the principal value is π/6. Since sin is positive in the first and second quadrants, the second solution is π – π/6 = 5π/6. There are no further solutions in 0 to 2π.
对于 sin θ = 0.5，主值是 π/6。由于 sine 在第一象限和第二象限为正，第二个解是 π – π/6 = 5π/6。在 0 到 2π 内没有更多解。
For cos θ = -0.5, the principal value is 2π/3. Cosine is negative in the second and third quadrants, so the solutions are 2π/3 and 4π/3.
对于 cos θ = -0.5，主值是 2π/3。余弦在第二象限和第三象限为负，因此解为 2π/3 和 4π/3。
For tan θ = 1, the principal value is π/4. Tangent has period π, so the general solution is θ = π/4 + nπ. In the interval 0 ≤ θ < 2π, the solutions are π/4 and 5π/4.
对于 tan θ = 1，主值是 π/4。正切函数的周期为 π，因此通解为 θ = π/4 + nπ。在 0 ≤ θ < 2π 内，解为 π/4 和 5π/4。
9. Quadratic Trigonometric Equations | 二次型三角方程
Many harder equations become quadratic after using an identity. For example, the equation 2 cos² x + 3 sin x = 0 can be rewritten using cos² x = 1 – sin² x:
很多较难的方程在代入恒等式后会变成二次方程。例如，方程 2 cos² x + 3 sin x = 0 可以用 cos² x = 1 – sin² x 改写为：
2(1 – sin² x) + 3 sin x = 0
2 sin² x – 3 sin x – 2 = 0
This factorises as (2 sin x + 1)(sin x – 2) = 0. Since sin x cannot equal 2, the only valid solution comes from sin x = -1/2.
该方程可以因式分解为 (2 sin x + 1)(sin x – 2) = 0。因为 sin x 不可能等于 2，所以唯一有效的解来自 sin x = -1/2。
Always check that any solution for sin, cos or tan lies within the allowed range. Discard impossible values such as sin x = 1.5 or cos x = -2.
一定要检查 sin、cos 或 tan 的解是否在允许的范围内。要舍去 sin x = 1.5 或 cos x = -2 这类不可能的值。
10. Using Identities to Simplify and Prove | 利用恒等式化简与证明
Proof questions are common in Edexcel A-Level papers. The safest approach is to start with the more complicated side of the identity and transform it into the simpler side.
证明题在 Edexcel A-Level 试卷中很常见。最稳妥的方法是从恒等式中较复杂的一边开始，把它变形为较简单的一边。
For example, to prove that (1 – cos 2x) / sin 2x = tan x, rewrite 1 – cos 2x as 2 sin² x and sin 2x as 2 sin x cos x.
例如，要证明 (1 – cos 2x) / sin 2x = tan x，可以把 1 – cos 2x 写成 2 sin² x，把 sin 2x 写成 2 sin x cos x。
(1 – cos 2x) / sin 2x = 2 sin² x / (2 sin x cos x) = sin x / cos x = tan x
When simplifying, look for common factors, reciprocal identities and double-angle substitutions. Present your steps clearly because method marks are awarded even if the final line is not reached.
化简时要寻找公因式、倒数关系以及倍角替换。要清楚地写出每一步，因为即使没有得出最后一行，过程分也会给。
11. Common Exam Pitfalls and Tips | 常见考试误区与技巧
One of the biggest mistakes is using degrees instead of radians. In A-Level trigonometry, unless a question explicitly uses °, every angle is in radians.
最常见的错误之一是使用角度而不是弧度。在 A-Level 三角学中，除非题目明确使用 °，否则所有角都是弧度。
Another common error is losing solutions when dividing both sides by a trigonometric term. For example, dividing sin x cos x = sin x by sin x loses the solutions where sin x = 0.
另一个常见错误是在方程两边同时除以一个三角函数项时丢失解。例如，把 sin x cos x = sin x 两边除以 sin x 会丢失 sin x = 0 的解。
Instead, bring all terms to one side and factorise: sin x (cos x – 1) = 0. This guarantees that every possible solution is found.
正确做法是把所有项都移到一边并因式分解：sin x (cos x – 1) = 0。这样可以保证求出所有可能的解。
Finally, always give solutions within the requested interval and check them by substitution. A small arithmetic slip can change a correct method into a wrong answer.
最后，一定要按题目要求的区间给出解，并通过代入检验。一个小的算术失误可能会把正确的方法变成错误答案。
12. Worked Mini-Example | 精讲例题
Worked example: Solve the equation 2 cos² x – sin x – 1 = 0 for 0 ≤ x < 2π.
精讲例题：解方程 2 cos² x – sin x – 1 = 0，其中 0 ≤ x < 2π。
Step 1: Replace cos² x with 1 – sin² x.
第一步：将 cos² x 替换为 1 – sin² x。
2(1 – sin² x) – sin x – 1 = 0
2 sin² x + sin x – 1 = 0
Step 2: Factorise the quadratic.
第二步：对二次式进行因式分解。
(2 sin x – 1)(sin x + 1) = 0
Step 3: Solve each factor.
第三步：分别解每个因式。
sin x = 1/2 ⇒ x = π/6, 5π/6
sin x = -1 ⇒ x = 3π/2
Therefore the full solution set is x = π/6, 5π/6, 3π/2. This example illustrates how an identity, factorisation and quadrant symmetry combine to solve a typical exam question.
因此完整解集为 x = π/6、5π/6、3π/2。这个例子说明如何综合运用恒等式、因式分解和象限对称性来解答典型考试题。
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