Trigonometric Equations and Identities for Edexcel A Level Maths | Edexcel A-Level 数学：三角函数方程与恒等式

📚 Trigonometric Equations and Identities for Edexcel A Level Maths | Edexcel A-Level 数学：三角函数方程与恒等式

through basic equations and the Pythagorean identity, and in A2 papers through compound angles, double angles, the harmonic form R sin(θ ± α), and rigorous identity proofs. This article reviews the essential identities, solving techniques, harmonic form conversions, graphical reasoning, and exam strategies you need to command.

三角函数方程与恒等式是 Edexcel A-Level 纯数学中考查最密集的板块之一。AS 阶段主要涉及基本方程与毕达哥拉斯恒等式，A2 阶段则进一步考查复合角、二倍角、辅助角形式 R sin(θ ± α) 以及严格的恒等式证明。本文梳理关键恒等式、解方程方法、辅助角转化、图像推理以及应试策略，帮助你系统掌握。

1. Pythagorean Identities and Reciprocal Functions | 毕达哥拉斯恒等式与倒数函数

you to replace one squared trigonometric function with the other, which is essential when an equation contains both sinθ and cosθ but only one squared term.

几乎所有三角函数方程的基础都是毕达哥拉斯恒等式 sin²θ + cos²θ = 1。它允许你把一个平方项替换为另一个平方项，当方程中同时含有 sinθ 和 cosθ 但只有一个平方项时，这一点至关重要。

sin²θ + cos²θ = 1

forms are especially useful for simplifying expressions involving tan, sec, cot, or cosec.

将 sin²θ + cos²θ = 1 除以 cos²θ 可得 1 + tan²θ = sec²θ；除以 sin²θ 可得 1 + cot²θ = cosec²θ。这两种形式在化简含有 tan、sec、cot 或 cosec 的表达式时特别有用。

1 + tan²θ = sec²θ,   1 + cot²θ = cosec²θ

ecognising these relationships quickly can turn a complicated equation into a standard quadratic in sinθ or cosθ.

你还应熟悉倒数定义：secθ = 1/cosθ，cosecθ = 1/sinθ，cotθ = 1/tanθ = cosθ/sinθ。快速识别这些关系，可以把复杂方程转化为关于 sinθ 或 cosθ 的标准二次方程。

When an exam question mixes sin²θ with cosθ, the standard move is to rewrite sin²θ as 1 − cos²θ so that the entire equation involves only one trigonometric ratio. This single substitution often converts an apparently unfamiliar equation into a routine quadratic.

当考题把 sin²θ 与 cosθ 混在一起时，标准做法是把 sin²θ 改写为 1 − cos²θ，使整个方程只涉及一个三角比。这一替换常把看似陌生的方程转化为常规二次方程。

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2. Compound Angle Identities | 复合角恒等式

es. For sine, sin(A ± B) = sin A cos B ± cos A sin B. For cosine, cos(A ± B) = cos A cos B ∓ sin A sin B. Notice that the cosine formula has a minus sign when the angle is A + B.

复合角恒等式用于展开角度和或差的三角函数。正弦公式为 sin(A ± B) = sin A cos B ± cos A sin B；余弦公式为 cos(A ± B) = cos A cos B ∓ sin A sin B。注意，当角为 A + B 时，余弦公式中出现负号。

sin(A ± B) = sin A cos B ± cos A sin B

cos(A ± B) = cos A cos B ∓ sin A sin B

The tangent form is tan(A ± B) = (tan A ± tan B)/(1 ∓ zero, so exam questions often ask for exact values such as tan 75°, where 75° = 45° + 30°.

正切公式为 tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)。分母不能为零，因此考题常要求精确值，例如 tan 75°，其中 75° = 45° + 30°。

tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)

sin(A + B)	sin A cos B + cos A sin B
sin(A − B)	sin A cos B − cos A sin B
cos(A + B)	cos A cos B − sin A sin B
cos(A − B)	cos A cos B + sin A sin B

These identities are not only for expansion. They also allow you to rewrite expressions such as sin x cos 2x + cos x sin 2x as sin 3x, which is a common first step in solving an equation that otherwise looks impossible.

这些恒等式不仅用于展开。它们还能把 sin x cos 2x + cos x sin 2x 这类表达式改写为 sin 3x，这通常是解看似不可能方程的第一步。

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3. Double Angle and Half Angle Formulae | 二倍角与半角公式

etting A = B. For sine, sin 2A = 2 sin A cos A. For tangent, tan 2A = 2 tan A/(1 − tan²A).

二倍角公式由复合角恒等式令 A = B 直接得到。正弦公式为 sin 2A = 2 sin A cos A；正切公式为 tan 2A = 2 tan A/(1 − tan²A)。

sin 2A = 2 sin A cos A

tan 2A = 2 tan A/(1 − tan²A)

− 2 sin²A. The choice depends on whether the rest of the equation contains sine or cosine.

余弦有三种等价形式：cos 2A = cos²A − sin²A = 2 cos²A − 1 = 1 − 2 sin²A。选择哪一种，取决于方程其余部分含有正弦还是余弦。

cos 2A = cos²A − sin²A = 2 cos²A − 1 = 1 − 2 sin²A

2A)/2. These half angle forms are useful for proving identities and for solving equations that contain squared trigonometric terms.

由 cos 2A = 1 − 2 sin²A 可得 sin²A = (1 − cos 2A)/2；由 cos 2A = 2 cos²A − 1 可得 cos²A = (1 + cos 2A)/2。这些半角形式在证明恒等式和求解含平方三角项的方程时非常有用。

sin²A = (1 − cos 2A)/2,   cos²A = (1 + cos 2A)/2

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4. The Harmonic Form R sin(θ ± α) and R cos(θ ± α) | 辅助角形式

An expression of the form a sinθ + b cosθ can be written as a single sine or cosine function. This is called the harmonic form, and it is a major A2 topic in Edexcel Pure Mathematics.

形如 a sinθ + b cosθ 的表达式可以写成单个正弦或余弦函数。这称为辅助角形式，是 Edexcel 纯数学 A2 的重要考点。

a sinθ + b cosθ = R sin(θ + α)

The amplitude is R = √a² + b² must check the quadrant of α so that the expansion matches the original signs of a and b.

振幅为 R = √a² + b²，相位角 α 满足 tan α = b/a。你必须检查 α 所在象限，使展开式与原式中 a、b 的符号一致。

R = √a² + b²,   tan α = b/a

A cosine version is also common: a sinθ + b cosθ = R cos(θ − α), where R = √a² + b² and tan α = a/b. The form you choose should make the resulting equation easiest to solve.

余弦形式也很常见：a sinθ + b cosθ = R cos(θ − α)，其中 R = √a² + b²，tan α = a/b。你应选择使最终方程最容易求解的形式。

a sinθ + b cosθ = R cos(θ − α)

you how many solutions an equation such as R sin(θ + α) = k can have in a given interval.

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化为辅助角形式后，最大值为 R，最小值为 −R。这能立刻告诉你方程 R sin(θ + α) = k 在给定区间内可能有多少个解。

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5. Solving Basic Trigonometric Equations in a Given Interval | 在给定区间内解基本三角方程

by finding the principal value using the inverse function. Then use the symmetry and periodicity of the graph to find every solution in the required interval.

对于 sinθ = k、cosθ = k 或 tanθ = k，先用反函数求主值，再利用图像的对称性和周期性找出指定区间内的所有解。

n α, θ = 2nπ ± α for cosθ = cos α, and θ = nπ + α for tanθ = tan α, where n is an integer.

正弦和余弦的周期为 2π，正切的周期为 π。通解为：sinθ = sin α 时 θ = nπ + (−1)ⁿα；cosθ = cos α 时 θ = 2nπ ± α；tanθ = tan α 时 θ = nπ + α，其中 n 为整数。

sinθ = sin α ⇒ θ = nπ + (−1)ⁿα

cosθ = cos α ⇒ θ = 2nπ ± α

tanθ = tan α ⇒ θ = nπ + α

which case there is one. A tangent equation has one solution in any interval of length π.

当区间为 0 ≤ θ < 2π 时，正弦或余弦方程通常有两个解；除非 k = ±1，此时只有一个解。正切方程在任何长度为 π 的区间内恰好有一个解。

Always substitute your final answers back into the original equation, especially when the equation was squared or when you divided by a trigonometric function during the solution.

务必把最终答案代回原方程检验，尤其是在解方程过程中进行了平方，或除以了某个三角函数时。

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6. General Solutions and Radian Measure | 通解与弧度制

Edexcel A Level frequently uses radians in trigonometric equations, especially in calculus-related questions. You must be comfortable switching between degrees and radians and must keep your calculator in the correct mode.

Edexcel A-Level 在三角方程中经常使用弧度，尤其是在与微积分相关的问题中。你必须熟练地在角度与弧度之间转换，并确保计算器处于正确模式。

π rad = 180°,   1 rad = 180°/π,   1° = π/180 rad

90° = π/2, and 180° = π. Recognising these values saves time when writing exact solutions.

常见精确弧度值包括 30° = π/6、45° = π/4、60° = π/3、90° = π/2、180° = π。记住这些值能在写精确解时节省时间。

If the interval is written as [0, 2π] or [−π, π], all answers must be in radians. Mixing degrees and radians in the same solution is a very common source of lost marks.

如果区间写成 [0, 2π] 或 [−π, π]，所有答案必须用弧度。在同一解答中混用角度与弧度是丢分的常见原因。

2θ over an interval that is twice as long, then divide every solution by 2 to return to θ.

对于含倍角的方程，如 sin 2θ = k，先在长度为两倍的区间内解出 2θ，再把每个解除以 2 得到 θ。

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7. Quadratic Equations in sin, cos or tan | 关于 sin、cos、tan 的二次方程

sin²θ + sinθ − 1 = 0, which can be solved by factoring or by using the quadratic formula.

许多三角方程可化为关于 sinθ、cosθ 或 tanθ 的二次方程。典型例子是 2 sin²θ + sinθ − 1 = 0，可用因式分解或求根公式求解。

Use the substitution u = sinθ, solve the quadratic in u, and then reject any root with |u| because sine and cosine values must lie between −1 and 1. Tangent has no such restriction.

令 u = sinθ，解关于 u 的二次方程，然后舍去 |u| > 1 的根，因为正弦和余弦的值必须在 −1 与 1 之间。正切没有这个限制。

becomes quadratic in cosθ. Similarly, use cos²θ = 1 − sin²θ when the mixed ratio is sine.

如果方程同时含 sin²θ 和 cosθ，把 sin²θ 改写为 1 − cos²θ，使方程变为关于 cosθ 的二次方程。类似地，当混合的是正弦时，用 cos²θ = 1 − sin²θ。

After solving for sinθ or cosθ, do not forget to find all angles in the given interval. A single quadratic root can generate two or more angle solutions because of periodicity.

求出 sinθ 或 cosθ 后，不要忘记找出给定区间内的所有角。由于周期性，一个二次根可能对应两个或更多角解。

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8. Equations Requiring Identities: Reducing to a Single Trigonometric Ratio | 需要恒等式化简的方程

Some equations mix different angles or different trigonometric functions. The key is to use identities to reduce everything to a single ratio or a single angle before solving.

有些方程混合了不同角或不同三角函数。关键是先用恒等式把所有项化为同一个三角比或同一个角，然后再求解。

2θ with 2 sinθ cosθ. This gives 2 sinθ cosθ = cosθ, which factors as cosθ(2 sinθ − 1) = 0.

对于 sin 2θ = cosθ 这类方程，把 sin 2θ 替换为 2 sinθ cosθ，得到 2 sinθ cosθ = cosθ，因式分解为 cosθ(2 sinθ − 1) = 0。

g by a trigonometric function can remove valid solutions and is one of the most frequently penalised errors in Edexcel exams.

在没有检查 cosθ = 0 的情况前，绝不要两边同时除以 cosθ。除以三角函数可能丢掉合法解，这是 Edexcel 考试中最常被扣分的错误之一。

oute. If it contains tanθ together with sinθ or cosθ, write tanθ = sinθ/cosθ and simplify.

如果方程中 sinθ 和 cosθ 都只是一次项，辅助角形式通常是最快的方法。如果方程同时含 tanθ 与 sinθ 或 cosθ，把 tanθ 写成 sinθ/cosθ 再化简。

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9. Proving Identities Rigorously | 严格证明恒等式

An identity proof requires you to show that one side of the equation is exactly equal to the other for all allowed values. Start from one side, usually the more complicated one, and manipulate it until it matches the other side.

恒等式证明要求你说明方程一侧对所有允许值都严格等于另一侧。从一侧开始，通常选择较复杂的一侧，逐步变形直到与另一侧一致。

in²θ + cos²θ = 1, secθ = 1/cosθ, and tanθ = sinθ/cosθ to rewrite expressions step by step.

证明过程中不要跨等号移项。相反，应使用已知恒等式，如 sin²θ + cos²θ = 1、secθ = 1/cosθ、tanθ = sinθ/cosθ，逐步改写表达式。

Common techniques include combining fractions over a common denominator, multiplying numerator

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