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Trigonometric Identities & Equations | 三角恒等式与方程

2026-08-02 17:43 · 本文部分内容由AI辅助创作

📚 Trigonometric Identities & Equations | 三角恒等式与方程

In A-Level Edexcel Mathematics, mastering trigonometric identities and equations is essential for success in Pure Mathematics. These identities allow you to simplify complicated expressions, prove relationships, and solve equations that model periodic behaviour. This article covers all key identities, effective solving strategies, common pitfalls, and exam-style questions to help you build confidence.

在 A-Level Edexcel 数学中，掌握三角恒等式与方程是纯数部分取得高分的关键。这些恒等式能帮助你化简复杂表达式、证明三角关系，并求解描述周期现象的方程。本文涵盖所有核心恒等式、高效求解策略、常见错误以及真题风格的练习，帮助你建立扎实的解题能力。

1. Reciprocal and Pythagorean Identities | 倒数恒等式与毕达哥拉斯恒等式

The foundational identities link the six trigonometric functions. The reciprocal identities define secant, cosecant, and cotangent:

最基本的恒等式将六个三角函数联系在一起。倒数恒等式定义了正割、余割和余切：

sec θ = 1/cos θ,    csc θ = 1/sin θ,    cot θ = 1/tan θ

sec θ = 1/cos θ,    csc θ = 1/sin θ,    cot θ = 1/tan θ

The Pythagorean identities are derived from the unit circle and are used frequently to rewrite expressions in terms of a single trigonometric ratio:

毕达哥拉斯恒等式源自单位圆，经常用于将表达式转化为只含一个三角比的形式：

sin2θ + cos2θ ≡ 1

1 + tan2θ ≡ sec2θ

1 + cot2θ ≡ csc2θ

Remember to use these identities when you need to replace sin2θ with 1 - cos2θ, or to turn a mix of tan2θ and sec2θ into a single function. The identity 1 + tan2θ ≡ sec2θ is especially useful for solving equations involving both tan and sec.

当你需要将 sin2θ 替换为 1 - cos2θ，或把 tan2θ 和 sec2θ 的混合表达式统一为一个函数时，要想到这些恒等式。恒等式 1 + tan2θ ≡ sec2θ 在同时包含 tan 和 sec 的方程中尤为有用。

2. Compound Angle Formulae | 和角公式与差角公式

al for solving equations like sin(x + 30°) = 0.5 and for deriving double-angle identities.

和角公式及差角公式可以将两个角度和或差的三角函数展开。它们在求解 sin(x + 30°) = 0.5 这样的方程以及推导二倍角公式时至关重要。

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)

Notice the subtle sign change in the cosine formula: for cos(A + B) it is minus, and for cos(A - B) it is plus. The tan formula has the same sign in the numerator as the original sign of the angle, but the denominator sign flips.

注意余弦公式中符号的变化：cos(A + B) 中间是减号，cos(A - B) 中间是加号。正切公式中，分子上的符号与角度间的符号一致，但分母上的符号相反。

t the full expansion and substitute known exact values to obtain a simplified surd answer.

应用这些公式时，你可能会被要求求精确值，例如 cos 15° = cos(45° - 30°)。需要写出完整的展开式，代入已知的精确值，最后化简成带根号的答案。

3. Double Angle Formulae | 二倍角公式

mpact but powerful tools for solving trig equations and for integration (in later topics).

二倍角恒等式可直接由和角公式令 A = B = θ 得到。它们形式简洁，但在求解三角方程和后续积分课题中应用广泛。

sin 2θ = 2 sin θ cos θ

cos 2θ = cos2θ - sin2θ

cos 2θ = 2 cos2θ - 1

cos 2θ = 1 - 2 sin2θ

tan 2θ = (2 tan θ) / (1 - tan2θ)

The three versions of cos 2θ are interchangeable. Choosing the right form is critical: if an equation involves sin2θ, use cos 2θ = 1 - 2 sin2θ; if it involves cos2θ, use cos 2θ = 2 cos2θ - 1. This technique reduces equations to a single trigonometric function, enabling straightforward solving.

cos 2θ 的三种形式可以互相转换。选择合适的形式至关重要：如果方程中出现 sin2θ，则用 cos 2θ = 1 - 2 sin2θ；如果出现 cos2θ，则用 cos 2θ = 2 cos2θ - 1。这一技巧可将方程化为只含一个三角比的形式，从而顺利求解。

4. Factor Formulae (Sum-to-Product) | 和差化积公式

roving identities and for solving equations that contain sums such as sin 5θ + sin 3θ = 0.

和差化积公式能将正弦或余弦的和或差转化为乘积。它们在证明恒等式和求解含有 sin 5θ + sin 3θ = 0 这样和式的方程时非常方便。

sin P + sin Q = 2 sin((P+Q)/2) cos((P-Q)/2)

sin P - sin Q = 2 cos((P+Q)/2) sin((P-Q)/2)

cos P + cos Q = 2 cos((P+Q)/2) cos((P-Q)/2)

cos P - cos Q = -2 sin((P+Q)/2) sin((P-Q)/2)

When you see a sum of two sines or two cosines, think of these formulae immediately. After factorising, you can set each factor to zero, converting a tricky sum into two simpler equations.

一旦看到两个正弦或两个余弦的和，应立刻想到这些公式。因式分解后，可以令每个因子为零，从而将一个复杂的和式转化为两个简单的方程。

ou to show that sin 3θ + sin θ = 2 sin 2θ cos θ, then solve sin 3θ + sin θ = 0 for 0 ≤ θ ≤ π.

这些恒等式收录在 Edexcel 公式手册中，但你必须能识别何时以及如何使用它们。典型的考题会要求你证明 sin 3θ + sin θ = 2 sin 2θ cos θ，然后求解 sin 3θ + sin θ = 0 在 0 ≤ θ ≤ π 内的所有解。

5. Solving Basic Trigonometric Equations | 基本三角方程的求解

e solution process relies on quadrant knowledge and the periodic nature of trig functions.

最简单的三角方程形式为 sin x = k、cos x = k 或 tan x = k。求解过程依赖于象限知识以及三角函数的周期性。

• For sin x = k: find the principal value α = sin-1 /2 ≤ α ≤ π/2). Then general solutions: x = α + 360°n or x = 180° - α + 360°n (in degrees).

• 对于 sin x = k：先求出主值 α = sin-1(k)，范围在 -90° ≤ α ≤ 90°（或 -π/2 ≤ α ≤ π/2）之间。通解为：x = α + 360°n 或 x = 180° - α + 360°n（角度制）。

• For cos x = k: α = cos-1(k) in 0 ≤ α ≤ 180° (0 ≤ α ≤ π). Then x = ±α + 360°n.

• 对于 cos x = k：α = cos-1(k) 在 0 ≤ α ≤ 180° (0 ≤ α ≤ π) 之间。通解为 x = ±α + 360°n。

• For tan x = k: α = tan-1 < α < 90° (-π/2 < α < π/2). Then x = α + 180°n, because the period of tan is 180° (π rad).

• 对于 tan x = k：α = tan-1(k) 在 -90° < α < 90° (-π/2 < α < π/2) 之间。因为 tan 的周期是 180° (π rad)，通解为 x = α + 180°n。

Always check the given interval and list all solutions that fall within it. Sketching the graph or using CAST diagram helps to avoid missing solutions.

务必检查给定的区间，列出落在该区间的所有解。快速画出函数图像或使用 CAST 图，有助于避免漏解。

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6. Using Identities to Solve Equations | 运用恒等式求解方程

Most A-Level equations require you to apply an identity to rewrite the equation in terms of a single trig function. For example, if you see 2 sin2x - cos x = 1, replace sin2x with 1 - cos2x to obtain a quadratic in cos x.

大多数 A-Level 方程需要先运用恒等式将方程化为仅含一个三角比的形式。例如，遇到 2 sin2x - cos x = 1，可将 sin2x 替换为 1 - cos2x，从而得到一个关于 cos x 的二次方程。

A systematic strategy works well:

一个系统性的策略非常有效：

Watch out for hidden identities: equations like sec2x - 3 tan x = 5 should prompt you to use sec2x = 1 + tan2x, creating a quadratic in tan x.

要注意隐藏的恒等式：比如 sec2x - 3 tan x = 5 这样的方程，应联想到 sec2x = 1 + tan2x，从而生成一个关于 tan x 的二次方程。

7. Quadratic Trigonometric Equations | 二次三角方程

Equations of the form a sin2θ + b sin θ + c = 0 or a cos2 ar regularly. Treat them like standard quadratics: factorise or use the quadratic formula.

形如 a sin2θ + b sin θ + c = 0 或 a cos2θ + b cos θ + c = 0 的方程经常出现。将其当作普通二次方程来处理：因式分解或使用求根公式。

Example: Solve 3 cos2x - 5 cos x + 2 = 0 for 0 ≤ x ≤ 360°.

例：解 3 cos2x - 5 cos x + 2 = 0，区间 0 ≤ x ≤ 360°。

Factorising: (3 cos x - 2)(cos x - 1) = 0 ⇒ cos x = 2/3 or cos x = 1.

因式分解：(3 cos x - 2)(cos x - 1) = 0 ⇒ cos x = 2/3 或 cos x = 1。

Then x = cos-1 0° - 48.2° = 311.8°. From cos x = 1, x = 0°, 360°. Keep all solutions within the interval.

然后 x = cos-1(2/3) ≈ 48.2°，由对称性得 x = 360° - 48.2° = 311.8°。由 cos x = 1 得 x = 0° 和 360°。保留所有落在区间内的解。

If factorisation is not obvious, use the formula: sin θ = [-b ± √b2 - 4ac] / (2a), but ensure the resulting values lie between -1 and 1 for sin or cos; discard any values outside this range.

如果难以因式分解，可用公式 sin θ = [-b ± √b2 - 4ac] / (2a)，但要确保得到的值介于 -1 和 1 之间（对 sin 和 cos 而言）；舍弃超出该范围的值。

8. Equations with Multiple Angles | 多倍角方程

When the argument is a multiple angle, such as sin 2x = 0.5 or tan(3x - 10°) = √3 he given interval for x is 0° ≤ x ≤ 360°, then for 2x the interval becomes 0° ≤ 2x ≤ 720°.

当角度为多倍角时，例如 sin 2x = 0.5 或 tan(3x - 10°) = √3，必须在列解之前调整区间。假设给定 x 的区间为 0° ≤ x ≤ 360°，那么对于 2x，区间变为 0° ≤ 2x ≤ 720°。

s for u in the expanded interval using the standard methods. Then divide by the coefficient to recover x. For 3x - 10°, first solve for 3x - 10°, add 10°, then divide by 3.

首先求解多倍角变量：令 u = 2x。在扩展后的区间上用标准方法求出 u 的所有解，再除以系数得到 x。对于 3x - 10°，先解出 3x - 10° 的值，加 10° 后再除以 3。

en x = 15°, 75°. No further solutions from extra periods because 30°+360° is out of range.

例：sin 2x = 0.5，0 ≤ x ≤ 180°。令 u = 2x，得 0 ≤ u ≤ 360°。sin u = 0.5 → u = 30°, 150°，所以 x = 15° 和 75°。因 30°+360° 超出范围，不会再有其他解。

Multiple-angle equations often yield more solutions hidden within the expanded interval. Extend your "solution search" by adding or subtracting the period as many times as needed while still within the expanded interval.

多倍角方程常常会在扩展区间内隐藏更多解。只要还在扩展区间内，可以通过不断加减周期来延伸“解搜索”的范围。

9. Proving Trigonometric Identities | 证明三角恒等式

Proof questions require you to show that one side of an equation is equivalent to the other. Start from the more complicated side and transform it step by step until it matches the simpler side.

证明题要求你展示等式一边与另一边等价。通常从较复杂的一边入手，逐步变形，直至与较简单的一边相同。

Common techniques include:

常用技巧包括：

Example: Prove that (1 - cos 2θ) / sin 2θ = tan θ. LHS: use cos 2θ = 1 - 2 sin2θ, sin 2θ = 2 sin θ cos θ. Numerator becomes 2 sin2 ominator 2 sin θ cos θ = 2 sin θ cos θ. Cancel 2 sin θ to get sin θ / cos θ = tan θ = RHS.

例：证明 (1 - cos 2θ) / sin 2θ = tan θ。左边：用 cos 2θ = 1 - 2 sin2θ，sin 2θ = 2 sin θ cos θ。分子变为 2 sin2θ，分母为 2 sin θ cos θ。约去 2 sin θ，得 sin θ / cos θ = tan θ = 右边。

Always write "LHS = ... = RHS" or "RHS = ... = LHS" w clear. Avoid treating the identity like an equation to be solved; work on one side only.

始终写出“左边 = ... = 右边”或“右边 = ... = 左边”，使逻辑脉络清晰。不要把恒等式当作方程去解；只能对一边进行操作。

10. Applications in Modelling | 三角学在建模中的应用

ations, and pendulum motion. A typical model might be h(t) = a sin(bt + c) + d or similar.

三角函数可用于模拟潮汐、温度变化和钟摆运动等周期现象。典型的模型可能是 h(t) = a sin(bt + c) + d 或类似形式。

To answer modelling questions:

解答建模题时：

ind multiple times within a given interval, reflecting the periodic nature. Use the general solution approach learned earlier, adjusting for the phase shift.

在求解 h(t) = k 时，可能需要在给定区间内找出多个时刻，这反映了周期性。运用之前学过的通解方法，并根据相位移进行调整即可。

Differentiation of trig functions is also used to find rates of change in these models, linking pure trigonometry with calculus.

三角函数的微分也常用于求这些模型中的变化率，将纯三角学与微积分紧密结合。

11. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Even strong students lose marks on trigonometric equations. Here are the most frequent pitfalls and how to sidestep them:

即使是程度较好的学生，也会在三角方程上丢分。以下是最常见的陷阱及应对策略：