Trigonometric Identities and Equations | 三角函数恒等式与方程

📚 Trigonometric Identities and Equations | 三角函数恒等式与方程

This revision guide covers the Edexcel A-Level Pure Mathematics topic of trigonometric identities and equations. You will learn how to use the unit circle, exact values, CAST diagrams, identities and harmonic form to solve equations and prove statements.

本复习指南涵盖 Edexcel A-Level 纯数学中的三角函数恒等式与方程主题。你将学习如何使用单位圆、特殊角精确值、CAST 图、恒等式和辅助角形式来解方程和证明三角命题。

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1. The Unit Circle and Quadrants | 单位圆与象限

The unit circle has radius 1 centred at the origin. A point on the circle making an angle θ with the positive x-axis has coordinates (cos θ, sin θ).

单位圆以原点为圆心、半径为 1。圆上一点与 x 轴正方向成角 θ，其坐标为 (cos θ, sin θ)。

Angles are measured anticlockwise as positive and clockwise as negative. The signs of sin θ, cos θ and tan θ depend on the quadrant.

角度以逆时针为正、顺时针为负。sin θ、cos θ 和 tan θ 的符号取决于象限。

In quadrant 1 all three ratios are positive. In quadrant 2 only sin θ is positive, in quadrant 3 only tan θ is positive, and in quadrant 4 only cos θ is positive.

在第一象限三种比值均为正；第二象限只有 sin θ 为正；第三象限只有 tan θ 为正；第四象限只有 cos θ 为正。

(cos θ, sin θ) = (x, y) on the unit circle

单位圆上的点坐标直接对应余弦和正弦值。

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2. Exact Values for Special Angles | 特殊角的精确值

You must know exact trig values for 0°, 30°, 45°, 60° and 90°, in both degrees and radians.

你必须掌握 0°、30°、45°、60° 和 90° 的精确三角值，包括角度制和弧度制。

sin 30° = ½, sin 45° = √2/2, sin 60° = √3/2, cos 30° = √3/2, cos 45° = √2/2, cos 60° = ½, tan 30° = 1/√3, tan 45° = 1, tan 60° = √3

例如 sin 30° = ½，cos 60° = ½，tan 45° = 1；这些值常用于求解方程和化简表达式。

A common method is to sketch the two special triangles: the 45° right triangle with legs 1 and 1, and the 30°-60° right triangle with sides 1, √3 and 2.

常用方法是画出两个特殊直角三角形：45° 直角三角形两直角边为 1 和 1；30°-60° 直角三角形三边为 1、√3 和 2。

You should also express angles in radians: 30° = π/6, 45° = π/4, 60° = π/3, 90° = π/2.

你还应能将角度转换为弧度：30° = π/6，45° = π/4，60° = π/3，90° = π/2。

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3. Core Trigonometric Identities | 核心三角恒等式

The Pythagorean identity is sin²θ + cos²θ ≡ 1. Dividing by cos²θ gives tan²θ + 1 ≡ sec²θ, and dividing by sin²θ gives 1 + cot²θ ≡ cosec²θ.

毕达哥拉斯恒等式为 sin²θ + cos²θ ≡ 1。将它除以 cos²θ 得 tan²θ + 1 ≡ sec²θ；除以 sin²θ 得 1 + cot²θ ≡ cosec²θ。

sin²θ + cos²θ ≡ 1

tan²θ + 1 ≡ sec²θ

1 + cot²θ ≡ cosec²θ

These identities are valid for all values of θ for which the functions are defined, so they use the identity symbol ≡ rather than =.

这些恒等式对所有使其有意义的 θ 都成立，因此使用恒等号 ≡ 而不是等号 =。

You will often use sin²θ = 1 – cos²θ or cos²θ = 1 – sin²θ to change a quadratic expression into one variable.

解题中常使用 sin²θ = 1 – cos²θ 或 cos²θ = 1 – sin²θ，将二次表达式化为单一变量的方程。

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4. Solving Simple Trigonometric Equations | 解简单三角方程

To solve sin θ = k, cos θ = k or tan θ = k, first find the principal value using your calculator, then use symmetry and periodicity to find all solutions in the required interval.

解 sin θ = k、cos θ = k 或 tan θ = k 时，先用计算器求出主值，再利用对称性和周期性求出指定区间内的所有解。

For sin θ, general solutions are θ = 180°n + (-1)ⁿα in degrees, or θ = nπ + (-1)ⁿα in radians, where α is the principal value.

对于 sin θ，通解为 θ = 180°n + (-1)ⁿα（度）或 θ = nπ + (-1)ⁿα（弧度），其中 α 是主值。

For cos θ, general solutions are θ = 360°n ± α in degrees, or θ = 2nπ ± α in radians.

对于 cos θ，通解为 θ = 360°n ± α（度）或 θ = 2nπ ± α（弧度）。

For tan θ, general solutions are θ = 180°n + α in degrees, or θ = nπ + α in radians.

对于 tan θ，通解为 θ = 180°n + α（度）或 θ = nπ + α（弧度）。

Always check the domain, such as 0° ≤ θ ≤ 360° or 0 ≤ θ < 2π, and list every solution in that range.

务必检查定义域，例如 0° ≤ θ ≤ 360° 或 0 ≤ θ < 2π，并列出该范围内的所有解。

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5. Using the CAST Diagram | 使用 CAST 图

The CAST diagram labels the quadrants in which cos, all, sin and tan are positive, moving anticlockwise from quadrant 4 to quadrant 1, 2 and 3.

CAST 图按逆时针从第四象限到第一、第二、第三象限，标出 cos、all、sin、tan 为正的区域。

After finding an acute related angle, place the angle in the correct quadrants according to the sign of the trig ratio.

求出锐角相关角后，根据三角比的符号把角放到正确象限中。

For example, to solve sin θ = 0.5 for 0° ≤ θ ≤ 360°, the calculator gives α = 30°. Sin is positive in quadrants 1 and 2, so θ = 30° and θ = 180° – 30° = 150°.

例如，解 sin θ = 0.5（0° ≤ θ ≤ 360°）时，计算器给出 α = 30°。sin 在第一、二象限为正，所以 θ = 30° 和 θ = 180° – 30° = 150°。

If the equation were sin θ = -0.5, sin is negative in quadrants 3 and 4, so θ = 180° + 30° = 210° and θ = 360° – 30° = 330°.

若方程是 sin θ = -0.5，sin 在第三、四象限为负，因此 θ = 180° + 30° = 210° 和 θ = 360° – 30° = 330°。

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6. Quadratic Trigonometric Equations | 二次三角方程

Equations such as 2sin²θ + 3sin θ – 2 = 0 can be solved by treating sin θ as a variable and factorising.

形如 2sin²θ + 3sin θ – 2 = 0 的方程，可把 sin θ 当作变量进行因式分解来解。

Example: Solve 2sin²θ – sin θ – 1 = 0 for 0° ≤ θ ≤ 360°. Factorise to get (2sin θ + 1)(sin θ – 1) = 0, so sin θ = -1/2 or sin θ = 1.

例如，解 2sin²θ – sin θ – 1 = 0（0° ≤ θ ≤ 360°）。因式分解得 (2sin θ + 1)(sin θ – 1) = 0，因此 sin θ = -1/2 或 sin θ = 1。

From sin θ = -1/2, the related angle is 30°, so θ = 210°, 330°. From sin θ = 1, θ = 90°. The full solution set is {90°, 210°, 330°}.

由 sin θ = -1/2，相关角为 30°，所以 θ = 210°、

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