📚 Solving Trigonometric Equations | 解三角函数方程
Trigonometric equations are a central topic in Edexcel A-Level Pure Mathematics. They require you to find all angles, in degrees or radians, that satisfy a given equation within a specified interval. Success depends on fluent use of identities, the CAST diagram, and algebraic manipulation. This guide distills the key methods and common pitfalls.
三角函数方程是Edexcel A-Level纯数中的核心内容，要求你找出在指定区间内所有满足方程的角（角度制或弧度制）。顺利解题的关键在于熟练运用恒等式、CAST图以及代数变形技巧。本文凝练了最重要的方法和常见误区。
1. The CAST Diagram and Signs of Ratios | CAST图与三角比的符号
The CAST diagram labels the quadrants of the unit circle to show which trigonometric ratios are positive. Moving anticlockwise: Quadrant I (All positive), Quadrant II (Sin positive), Quadrant III (Tan positive), Quadrant IV (Cos positive). The word 'CAST' helps you remember All, Sin, Tan, Cos.
CAST图通过标记单位圆的象限来展示哪些三角比为正值。逆时针方向依次为：第一象限（全部为正），第二象限（正弦为正），第三象限（正切为正），第四象限（余弦为正）。单词“CAST”可以帮助记忆All, Sin, Tan, Cos的顺序。
| Quadrant | sin θ | cos θ | tan θ |
|---|---|---|---|
| I (0°–90°) | + | + | + |
| II (90°–180°) | + | – | – |
| III (180°–270°) | – | – | + |
| IV (270°–360°) | – | + | – |
nce sine is positive in QI and QII, the solutions between 0° and 360° are θ = 30° and 180° – 30° = 150°.
要解 sin θ = 0.5 这类方程，先求出主值 α = sin⁻¹(0.5) = 30°。由于正弦在第一、二象限为正，在0°到360°之间解为 θ = 30° 和 180° – 30° = 150°。
The general solution for sine uses the pattern θ = 180°n + (−1)ⁿα, where n is an integer. This compact form generates all coterminal solutions.
正弦的通解公式为 θ = 180°n + (−1)ⁿα，其中 n 为整数。这个紧凑形式可以生成所有终边相同的解。
For cosine, the symmetry about the x-axis gives θ = 360°n ± α. For tangent, which repeats every 180°, θ = 180°n + α.
对于余弦，利用关于x轴的对称性可得 θ = 360°n ± α；正切则以180°为周期，通解为 θ = 180°n + α。
2. Principal Values and Setting the Right Interval | 主值与确定正确区间
Your calculator gives the principal value of an inverse trigonometric function: sin⁻¹, cos⁻¹ return angles in [−90°, 90°] and [0°, 180°] respectively; tan⁻¹ gives a value in (−90°, 90°). Always interpret this result using the CAST diagram before generating additional solutions.
计算器给出的反三角函数值为主值：sin⁻¹ 的结果在 [−90°, 90°] 区间，cos⁻¹ 在 [0°, 180°]，tan⁻¹ 在 (−90°, 90°)。拿到主值后，务必结合CAST图判断其他解。
Edexcel exam questions typically specify an interval such as 0° ⩽ θ ⩽ 360° or 0 ⩽ θ < 2π. When working in radians, use the equivalent general forms: sin: θ = nπ + (−1)ⁿα, cos: θ = 2nπ ± α, tan: θ = nπ + α. Always check whether the interval is open or closed at the endpoints.
Edexcel考试题通常会给出区间，如0° ⩽ θ ⩽ 360°或0 ⩽ θ < 2π。处理弧度制时，通解公式相应地变为：sin: θ = nπ + (−1)ⁿα，cos: θ = 2nπ ± α，tan: θ = nπ + α。务必注意区间端点是否包含。
Write down all values of n that keep θ inside the required interval. Many marks are lost by stopping after finding only the principal value.
列出所有使 θ 保持在要求区间内的整数 n。很多学生仅求出主值就停笔，白白丢分。
3. Equations of the Form sin θ = k, cos θ = k, tan θ = k | sin θ = k 等基本方程
alculate α = sin⁻¹(0.7) ≈ 44.4°. Since sine is positive in QI and QII, the solutions in 0°–360° are θ ≈ 44.4° and 180° – 44.4° = 135.6°.
解 sin θ = 0.7 时，先算 α = sin⁻¹(0.7) ≈ 44.4°。正弦在第一、二象限为正，因此在0°–360°内解为 θ ≈ 44.4° 和 180° – 44.4° = 135.6°。
an⁻¹(1) = 45° (ignore the sign initially). Tangent is negative in QII and QIV, so θ = 180° – 45° = 135° and 360° – 45° = 315°. Using symmetry always gives correct solutions.
对于 tan θ = −1，先不考虑符号，求出 α = tan⁻¹(1) = 45°。正切在第二、四象限为负，因此 θ = 180° – 45° = 135° 和 360° – 45° = 315°。利用对称性能保证解的正确性。
cos θ = −0.8, α = cos⁻¹(0.8) ≈ 36.9°. Cosine is negative in QII and QIII, giving θ ≈ 180° – 36.9° = 143.1° and 180° + 36.9° = 216.9°.
对于 cos θ = −0.8，α = cos⁻¹(0.8) ≈ 36.9°。余弦在第二、三象限为负，因此 θ ≈ 180° – 36.9° = 143.1° 和 180° + 36.9° = 216.9°。
4. Using the Pythagorean Identity sin²θ + cos²θ = 1 | 使用平方恒等式
The identity sin²θ + cos²θ ≡ 1 allows you to replace sin²θ with 1 – cos²θ, or cos²θ with 1 – sin²θ. This converts an equation with mixed ratios into a quadratic in a single trig function.
恒等式 sin²θ + cos²θ ≡ 1 让我们可以将 sin²θ 替换为 1 – cos²θ，或将 cos²θ 替换为 1 – sin²θ，从而将含有混合三角比的方程转化为只含单一三角函数的二次方程。
Example: Solve 3sin²θ + 2cos θ – 3 = 0 for 0° ⩽ θ ⩽ 360°. Replace sin²θ with 1 – cos²θ: 3(1 – cos²θ) + 2cos θ – 3 = 0 → −3cos²θ + 2cos θ = 0 → cos θ (3cos θ – 2) = 0. So cos θ = 0 or cos θ = 2/3.
例：解 3sin²θ + 2cos θ – 3 = 0，区间 0° ⩽ θ ⩽ 360°。将 sin²θ 替换为 1 – cos²θ：3(1 – cos²θ) + 2cos θ – 3 = 0 → −3cos²θ + 2cos θ = 0 → cos θ (3cos θ – 2) = 0。于是 cos θ = 0 或 cos θ = 2/3。
θ = 2/3, α ≈ 48.2° gives θ ≈ 48.2°, 311.8°. Always list all solutions within the interval.
由 cos θ = 0 得 θ = 90°, 270°；由 cos θ = 2/3，α ≈ 48.2° 得 θ ≈ 48.2°, 311.8°。永远在区间内列出全部解。
5. Exploiting the tanθ = sinθ / cosθ Identity | 利用正切恒等式
≠ 0), yielding 2 tan θ = 3, so tan θ = 1.5. This avoids juggling two different functions.
像 2sin θ = 3cos θ 这样的方程，只需两边除以 cos θ（确保 cos θ ≠ 0），即得 2 tan θ = 3，tan θ = 1.5。这能避免同时处理两种不同的三角函数。
nal equation becomes 2(±1) = 3(0) → ±2 = 0, it is impossible, so no extra solutions arise.
务必单独验证 cos θ = 0 是否为原方程的解。代入原方程得 2(±1) = 3(0) → ±2 = 0，不可能成立，因此没有额外解。
solving equations that appear in the form a sin θ = b cos θ + c after using harmonic form.
该恒等式在证明其他关系和求解形如 a sin θ = b cos θ + c 的方程时也非常有用，尤其是在结合辅助角公式时。
6. Double Angle Formulas | 倍角公式
The double angle formulas you must know for Edexcel are:
Edexcel考纲要求你掌握以下倍角公式：
sin 2θ ≡ 2 sin θ cos θ
cos 2θ ≡ cos²θ − sin²θ ≡ 2cos²θ − 1 ≡ 1 − 2sin²θ
tan 2θ ≡ 2 tan θ / (1 − tan²θ)
le to a single angle or to factorise expressions. For instance, solve sin 2θ = cos θ for 0 ⩽ θ < 2π. Use sin 2θ = 2 sin θ cos θ: 2 sin θ cos θ = cos θ → cos θ (2 sin θ − 1) = 0.
倍角公式可将多倍角化为单角或进行因式分解。例如，解 sin 2θ = cos θ，区间 0 ⩽ θ < 2π。利用 sin 2θ = 2 sin θ cos θ：2 sin θ cos θ = cos θ → cos θ (2 sin θ − 1) = 0。
Solutions: cos θ = 0 gives θ = π/2, 3π/
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