Solving Trigonometric Equations | 求解三角方程

📚 Solving Trigonometric Equations | 求解三角方程

Trigonometric equations play a central role in A-level mathematics, linking geometry, algebra, and periodic functions. Whether you are modelling wave motion, analysing vectors, or preparing for an exam, the ability to find all angles that satisfy an equation within a given interval is essential. This revision guide walks you through the core techniques required by the Edexcel specification, from basic sine and cosine equations to harmonic forms, ensuring you can approach any problem with confidence.

三角方程是A-level数学的核心内容，它连接了几何、代数与周期函数。无论你是模拟波动、分析向量还是备战考试，能够找出在给定区间内满足方程的所有角都是至关重要的能力。本复习指南将带你系统学习Edexcel考试大纲所要求的各种解题技巧，从最基本的正弦、余弦方程到辅助角公式，帮助你在面对任何题型时都能充满信心。

1. Introduction to Trigonometric Equations | 三角方程导论

cosθ − 1 = 0, or 3sinθ + 4cosθ = 2. At A-level, you are usually asked to find all solutions within a specific interval, most often 0 ≤ θ ≤ 2π (or 0° ≤ θ ≤ 360°). The periodic nature of trigonometric functions means that equations often have infinitely many solutions unless the domain is restricted, so always pay close attention to the interval given in the question.

三角方程是指包含未知角度的三角函数的方程。常见的例子有 sinθ = 0.5、2cos²θ − cosθ − 1 = 0 或 3sinθ + 4cosθ = 2。在A-level考试中，通常会要求你找出给定区间内的所有解，最常用的是 0 ≤ θ ≤ 2π（或 0° 至 360°）。由于三角函数具有周期性，如果不限制定义域，方程往往有无穷多个解。因此，务必仔细审题并明确题目给定的求解区间。

2. Basic Principles and Principal Values | 基本原理与主值

Every inverse trigonometric function gives a single principal value. For sin⁻¹(k) the principal value lies in [−π/2, π/2]; for cos⁻¹(k) it is in [0, π]; and for tan⁻¹(k) it is in (−π/2, π/2). This principal value is the one your calculator returns, but it is rarely the only solution. You must then use symmetry and periodicity to generate all other solutions that fall within the required interval. The CAST diagram and unit circle are the best tools for visualising where the required angle lies.

每一个反三角函数只会给出一个主值。sin⁻¹(k) 的主值范围是 [−π/2, π/2]；cos⁻¹(k) 的主值范围是 [0, π]；tan⁻¹(k) 的主值范围是 (−π/2, π/2)。计算器直接给出的就是主值，但这很少是唯一的解。你需要利用对称性和周期性，在所需区间内找出所有其他解。CAST 图与单位圆是帮助判断角度所在位置的最佳工具。

3. Using the Unit Circle and CAST Diagram | 使用单位圆与CAST图

The CAST diagram divides the coordinate plane into four quadrants and reminds you which trigonometric ratios are positive in each: All in Quadrant I (0 to π/2), Sin only in Quadrant II (π/2 to π), Tan only in Quadrant III (π to 3π/2), and C 0.7, you know that θ must be in Quadrant I or II. The principal value from your calculator gives the Quadrant I solution; subtracting from π gives the Quadrant II solution. For negative values, adjust symmetrically.

CAST 图将坐标平面分成四个象限，并标出了每个象限中哪些三角函数值为正：第一象限（0 至 π/2）All（全正），第二象限（π/2 至 π）仅 Sin（正弦）为正，第三象限（π 至 3π/2）仅 Tan（正切）为正，第四象限（3π/2 至 2π）仅 Cos（余弦）为正。当你求解如 sinθ = +0.7 的方程时，θ 必须位于第一或第二象限。计算器给出的主值对应第一象限的解；用 π 减去主值即可得到第二象限的解。对于负值，则利用对称性进行调整。

4. Solving sinθ = k Equations | 求解 sinθ = k 型方程

To solve sinθ = k for 0 ≤ θ ≤ 2π, first let α = sin⁻¹(|k| alue β = sin⁻¹(k); then add 2π to β to get the first solution in the interval, and π − β gives the second. Alternatively, work with the reference angle α and place solutions into Quadrants III and IV. The periodic general solutions are θ = nπ + (−1)ⁿα for integer n.

要在 0 ≤ θ ≤ 2π 内求解 sinθ = k，首先令 α = sin⁻¹(|k|)（取正值），然后判断正确的象限。若 k ≥ 0，解为 θ = α 和 θ = π − α。若 k < 0，可以求出负数主值 β = sin⁻¹(k)，然后给 β 加上 2π 得到区间内的第一个解，π − β 则给出第二个解。也可以利用参考角 α，将解放置在第三、第四象限。正弦函数的周期性通解公式为 θ = nπ + (−1)ⁿα，其中 n 为整数。

Solve sinθ = 0.5 for 0 ≤ θ ≤ 2π. α = sin⁻¹(0.5) = π/6. Solutions: θ = π/6, π − π/6 = 5π/6.

示例：在 0 ≤ θ ≤ 2π 内解 sinθ = 0.5。α = sin⁻¹(0.5) = π/6。解为 θ = π/6 和 5π/6。

0 ≤ θ ≤ 2π. β = sin⁻¹(−0.5) = −π/6. The two solutions are β + 2π = 11π/6 and π − β = 7π/6.

示例：在 0 ≤ θ ≤ 2π 内解 sinθ = −0.5。β = sin⁻¹(−0.5) = −π/6。两个解为 11π/6 和 7π/6。

5. Solving cosθ = k Equations | 求解 cosθ = k 型方程

2π] are θ = α and θ = 2π − α (provided α ≠ 0 and α ≠ π). If k is negative, α will be in Quadrant II, and again the solutions are α and 2π − α. The general solution is θ = 2nπ ± α, where n ∈ ℤ.

对于 cosθ = k，主值为 α = cos⁻¹(k)，这个值总是在 [0, π] 内。因为余弦在第一和第四象限为正，所以 [0, 2π] 内的两个解为 θ = α 和 θ = 2π − α（前提是 α ≠ 0 且 α ≠ π）。如果 k 为负值，α 将位于第二象限，解仍然为 α 和 2π − α。余弦方程的普遍解公式为 θ = 2nπ ± α，其中 n 为整数。

广告 / Sponsored

Solve cosθ = 0.5 for 0 ≤ θ ≤ 2π. α = cos⁻¹(0.5) = π/3. Solutions: π/3 and 2π − π/3 = 5π/3.

示例：在 0 ≤ θ ≤ 2π 内解 cosθ = 0.5。α = π/3。解为 π/3 和 5π/3。

Example: Solve cosθ = −1/2. α = cos⁻¹(−1/2) = 2π/3. Solutions: 2π/3 and 2π − 2π/3 = 4π/3.

示例：解 cosθ = −1/2。α = 2π/3。解为 2π/3 和 4π/3。

6. Solving tanθ = k Equations | 求解 tanθ = k 型方程

], take α and then add π repeatedly as long as the result remains inside the interval. Thus the solutions are θ = α and θ = α + π (or subtract π if α is negative and adding π still falls below 2π). The general solution is θ = nπ + α.

正切函数的周期为 π，即图像每隔 π 弧度重复一次。主值为 α = tan⁻¹(k)，落在 (−π/2, π/2) 区间内。要找出 [0, 2π] 内的所有解，首先取 α，然后反复加 π，只要结果仍在区间内。因此解为 θ = α 和 θ = α + π。（若 α 为负值，可先加 π 使其落在区间内，然后视情况再加 π。）正切方程的普遍解为 θ = nπ + α。

ple: Solve tanθ = 1 for 0 ≤ θ ≤ 2π. α = tan⁻¹(1) = π/4. Solutions: π/4 and π/4 + π = 5π/4.

示例：在 0 ≤ θ ≤ 2π 内解 tanθ = 1。α = π/4。解为 π/4 和 5π/4。

7. Equations Involving Multiple Angles | 含倍角的方程

e the new interval for u based on the original interval for θ. Solve the equation for u by finding all solutions within the expanded interval, then divide each solution by n to recover θ. This method ensures you do not miss any solutions that arise from the compressed period of the multiple-angle function.

若方程中的角度是倍角形式，例如 sin(2θ) = 0.5 或 tan(3θ) = 1，可以通过代换 u = nθ 来处理。首先根据 θ 的原始区间确定 u 的新区间，然后在扩大后的区间内求出所有关于 u 的解，最后将每个解除以 n 得到 θ。这种方法能确保你不会遗漏因倍角函数周期压缩而产生的任何解。

Example: Solve sin(2θ) = 0.5 for 0 ≤ θ ≤ 2π. Let u = 2θ, then u ∈ ve sin u = 0.5: u = π/6, 5π/6, 13π/6, 17π/6. Divide by 2: θ = π/12, 5π/12, 13π/12, 17π/12.

示例：在 0 ≤ θ ≤ 2π 内解 sin(2θ) = 0.5。令 u = 2θ，则 u ∈ [0, 4π]。解 sin u = 0.5 得 u = π/6, 5π/6, 13π/6, 17π/6。除以 2 得 θ = π/12, 5π/12, 13π/12, 17π/12。

8. Using Trigonometric Identities | 利用三角恒等式

s. The most frequently needed identities are sin²θ + cos²θ = 1, tanθ = sinθ / cosθ, and the double-angle identities. For example, to solve 3sin²θ + 2cosθ − 2 = 0, replace sin²θ with 1 − cos²θ to obtain a quadratic in cosθ. Always check that the transformed expression remains valid for all values in the domain.

许多方程会同时包含不同的函数，如 sin²θ 和 cosθ。关键是要利用恒等式将所有项化为同一个三角函数。最常用的恒等式有 sin²θ + cos²θ = 1、tanθ = sinθ / cosθ 以及二倍角公式。例如，要解 3sin²θ + 2cosθ − 2 = 0，可将 sin²θ 替换为 1 − cos²θ，得到一个关于 cosθ 的二次方程。务必检查变换后的表达式在定义域内是否始终成立。

9. Quadratic Trigonometric Equations | 二次三角方程

0 factorises to (2sinθ + 1)(sinθ − 1) = 0. Thus sinθ = −1/2 or sinθ = 1. Find all solutions for each simple equation within the required interval. Not all algebraic solutions are admissible if the value falls outside the range [−1, 1] for sine and cosine; discard any such extraneous values immediately.

如果一个方程可以写成关于 sinθ、cosθ 或 tanθ 的二次形式，可先进行因式分解，再分别求解每一个线性因子。例如 2sin²θ − sinθ − 1 = 0 可分解为 (2sinθ + 1)(sinθ − 1) = 0，因此 sinθ = −1/2 或 sinθ = 1。然后在给定区间内分别求出每个简单方程的所有解。需要注意的是，对于正弦和余弦，若方程给出的值超出 [−1, 1] 范围，则该解无效，应立即舍去。

10. Equations Requiring Factorisation | 需要因式分解的方程

side and factorise. For example, in sinθ tanθ = sinθ, moving terms gives sinθ tanθ − sinθ = 0, which factorises to sinθ (tanθ − 1) = 0. This yields sinθ = 0 or tanθ = 1, giving all solutions without loss. Always factorise rather than cancel.

千万不要随意地将方程两边同时除以一个三角函数表达式，因为这样可能会漏掉有效解。正确的做法是把所有项移到同一边，然后进行因式分解。例如，对于 sinθ tanθ = sinθ，移项得 sinθ tanθ − sinθ = 0，提取公因式 sinθ (tanθ − 1) = 0，从而得到 sinθ = 0 或 tanθ = 1，这样就能毫无遗漏地求出所有解。记住，能用因式分解就不要约分。

11. Solving a sinθ + b cosθ = c Using Harmonic Form | 利用辅助角公式求解 a sinθ + b cosθ = c

le sine (or cosine) function. The harmonic form is R sin(θ ± α) or R cos(θ ± α), where R = √a² + b² sθ = 2 becomes 5 sin(θ + 0.927) = 2, so sin(θ + α) = 0.4. Solve for θ + α, then subtract α

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）