📚 Solving Trigonometric Equations and Identities for Edexcel A-Level Pure Maths | 攻克 Edexcel A-Level 纯数学：三角方程与恒等式
Trigonometric equations and identities are central to Edexcel A-Level Pure Mathematics, appearing in Paper 1 and Paper 2 as well as in applied contexts such as simple harmonic motion. This revision guide works through the core methods from radian measure to the R-formula, with paired explanations to help you build both fluency and exam confidence.
三角方程与恒等式是 Edexcel A-Level 纯数学的核心内容，出现在 Paper 1 和 Paper 2 以及简谐运动等应用情境中。本复习指南从弧度制到 R 公式，系统梳理核心方法，并配有中英对照讲解，帮助你提升熟练度与应试信心。
1. Radian Measure and the Unit Circle | 弧度制与单位圆
Before solving any trigonometric equation, you must be comfortable with radians. Edexcel expects exact answers in terms of π unless a question explicitly asks for degrees. The unit circle gives a geometric way to read off sine, cosine and tangent values, and it helps you understand why equations have multiple solutions in a given interval.
在解任何三角方程之前，你必须熟练掌握弧度制。除非题目明确要求用角度制，Edexcel 通常要求以 π 给出精确答案。单位圆提供了一种几何方式来读取正弦、余弦和正切的值，也有助于你理解为什么方程在给定区间内会有多个解。
180° = π rad
On the unit circle, the x-coordinate of a point is cos θ and the y-coordinate is sin θ. Because the circle repeats every full turn, adding 2π to any angle gives the same trigonometric values.
在单位圆上，一个点的 x 坐标是 cos θ，y 坐标是 sin θ。由于圆每转一圈重复一次，因此任何角度加上 2π 后，三角函数值都相同。
2. Standard Trigonometric Values and CAST Diagram | 标准三角值与 CAST 图
Learn the exact values for 0, π/6, π/4, π/3 and π/2. The CAST diagram tells you which trigonometric functions are positive in each quadrant: Cosine in the fourth, All in the first, Sine in the second, Tangent in the third. This is essential for finding all solutions in a given range.
熟记 0、π/6、π/4、π/3 和 π/2 的精确值。CAST 图告诉你每个象限中哪些三角函数为正：第四象限余弦为正，第一象限全为正，第二象限正弦为正，第三象限正切为正。这对于找出给定范围内的所有解至关重要。
- Quadrant I: All positive | 第一象限：全部为正
- Quadrant II: Sin positive | 第二象限：正弦为正
- Quadrant III: Tan positive | 第三象限：正切为正
- Quadrant IV: Cos positive | 第四象限：余弦为正
| θ | 0 | π/6 | π/4 | π/3 | π/2 |
|---|---|---|---|---|---|
| sin θ | 0 | ½ | √2/2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | √2/2 | ½ | 0 |
| tan θ | 0 | 1/√3 | 1 | √3 | undefined |
Being able to recall these values instantly saves time and reduces sign errors, especially when you are working through CAST quadrant logic under exam pressure.
能够迅速回忆这些数值可以节省时间并减少符号错误，尤其是在考试压力下应用 CAST 象限逻辑时尤为重要。
3. Solving Basic Sine Equations | 解基本正弦方程
To solve sin θ = k, first find the principal value θ₁ = arcsin k. Then use symmetry: the second solution in one full turn is θ₂ = π − θ₁. After that, add multiples of 2π to write the general solution or list all values inside the requested interval.
要解 sin θ = k，先求主值 θ₁ = arcsin k。然后利用对称性：一个完整周期内的第二个解是 θ₂ = π − θ₁。之后加上 2π 的整数倍即可写出通解，或列出指定区间内的所有值。
sin θ = k ⇒ θ = θ₁ + 2nπ or θ = π − θ₁ + 2nπ
Always draw a sketch or mark the CAST diagram before listing solutions. This helps you avoid missing the second solution, which is one of the most common errors in A-Level trigonometry.
在列出解之前，一定要画出草图或标记 CAST 图。这可以帮助你避免漏掉第二个解，这是 A-Level 三角学中最常见的错误之一。
4. Solving Basic Cosine and Tangent Equations | 解基本余弦与正切方程
For cos θ = k, the two solutions in one period are θ₁ and −θ₁, which can also be written as 2π − θ₁. The cosine graph is symmetric about the y-axis and about the line θ = π, so this sign pattern is very reliable.
对于 cos θ = k，一个周期内的两个解是 θ₁ 和 −θ₁，也可以写成 2π − θ₁。余弦图像关于 y 轴以及直线 θ = π 对称，因此这种符号规律非常可靠。
cos θ = k ⇒ θ = ±θ₁ + 2nπ
For tan θ = k, the graph repeats every π, so once you have one solution θ₁, the general solution is θ = θ₁ + nπ. There is no need to find a second solution using symmetry because tangent has a period of π, not 2π.
对于 tan θ = k，图像每 π 重复一次，因此只要有一个解 θ₁，通解就是 θ = θ₁ + nπ。由于正切函数的周期是 π 而不是 2π，因此不需要利用对称性求第二个解。
tan θ = k ⇒ θ = θ₁ + nπ
5. Using Pythagorean Identities | 使用勾股恒等式
When an equation mixes sin θ and cos θ, the identity sin² θ + cos² θ = 1 often lets you rewrite everything in terms of one function. The rearranged forms sin² θ = 1 − cos² θ and cos² θ = 1 − sin² θ are especially useful in quadratic equations.
当方程同时含有 sin θ 和 cos θ 时，恒等式 sin² θ + cos² θ = 1 通常可以让你把所有项都化为同一个函数。变形形式 sin² θ = 1 − cos² θ 和 cos² θ = 1 − sin² θ 在二次方程中尤其有用。
sin² θ + cos² θ = 1
There are also two derived identities that appear in harder questions: 1 + tan² θ = sec² θ and 1 + cot² θ = cosec² θ. These are obtained by dividing the main identity by cos² θ or sin² θ respectively, and they are useful when a question involves sec, cosec or cot.
还有两个衍生恒等式会出现在较难的题目中：1 + tan² θ = sec² θ 和 1 + cot² θ = cosec² θ。它们分别由主恒等式除以 cos² θ 或 sin² θ 得到，在涉及 sec、cosec 或 cot 的题目中非常有用。
6. Double Angle Identities | 二倍角恒等式
The double angle formulas are essential for many exam questions. Remember the three forms of cos 2θ: cos² θ − sin² θ, 2cos² θ − 1 and 1 − 2sin² θ. Choosing the right form often depends on whether you want to express the equation in terms of cos θ or sin θ.
二倍角公式在许多考试题中都必不可少。记住 cos 2θ 的三种形式：cos² θ − sin² θ、2cos² θ − 1 和 1 − 2sin² θ。选择哪种形式通常取决于你想把方程表示成 cos θ 还是 sin θ 的函数。
sin 2θ = 2 sin θ cos θ
cos 2θ = cos² θ − sin² θ = 2cos² θ − 1 = 1 − 2sin² θ
For example, if an equation contains cos 2θ and sin θ, writing cos 2θ as 1 − 2sin² θ will turn the whole equation into a quadratic in sin θ. This is a standard Edexcel technique.
例如，如果一个方程同时含有 cos 2θ 和 sin θ，把 cos 2θ 写成 1 − 2sin² θ 就能将整个方程转化为关于 sin θ 的二次方程。这是 Edexcel 的标准解题技巧。
7. Quadratic Trigonometric Equations | 二次三角方程
A typical exam problem asks you to solve something like 2cos² θ + 3sin θ = 3. Use an identity to rewrite cos² θ in terms of sin θ, rearrange into a quadratic in sin θ, then factorise or use the quadratic formula. Always check that each root lies between −1 and 1 before proceeding.
典型的考试题会要求你解类似 2cos² θ + 3sin θ = 3 的方程。先用恒等式把 cos² θ 用 sin θ 表示，整理成关于 sin θ 的二次方程，再进行因式分解或使用求根公式。继续求解前，务必检查每个根是否在 −1 到 1 之间。
2cos² θ + 3sin θ = 3 → 2(1 − sin² θ) + 3sin θ = 3
After simplifying, you might obtain something like 2sin² θ − 3sin θ + 1 = 0. Factorising gives (2sin θ − 1)(sin θ − 1) = 0, so sin θ = ½ or sin θ = 1. Then solve each simple equation using the CAST diagram.
化简后，你可能会得到类似 2sin² θ − 3sin θ + 1 = 0 的方程。因式分解得到 (2sin θ − 1)(sin θ − 1) = 0，因此 sin θ = ½ 或 sin θ = 1。然后使用 CAST 图分别求解这两个简单方程。
8. Equations Involving sin θ and cos θ: The R-Formula | 含 sin θ 与 cos θ 的方程：R 公式
Equations of the form a sin θ + b cos θ = c are easier to solve after rewriting the left side as R sin(θ ± α) or R cos(θ ± α). Here R = √(a² + b²), and α is chosen so that cos α = a/R and sin α = b/R, or the equivalent form for cosine.
形如 a sin θ + b cos θ = c 的方程，在把左边改写为 R sin(θ ± α) 或 R cos(θ ± α) 后会更容易求解。其中 R = √(a² + b²)，α 的选取应满足 cos α = a/R 和 sin α = b/R，或相应余弦形式的等价条件。
a sin θ + b cos θ = R sin(θ + α), R = √(a² + b²)
To find α, remember that tan α = b/a if you are using R sin(θ + α). The quadrant of α must be chosen carefully using the signs of a and b, exactly as you would when converting Cartesian coordinates to polar form.
求 α 时，如果使用 R sin(θ + α)，记住 tan α = b/a。α 所在的象限必须根据 a 和 b 的符号仔细选择，这与将直角坐标转换为极坐标时的做法完全相同。
9. Harmonic Form and Maximum/Minimum Values | 谐振形式与最值
Writing an expression in harmonic form also reveals its maximum and minimum values instantly. For example, 3 sin θ + 4 cos θ can be written as 5 sin(θ + 53.13°), so the maximum value is 5 and the minimum value is −5. The angle at which the maximum occurs is found by setting the argument equal to π/2 plus multiples of 2π.
把表达式写成谐振形式还能立即看出它的最大值和最小值。例如 3 sin θ + 4 cos θ 可以写成 5 sin(θ + 53.13°)，因此最大值是 5，最小值是 −5。最大值出现时对应的角度可通过令括号内的角度等于 π/2 加上 2π 的整数倍来求得。
3 sin θ + 4 cos θ = 5 sin(θ + 53.13°)
For R sin(θ + α), the maximum value R occurs when θ + α = π/2 + 2nπ, so θ = π/2 − α + 2nπ. The minimum value −R occurs when θ + α = 3π/2 + 2nπ. This is a very common exam question, often set in the context of temperature, tides or mechanical motion.
对于 R sin(θ + α)，最大值 R 出现在 θ + α = π/2 + 2nπ 时，因此 θ = π/2 − α + 2nπ。最小值 −R 出现在 θ + α = 3π/2 + 2nπ 时。这是一种非常常见的考试题型，通常以温度、潮汐或机械运动为背景。
10. Harder Mixed Equations
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