📚 Mastering Trigonometric Equations | 精通三角函数方程
Trigonometric equations appear throughout Edexcel A-Level Pure Mathematics, especially in AS Pure 1 and A2 Pure 2. Many students lose marks not because they cannot solve the basic equations, but because they forget the periodic nature of sine, cosine and tangent, or they fail to select the correct solutions within a given interval. This revision guide builds the topic from the unit circle upwards, then moves through identities, general solutions, compound angles, quadratic forms and reciprocal ratios, finishing with examination technique and the errors that cost the most marks.
三角函数方程贯穿 Edexcel A-Level 纯数学的各个部分，尤其是在 AS Pure 1 和 A2 Pure 2 中。许多学生失分并不是因为不会解基本方程，而是因为他们忘记了正弦、余弦和正切的周期性，或者未能在指定区间内选出正确的解。本篇复习指南从单位圆出发，逐步延伸到三角恒等式、通解、复合角、二次型方程与倒数比值方程，最后总结考试技巧与最容易丢分的错误。
1. Understanding the Unit Circle | 理解单位圆
e reads the cosine and the y-coordinate reads the sine. Because the radius is 1, every point satisfies x² + y² = 1, which is exactly why sin² θ + cos² θ = 1 holds for every angle.
单位圆是所有三角函数方程的基础。圆上角 θ 处的点坐标为 (cos θ, sin θ)，因此 x 坐标表示余弦，y 坐标表示正弦。由于半径等于 1，圆上每一点都满足 x² + y² = 1，这正是 sin² θ + cos² θ = 1 对任意角都成立的根本原因。
In quadrant I all three ratios are positive. In quadrant II only sin is positive, in quadrant III only tan is positive, and in quadrant IV only cos is positive. This sign pattern is often summarised as 'CAST' moving anticlockwise from the fourth quadrant.
在第一象限，三个比值均为正。第二象限只有 sin 为正，第三象限只有 tan 为正，第四象限只有 cos 为正。这个符号规律通常用 'CAST' 来表示，从第四象限起按逆时针方向读取。
Angles are measured anticlockwise from the positive x-axis, and negative angles are measured clockwise. One complete revolution is 360° or 2π radians, and this periodicity is the single most important idea in the whole topic.
角度从 x 轴正方向起按逆时针方向度量，负角则按顺时针方向度量。完整一圈为 360° 或 2π 弧度，这种周期性是整个专题中最重要的概念。
When solving equations, draw or imagine the CAST diagram to decide which quadrants the required angle lies in after taking the inverse trigonometric function.
解方程时，可画出或想象 CAST 图，在求反三角函数后判断所需角位于哪些象限。
2. Key Trigonometric Identities | 关键三角恒等式
Before rearranging an equation, you often need to rewrite it using identities. The most important identity is shown below.
在移项变形之前，你通常需要先用恒等式改写方程。最重要的恒等式如下所示。
sin² θ + cos² θ = 1
rearrangements are used constantly in A2 papers when a question mixes squared ratios with reciprocal functions.
将该恒等式两边同除以 cos² θ，可得 1 + tan² θ = sec² θ；两边同除以 sin² θ，可得 1 + cot² θ = cosec² θ。当题目把平方比值与倒数函数混在一起时，A2 试卷中会频繁用到这两个变形。
h the reciprocal definitions sec θ = 1 ÷ cos θ, cosec θ = 1 ÷ sin θ and cot θ = 1 ÷ tan θ.
同时你必须对商数恒等式 tan θ = sin θ ÷ cos θ 与 cot θ = cos θ ÷ sin θ 非常熟练，并牢记倒数定义 sec θ = 1 ÷ cos θ、cosec θ = 1 ÷ sin θ、cot θ = 1 ÷ tan θ。
θ − 1 are the versions you actually substitute into equations. Whenever a question contains both a squared term and a first-power term of a different ratio, this substitution is almost always the intended first step.
三个常用变形 sin² θ = 1 − cos² θ、cos² θ = 1 − sin² θ、tan² θ = sec² θ − 1 才是你真正代入方程的形式。只要题目中同时出现某个比值的平方项与另一个比值的一次项，这一步代换几乎总是命题人预设的第一步。
3. The CAST Diagram and Sign Patterns | CAST 图与符号规律
The CAST diagram encodes the quadrant signs in a single picture. Starting in the fourth quadrant and moving anticlockwise, C means cos is positive, A means all are positive, S means sin is positive and T means tan is positive. Equivalently, reading the quadrants in the usual order I, II, III, IV, the positive ratios are all, sin, tan, cos.
CAST 图把四个象限的符号规律浓缩在一张图中。从第四象限开始逆时针看，C 表示 cos 为正，A 表示全部为正，S 表示 sin 为正，T 表示 tan 为正。等价地说，按常见的 I、II、III、IV 顺序读，各象限的正比值依次为全部、sin、tan、cos。
The safest routine is a reference angle method. First find the acute reference angle α using the inverse function and ignore the sign. Then use the CAST diagram to place α in the correct quadrants: for a negative sine, the answers lie in the third and fourth quadrants; for a negative cosine, in the second and third; for a positive tangent, in the first and third.
最稳妥的做法是"参考角法"。先用反函数求出锐角参考角 α，先不管符号；再用 CAST 图把 α 放到正确的象限中：sin 为负时解在第三、第四象限；cos 为负时解在第二、第三象限；tan 为正时解在第一、第三象限。
You must also know the exact values of sin, cos and tan at the special angles. They appear in almost every trigonometric equation question, and a calculator decimal will often not earn the accuracy mark.
你还必须掌握特殊角的 sin、cos、tan 精确值。它们几乎出现在每一道三角函数方程题中，而计算器给出的小数常常拿不到精确值分。
| θ | 0 | π ÷ 6 (30°) | π ÷ 4 (45°) | π ÷ 3 (60°) | π ÷ 2 (90°) |
| sin θ | 0 | 1 ÷ 2 | √2 ÷ 2 | √3 ÷ 2 | 1 |
| cos θ | 1 | √3 ÷ 2 | √2 ÷ 2 | 1 ÷ 2 | 0 |
| tan θ | 0 | 1 ÷ √3 | 1 | √3 | undefined |
4. Principal Values and General Solutions | 主值与通解
.5 has infinitely many solutions. The principal value is the one your calculator returns, and the general solution describes every other solution family.
由于正弦、余弦和正切都是周期函数，像 sin θ = 0.5 这样的方程有无数个解。主值是计算器给出的那个值，而通解则把所有解族完整描述出来。
Let α be the principal value in the appropriate range and let n be any integer. Then the general solutions take the following forms.
设 α 为相应范围内的主值，n 为任意整数，则通解形式如下。
sin θ = k: θ = 180°n + (−1)ⁿ α or θ = nπ + (−1)ⁿ α
cos θ = k: θ = 360°n ± α or θ = 2nπ ± α
tan θ = k: θ = 180°n + α or θ = nπ + α
integer values of n and keep the answers that fall in range; if a solution equals the upper endpoint of a closed-open interval, reject it.
注意正切的周期是 180°，而正弦和余弦的周期是 360°。在 0 ≤ θ < 360° 这样的固定区间内，只需代入整数 n，保留落在区间内的解；如果某个解等于区间上端点而区间是左闭右开，就要舍去。
5. Solving sin θ = k and cos θ = k | 解 sin θ = k 与 cos θ = k
in the first and second quadrants, so the solutions are θ = 30° and θ = 180° − 30° = 150°.
例：在 0 ≤ θ < 360° 内解 sin θ = 0.5。主值为 30°，故参考角为 30°。sin 在第一、第二象限为正，因此解为 θ = 30° 与 θ = 180° − 30° = 150°。
uadrants, so the solutions are θ = 120° and θ = 240°. Equivalently, use 180° ± 60° with the CAST diagram as a cross-check.
例：在 0 ≤ θ < 360° 内解 cos θ = −0.5。主值为 120°，参考角为 60°。cos 在第二、第三象限为负，因此解为 θ = 120° 与 θ = 240°。也可以用 180° ± 60° 计算，再用 CAST 图复核。
n the first quadrant, then the second-quadrant solution is 360° − α for sine-type reflections and 180° − α or 180° + α for cosine-type reflections. Practising both the CAST method and the formula method keeps you from depending on a single route under exam pressure.
一个值得记住的可靠技巧是：若 α 是 cos θ = k 在第一象限的解，则与正弦型对称相关的第二象限解为 360° − α，与余弦型对称相关的解为 180° − α 或 180° + α。同时练习 CAST 法与公式法，可以避免在考试压力下只依赖单一思路。
6. Solving tan θ = k | 解 tan θ = k
ipal value is 45°, and tangent is also positive in the third quadrant, so the solutions are 45° and 45° + 180° = 225°.
正切的周期为 180°，因此只要得到一个解，就可以不断加或减 180°。例：在 0 ≤ θ < 360° 内解 tan θ = 1。主值为 45°，且 tan 在第三象限也为正，所以解为 45° 与 45° + 180° = 225°。
second and fourth quadrants, so the solutions are 180° − 63.4° = 116.6° and 360° − 63.4° = 296.6°.
例：在 0 ≤ θ < 360° 内解 tan θ = −2。计算器给出约 −63.4°，其参考角为 63.4°。tan 在第二、第四象限为负，因此解为 180° − 63.4° = 116.6° 与 360° − 63.4° = 296.6°。
not be solved by dividing by cos θ. If a question requires you to divide by a trigonometric ratio, always state the condition under which that ratio is non-zero.
一个容易让学生失手的细节是：tan 在 90° 与 270° 处无定义，所以像 cos θ = 0 这样的方程不能通过除以 cos θ 来解。如果题目中需要除以某个三角比值，务必先说明该比值不为零的条件。
7. Equations with Compound and Multiple Angles | 多倍角与复合角方程
When the unknown appears as 2θ, 3θ or θ + 45°, the golden rule is to solve for the whole bracket first, then divide or subtract at the end. Crucially, the interval must be transformed at the same time.
当未知量以 2θ、3θ 或 θ + 45° 的形式出现时，黄金法则是：先对整体括号求解，最后再除系数或减去常数。关键是，区间必须同步变换。
195° and 255°. Forgetting to double the interval is the single most common error in this type of question, and it usually costs two of the four solutions.
例：在 0 ≤ θ < 360° 内解 sin 2θ = 0.5。此时 0 ≤ 2θ < 720°，2θ 的解为 30°、150°、390° 与 510°，除以 2 得 θ = 15°、75°、195° 与 255°。忘记把区间翻倍是这类题目中最常见的错误，通常会丢掉四个解中的两个。
= 0.5 would be handled in exactly the same way with the interval rewritten as an inequality in 2x + π ÷ 3.
例：在 0 ≤ θ < 360° 内解 cos(θ − 45°) = 0.5。此时 −45° ≤ θ − 45° < 315°，故 θ − 45° = 60° 或 300°，得 θ = 105° 与 345°。在弧度制下，cos(2x + π ÷ 3) = 0.5 的处理方式完全相同，只需把区间改写成关于 2x + π ÷ 3 的不等式。
If the bracket is negative for part of its range, for example −45° to 315°, you must still check whether a further solution exists in that negative portion.
如果括号在部分区间内取负值，例如 −45° 到 315°，你仍然必须检查负值部分是否还存在另一个解。
8. Quadratic Trigonometric Equations | 二次型三角方程
not. Then reject any root outside the range of the ratio: sine and cosine are restricted to −1 ≤ value ≤ 1, while tangent can take any real value.
许多考题中的方程是关于 sin θ、cos θ 或 tan θ 的二次方程。把该比值当作一个未知量，能因式分解就先分解，不能再用求根公式。之后必须舍去超出比值取值范围的根：正弦与余弦限制在 −1 ≤ 取值 ≤ 1，而正切可以取任意实数。
2 gives 30° and 150°, while sin θ = −1 gives 270°. The full solution set is therefore 30°, 150° and 270°.
例：在 0 ≤ θ < 360° 内解 2sin² θ + sin θ − 1 = 0。因式分解得 (2sin θ − 1)(sin θ + 1) = 0，故 sin θ = 1 ÷ 2 或 sin θ = −1，两根均有效。sin θ = 1 ÷ 2 给出 30° 与 150°，sin θ = −1 给出 270°，故全部解为 30°、150° 与 270°。
he second gives 180° − 33.7° ≈ 146.3° and 360° − 33.7° ≈ 326.3°. Four solutions, all distinct.
例：在 0 ≤ θ < 360° 内解 3tan² θ − tan θ − 2 = 0。因式分解得 (3tan θ + 2)(tan θ − 1) = 0，故 tan θ = 1 或 tan θ = −2 ÷ 3。前者给出 45° 与 225°；后者给出 180° − 33.7° ≈ 146.3° 与 360° − 33.7° ≈ 326.3°。四个解互不相同。
t explicitly state that the root is rejected because sine cannot exceed 1, and examiners award a mark for that statement.
还有一种特殊情况：某个根落在可行范围之外，例如 sin θ = 1.5。此时必须明确写出该根被舍去，因为正弦不可能大于 1，阅卷人会给这一步记分。
9. Using Identities to Reduce to a Single Ratio | 用恒等式化为单一比值
one ratio using sin² θ + cos² θ = 1, producing a quadratic you already know how to solve.
有些方程混合了两种比值，例如 2cos² θ + sin θ = 1。策略是利用 sin² θ + cos² θ = 1 把所有项化为同一种比值，从而得到一个你已经会解的二次方程。
= 0. Factorising gives (2sin θ + 1)(sin θ − 1) = 0, so sin θ = −1 ÷ 2 or sin θ = 1. The valid solutions are θ = 90°, 210° and 330°.
例题：在 0 ≤ θ < 360° 内解 2cos² θ + sin θ = 1。用
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