📚 Trigonometry for IGCSE Mathematics | IGCSE数学三角学全解析
Trigonometry is one of the most heavily examined topics in IGCSE Mathematics (0580 Extended), appearing in both Paper 2 and Paper 4. It connects geometry, algebra and real-world applications, and mastering it can secure substantial marks in your exam. This revision guide covers every essential area: right-angled triangle ratios, exact values, the sine rule, the cosine rule, the area formula, trigonometric graphs, bearings and common pitfalls.
三角学是IGCSE数学（0580 Extended）中考查频率最高的知识点之一，在Paper 2和Paper 4中都会出现。它把几何、代数与实际应用紧密联系在一起，掌握好三角学能在考试中稳稳拿下大量分数。本篇复习指南涵盖所有核心考点：直角三角形比值、精确值、正弦定理、余弦定理、面积公式、三角函数图像、方位角以及常见易错点。
1. Right-Angled Triangles | 直角三角形
For any right-angled triangle, the three primary trigonometric ratios relate an angle to the lengths of its sides. The mnemonic SOH CAH TOA is essential: sin θ = opposite ÷ hypotenuse, cos θ = adjacent ÷ hypotenuse, tan θ = opposite ÷ adjacent.
对于任意直角三角形，三个基本三角比值将一个角与三边长度联系起来。助记口诀SOH CAH TOA至关重要：sin θ = 对边 ÷ 斜边，cos θ = 邻边 ÷ 斜边，tan θ = 对边 ÷ 邻边。
- sin θ is used when you know the opposite side and hypotenuse | 已知对边和斜边时使用sin θ
- cos θ is used when you know the adjacent side and hypotenuse | 已知邻边和斜边时使用cos θ
- tan θ is used when you know the opposite side and adjacent side | 已知对边和邻边时使用tan θ
To find an unknown side, rearrange the ratio. To find an unknown angle, use the inverse function on your calculator, such as θ = sin⁻¹(opposite ÷ hypotenuse).
求未知边长时，重新整理比值公式即可；求未知角度时，在计算器上使用反函数，例如 θ = sin⁻¹(对边 ÷ 斜边)。
Example: In triangle ABC, angle C = 90°, angle A = 30°, and the hypotenuse AC = 10 cm. Find the length BC. Since BC is opposite angle A, we use sin 30° = BC ÷ 10, so BC = 10 × sin 30° = 10 × 0.5 = 5 cm.
例题：在三角形ABC中，∠C = 90°，∠A = 30°，斜边AC = 10 cm，求BC的长度。因为BC是∠A的对边，使用sin 30° = BC ÷ 10，所以BC = 10 × sin 30° = 10 × 0.5 = 5 cm。
2. Exact Trigonometric Values | 三角精确值
IGCSE requires you to know the exact values of sin, cos and tan for special angles: 0°, 30°, 45°, 60° and 90°. These values appear in non-calculator questions, so memorising them is non-negotiable.
IGCSE要求你熟记特殊角0°、30°、45°、60°和90°的正弦、余弦与正切精确值。这些值常出现在不允许使用计算器的题目中，因此必须牢记。
| θ | 0° | 30° | 45° | 60° | 90° |
| sin θ | 0 | ½ | √2/2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | √2/2 | ½ | 0 |
| tan θ | 0 | √3/3 | 1 | √3 | undefined |
Notice the pattern: sin increases from 0 to 1, cos decreases from 1 to 0, and tan is undefined at 90° because the adjacent side approaches zero. A neat trick is to write the fractions as √n/2 for sin: √0/2, √1/2, √2/2, √3/2, √4/2.
观察规律：sin从0递增到1，cos从1递减到0，而tan在90°处无定义，因为此时邻边趋近于零。记忆技巧：把sin写成√n/2的形式，依次为√0/2、√1/2、√2/2、√3/2、√4/2。
3. The Sine Rule | 正弦定理
The sine rule works for any triangle, not just right-angled ones. It states that the ratio of a side to the sine of its opposite angle is constant:
正弦定理适用于任意三角形，而不仅仅是直角三角形。它指出：边长与其对角正弦之比为常数：
a ÷ sin A = b ÷ sin B = c ÷ sin C
Use the sine rule when you know either two angles and any side (AAS) or two sides and a non-included angle (SSA). In the AAS case, you can immediately find the third angle using the fact that angles sum to 180°, then solve for the missing side.
当已知两角及任意一边（AAS），或已知两边及其中一边的对角（SSA）时，使用正弦定理。在AAS情形下，先用内角和为180°求出第三个角，再求缺失的边长。
Example: In triangle ABC, angle A = 40°, angle B = 60°, and side a = 8 cm. Find side b. First, angle C = 180° − 40° − 60° = 80°. Using the sine rule: b ÷ sin 60° = 8 ÷ sin 40°, so b = 8 × sin 60° ÷ sin 40° ≈ 8 × 0.8660 ÷ 0.6428 ≈ 10.8 cm.
例题：在三角形ABC中，∠A = 40°，∠B = 60°，边a = 8 cm，求边b。先求∠C = 180° − 40° − 60° = 80°。由正弦定理：b ÷ sin 60° = 8 ÷ sin 40°，所以b = 8 × sin 60° ÷ sin 40° ≈ 8 × 0.8660 ÷ 0.6428 ≈ 10.8 cm。
When using the sine rule to find an angle, invert the formula. The ambiguous case (two possible angles) is an important warning: a triangle with SSA data may have one valid answer, two valid answers, or none, so always check whether both candidate angles produce a triangle whose angles sum to 180°.
用正弦定理求角度时，需要将公式取倒数。这里有一个重要的“双解”警告：已知SSA时，三角形可能有一个解、两个解或无解。求出候选角后，务必验证两个候选角是否能分别与已知角构成内角和为180°的三角形。
4. The Cosine Rule | 余弦定理
The cosine rule is used for triangles where you know two sides and the included angle (SAS), or all three sides (SSS). In its side-finding form:
余弦定理适用于已知两边及其夹角（SAS），或已知三边（SSS）的三角形。求边长时的形式为：
a² = b² + c² − 2bc cos A
Here, side a is opposite angle A, and b and c are the two sides adjacent to angle A. To find a side, substitute the known lengths and the included angle directly. The formula resembles Pythagoras’ theorem, but with the extra term −2bc cos A correcting for the non-right angle.
其中边a是角A的对边，b和c是角A的两条邻边。求边长时，直接代入已知边长和夹角即可。该公式形似勾股定理，但多出的项−2bc cos A用于修正非直角的影响。
To find an angle, rearrange the formula:
求角度时，将公式变形：
cos A = (b² + c² − a²) ÷ 2bc
Example: Find side a given b = 5 cm, c = 7 cm and angle A = 50°. Using the formula: a² = 5² + 7² − 2 × 5 × 7 × cos 50° = 25 + 49 − 70 × 0.6428 = 74 − 45.0 = 29.0, so a = √29.0 ≈ 5.39 cm.
例题：已知b = 5 cm，c = 7 cm，∠A = 50°，求边a。代入公式：a² = 5² + 7² − 2 × 5 × 7 × cos 50° = 25 + 49 − 70 × 0.6428 = 74 − 45.0 = 29.0，所以a = √29.0 ≈ 5.39 cm。
5. Area of a Triangle | 三角形面积公式
Beyond the basic ½ × base × height, IGCSE requires the sine-area formula for any triangle where two sides and the included angle are known:
除了基本公式½ × 底 × 高，IGCSE还要求掌握适用于任意三角形（已知两边及夹角）的正弦面积公式：
Area = ½ ab sin C
Here, a and b are any two sides of the triangle, and C is the angle between them. This formula is particularly useful in problems involving vector diagrams, surveying and compound shapes.
其中a和b是三角形的任意两条边，C是它们之间的夹角。该公式特别适用于涉及向量图、测量以及复合图形的题目。
Example: Two sides of a triangle are 6 cm and 8 cm, with an included angle of 55°. The area is ½ × 6 × 8 × sin 55° = 24 × 0.8192 ≈ 19.7 cm². Note that you must use the angle between the two given sides, not any other angle.
例题：三角形两条边分别为6 cm和8 cm，夹角为55°。面积 = ½ × 6 × 8 × sin 55° = 24 × 0.8192 ≈ 19.7 cm²。注意必须使用两条已知边之间的夹角，而非其他角。
6. Trigonometric Graphs | 三角函数图像
You must be able to sketch and interpret the graphs of y = sin x, y = cos x and y = tan x for angles from 0° to 360°. Key features include the amplitude (maximum height from the centre line), the period (how many degrees for one complete cycle) and any asymptotes.
你必须能够画出并解读y = sin x、y = cos x和y = tan x在0°到360°范围内的图像。关键特征包括振幅（离开中心轴的最大高度）、周期（完成一个完整循环所需的角度数）以及渐近线。
- y = sin x: amplitude 1, period 360°, passes through (0, 0), maximum at x = 90°, minimum at x = 270° | y = sin x：振幅1，周期360°，过(0, 0)，在x = 90°取最大值，在x = 270°取最小值
- y = cos x: amplitude 1, period 360°, passes through (0, 1), maximum at x = 0° and 360°, minimum at x = 180° | y = cos x：振幅1，周期360°，过(0, 1)，在x = 0°和360°取最大值，在x = 180°取最小值
- y = tan x: no amplitude (unbounded), period 180°, passes through (0, 0), vertical asymptotes at x = 90° and x = 270° | y = tan x：无振幅（无界），周期180°，过(0, 0)，在x = 90°和270°处有垂直渐近线
Transformations are also tested. The graph of y = a sin bx + c has amplitude a, period 360° ÷ b, and a vertical shift of c. A negative value of a reflects the graph in the x-axis, while a negative value of c shifts it downward.
图像变换同样是考点。y = a sin bx + c的图像振幅为a，周期为360° ÷ b，垂直平移量为c。若a为负数，图像关于x轴翻转；若c为负数，则向下平移。
7. Bearings and Problem Solving | 方位角与综合应用
A bearing is an angle measured clockwise from north, always written as three digits, such as 047° or 135°. Bearings are used to describe directions in navigation and surveying problems, and they frequently appear in multi-step trigonometry questions.
方位角是从正北方向开始顺时针测量的角度，始终用三位数字表示，如047°或135°。方位角用于描述航行和测量问题中的方向，经常出现在多步骤三角综合题中。
Worked example: A ship sails 20 km from point P to point Q on a bearing of 060°, then 15 km from Q to R on a bearing of 150°. Find the distance PR. Sketch the diagram: at Q, the angle between the north line and QP is 60°, and the angle between north and QR is 150°, so angle PQR = 150° − 60° = 90°. Triangle PQR is right-angled at Q, so PR² = 20² + 15² = 400 + 225 = 625, and PR = 25 km.
例题：一艘船从P点沿方位角060°航行20 km到达Q点，再从Q沿方位角150°航行15 km到达R点。求PR的距离。先画示意图：在Q点，Q点正北方向与QP的夹角为60°，与QR的夹角为150°，所以∠PQR = 150° − 60° = 90°。三角形PQR在Q处为直角，所以PR² = 20² + 15² = 400 + 225 = 625，故PR = 25 km。
In more complex problems, you may need to find the bearing of P from R or vice versa. This requires calculating an angle inside the triangle, then converting it to a bearing by adding or subtracting from 180° or 360°. Always draw a clear diagram with the north line at each vertex before applying the sine or cosine rule.
在更复杂的问题中，你可能需要求R相对于P（或P相对于R）的方位角。这需要先求出三角形内角，再通过加减180°或360°将其转换为方位角。在套用正弦或余弦定理之前，务必在每个顶点画出清晰的正北方向线。
8. Common Pitfalls and Exam Tips | 常见误区与考试技巧
Below are the most frequent mistakes students make in IGCSE trigonometry, with practical advice to avoid each one.
下面是学生在IGCSE三角学中最常犯的错误，以及避免这些错误的实用建议。
- Calculator mode: Always set your calculator to degree mode (DEG), not radians (RAD). A wrong mode changes every answer silently.
- 计算器模式：始终将计算器设为角度制（DEG），而非弧度制（RAD）。选错模式会悄无声息地改变所有答案。
- Label the triangle first: Before using the sine or cosine rule, label sides a, b, c with opposite angles A, B, C. Substitution mistakes are the leading cause of lost marks.
- 先标注三角形：使用正弦或余弦定理前，将边a、b、c与其对角A、B、C一一对应标注。代入错误是失分的最主要原因。
- Ambiguous case warnings: When using the sine rule to find an angle from SSA data, your calculator gives only the acute angle. Check whether 180° − θ is also a valid solution.
- 警惕双解情形：用正弦定理由SSA求角时，计算器只给出锐角。要检查180° − θ是否也是有效解。
- Don’t confuse the rules: Use the sine rule for known angles, the cosine rule for two sides and an included angle, and ½ab sin C for area. Applying the wrong rule is a common but entirely avoidable error.
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