IGCSE Trigonometry Masterclass: Sine Rule, Cosine Rule & Graphs | IGCSE 三角函数精讲：正弦定理、余弦定理与图像

📚 IGCSE Trigonometry Masterclass: Sine Rule, Cosine Rule & Graphs | IGCSE 三角函数精讲：正弦定理、余弦定理与图像

Trigonometry is one of the most important topics in IGCSE Mathematics (0580/0606). It appears in at least three exam questions on Paper 2 and Paper 4, covering right-angled triangles, non-right-angled triangles, graphs, and bearings. Mastering this topic can easily secure 15–25 marks.

三角函数是 IGCSE 数学（0580/0606）中最重要的考点之一。在 Paper 2 和 Paper 4 中至少会出现三道相关题目，涉及直角三角形、非直角三角形、函数图像与方位角。掌握好本章内容，轻松拿下 15–25 分并非难事。

1. The Three Basic Ratios | 三大基本三角函数比

For a right-angled triangle with angle θ, we define three ratios using the labels ‘Opposite’ (O), ‘Adjacent’ (A) and ‘Hypotenuse’ (H). The memory aid SOH CAH TOA is essential.

在直角三角形中，对于角 θ，我们利用“对边”（O）、“邻边”（A）和“斜边”（H）来定义三个比值。记忆口诀 SOH CAH TOA 至关重要。

sin θ = Opposite ÷ Hypotenuse (SOH)
cos θ = Adjacent ÷ Hypotenuse (CAH)
tan θ = Opposite ÷ Adjacent (TOA)

sin θ = O / H cos θ = A / H tan θ = O / A

These ratios allow you to find missing sides or angles in a right-angled triangle. To find an angle, use the inverse functions: sin⁻¹, cos⁻¹, tan⁻¹.

利用这些比值，可以求出直角三角形中的未知边或未知角。若需求角度，则使用反函数：sin⁻¹、cos⁻¹、tan⁻¹。

Example: In a right triangle, the hypotenuse is 10 cm and one angle is 30°. Find the length of the side opposite the 30° angle.

示例：直角三角形中，斜边长为 10 厘米，一个角为 30°。求 30° 角所对的边长。

sin 30° = Opposite / 10 → Opposite = 10 × sin 30° = 10 × 0.5 = 5 cm

2. Exact Values You Must Memorise | 必背特殊角精确值

The following table of exact values is required for IGCSE. You must recall these instantly without a calculator for Paper 2 (non-calculator paper).

下表是 IGCSE 必考的特殊角精确值。在 Paper 2（不可使用计算器的试卷）中，你必须不假思索地写出这些结果。

Angle 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	1 / √3	1	√3	undefined

A neat pattern: sin values are √0/2, √1/2, √2/2, √3/2, √4/2. Cosine values are the reverse. Tan values come from sin ÷ cos.

记忆技巧：sin 值依次为 √0/2、√1/2、√2/2、√3/2、√4/2；cos 值顺序相反；tan 值由 sin ÷ cos 得到。

3. The Sine Rule | 正弦定理

For any triangle ABC with sides a, b, c opposite angles A, B, C respectively, the sine rule states:

对于任意三角形 ABC，设边 a、b、c 分别对角 A、B、C，则正弦定理表述为：

a / sin A = b / sin B = c / sin C

Use the sine rule when you know: (i) two angles and one side (AAS or ASA), or (ii) two sides and a non-included angle (SSA).

正弦定理适用于以下两种情况：(i) 已知两个角与一条边（AAS 或 ASA）；(ii) 已知两边与其中一边的对角（SSA）。

Example: In triangle ABC, angle A = 40°, angle B = 60°, side a = 8 cm. Find side b.

示例：在三角形 ABC 中，∠A = 40°，∠B = 60°，边 a = 8 厘米。求边 b。

8 / sin 40° = b / sin 60° → b = 8 × sin 60° / sin 40° = 8 × 0.866 / 0.643 ≈ 10.8 cm

When using the sine rule to find an angle with SSA information, be aware of the ambiguous case: there may be two possible answers. Check whether both fit the triangle’s angle sum of 180°.

当利用 SSA 条件用正弦定理求角时，要注意“两解”情况：可能存在两个符合条件的答案。务必检验两个答案是否满足三角形内角和为 180°。

4. The Cosine Rule | 余弦定理

The cosine rule connects three sides and one angle. Use it when you know: (i) two sides and the included angle (SAS), or (ii) all three sides (SSS).

余弦定理建立了三边与一角之间的关联。它适用于：(i) 已知两边及其夹角（SAS）；(ii) 已知三边（SSS）。

a² = b² + c² − 2bc × cos A

To find an angle when all three sides are known, rearrange the formula:

当已知三边求角度时，将公式变形：

cos A = (b² + c² − a²) / (2bc)

Example: Sides b = 5 cm, c = 7 cm and included angle A = 50°. Find side a.

示例：边 b = 5 厘米，c = 7 厘米，夹角 A = 50°。求边 a。

a² = 5² + 7² − 2 × 5 × 7 × cos 50° = 25 + 49 − 70 × 0.643 = 25.0 → a ≈ 5.0 cm

Notice that when A = 90°, cos 90° = 0, so the cosine rule simplifies to the Pythagorean theorem: a² = b² + c². This is a good sanity check.

注意：当 A = 90° 时，cos 90° = 0，余弦定理便退化为勾股定理 a² = b² + c²。这是一个很好的检验方法。

5. Area of a Triangle | 三角形面积公式

The standard area formula ½ × base × height only works for right-angled or known-perpendicular cases. For any triangle, if you know two sides and the included angle, use:

标准面积公式 ½ × 底 × 高仅适用于直角三角形或已知垂直关系的情形。对于任意三角形，若已知两边及其夹角，应使用：

Area = ½ × a × b × sin C

Here, a and b are any two sides, and C is the angle between them. The result is given in square units.

其中 a 与 b 为任意两条边，C 为它们的夹角。结果单位为平方单位。

Example: A triangle has sides 6 cm and 9 cm with included angle 35°. Find its area.

示例：三角形两条边分别为 6 厘米和 9 厘米，夹角为 35°。求其面积。

Area = ½ × 6 × 9 × sin 35° = 27 × 0.574 ≈ 15.5 cm²

This formula is especially useful in IGCSE questions involving bearings or vector diagrams, where heights are not obvious.

在处理方位角或向量图相关的 IGCSE 题目时，该公式尤为实用，因为此时三角形的高往往不直观。

6. Bearings and Angles of Elevation & Depression | 方位角与仰角/俯角

A bearing is an angle measured clockwise from north, written as a three-digit number (e.g., 047°, 132°, 300°). Bearings are always measured from north and always clockwise.

方位角（罗盘方位）是从正北方向顺时针测量的角度，以三位数表示（例如 047°、132°、300°）。方位角始终从北方向顺时针度量。

Angle of elevation: the angle between the horizontal and the line of sight looking up at an object.
仰角（视角向上）：从水平线向上仰望物体时，视线与水平线的夹角。
Angle of depression: the angle between the horizontal and the line of sight looking down at an object.
俯角（视角向下）：从水平线向下俯视物体时，视线与水平线的夹角。

tan θ = (vertical height) / (horizontal distance)

In bearing problems, always draw a diagram with north arrows. Break the journey into right-angled triangles using north-south or east-west components.

在方位角问题中，务必画出带有指北箭头的示意图。将路径分解为南北或东西方向的分量，构造直角三角形求解。

Exam tip: If a boat travels 20 km on a bearing of 060°, its eastward component = 20 × sin 60° and its northward component = 20 × cos 60°. Memorise this decomposition!

考试技巧：若船沿 060° 方位角行驶 20 千米，则其向东分量为 20 × sin 60°，向北分量为 20 × cos 60°。务必牢记这一分解方法！

7. Trigonometric Graphs | 三角函数图像

You must know the shapes and key features of y = sin x, y = cos x and y = tan x for 0° ≤ x ≤ 360°.

你必须掌握在 0° ≤ x ≤ 360° 范围内 y = sin x、y = cos x 和 y = tan x 的图像形状与关键特征。

y = sin x: passes through (0°, 0), maximum 1 at 90°, zero at 180°, minimum −1 at 270°, zero at 360°. Period = 360°.
y = sin x：过 (0°, 0)，在 90° 处取最大值 1，在 180° 处为零，在 270° 处取最小值 −1，在 360° 处为零。周期 = 360°。
y = cos x: starts at (0°, 1), zero at 90°, minimum −1 at 180°, zero at 270°, back to 1 at 360°. Period = 360°. It is a sine graph shifted left by 90°.
y = cos x：从 (0°, 1) 开始，在 90° 处为零，在 180° 处取最小值 −1，在 270° 处为零，在 360° 处回到 1。周期 = 360°。它是正弦图像向左平移 90° 的结果。
y = tan x: passes through (0°, 0), rises to +∞ at 90° (vertical asymptote), crosses zero at 180°, falls from −∞ toward 0 at 360°. Period = 180°.
y = tan x：过 (0°, 0)，在 90° 处升至正无穷（垂直渐近线），在 180° 处过零，在 360° 处从负无穷回到 0。周期 = 180°。

Examiners often ask you to read values from a graph, estimate solutions to equations, or identify the equation of a transformed graph. Label axes, maximum/minimum points, and asymptotes clearly.

考官常要求你从图像读取数值、估算方程的解，或辨别变换后图像所对应的方程。请在图上清晰标注坐标轴、最大值/最小值点以及渐近线。

8. Transformations of Trig Graphs | 三角函数的图像变换

The general forms below allow you to describe any sinusoidal graph:

以下一般形式可用于描述任何正弦型曲线：

y = a × sin(bx) + c or y = a × cos(bx) + c

|a| is the amplitude (vertical stretch). If a is negative, the graph is reflected in the x-axis.
|a| 为振幅（垂直伸缩）。若 a 为负，图像关于 x 轴对称翻转。
b affects the period: new period = 360° ÷ b (for sin/cos) or 180° ÷ b (for tan).
b 影响周期：新周期 = 360° ÷ b（对 sin/cos），或 180° ÷ b（对 tan）。
c shifts the graph vertically: the midline becomes y = c.
c 使图像垂直平移：中线变为 y = c。

Example: y = 3 sin 2x has amplitude 3 and period 360° ÷ 2 = 180°. Its maximum is 3 and minimum is −3.

示例：y = 3 sin 2x 的振幅为 3，周期为 360° ÷ 2 = 180°。最大值为 3，最小值为 −3。

Example: y = 2 cos x − 1 oscillates between 2 − 1 = 1 and −2 − 1 = −3, with midline y = −1.

示例：y = 2 cos x − 1 在最大值 2 − 1 = 1 与最小值 −2 − 1 = −3 之间波动，中线为 y = −1。

When sketching transformed graphs, first draw the midline, then mark the max and min points, and finally locate the zeros from the original graph’s pattern.

绘制变换后的图像时，先画出中线，再标记最大与最小点，最后根据原图像的规律确定零点位置。

9. Solving Trigonometric Equations | 解三角方程

IGCSE requires solving equations like sin x = 0.5 or 2 cos x + 1 = 0 within a given interval (usually 0° ≤ x ≤ 360°). Follow this systematic approach:

IGCSE 要求解形如 sin x = 0.5 或 2 cos x + 1 = 0 的方程，通常在限定区间 0° ≤ x ≤ 360° 内。请按以下系统方法操作：

Rearrange the equation to isolate a single trigonometric ratio.
将方程变形，分离出单一的三角函数值。
Use a calculator to find the acute reference angle (the principal value).
利用计算器求出锐角参考角（主值）。
Use the CAST diagram (or quadrants) to find all solutions in the required interval.
利用 CAST 图（或象限判断）找出区间内的所有解。

Example: Solve 2 sin x − 1 = 0 for 0° ≤ x ≤ 360°.

示例：解方程 2 sin x − 1 = 0，其中 0° ≤ x ≤ 360°。

sin x = ½ → x = 30° or x = 180° − 30° = 150°

Remember the CAST rule: Sine is positive in Quadrants I and II; Cosine is positive in Quadrants I and IV; Tangent is positive in Quadrants I and III.

牢记 CAST 规则：正弦在一、二象限为正；余弦在一、四象限为正；正切在一、三象限为正。

Common trap: For tan x = k, the second solution is 180° + reference angle, not 180° − reference angle.

常见陷阱：对于 tan x = k，第二个解应为 180° + 参考角，而非 180° − 参考角。

10. Common Mistakes and How to Avoid Them | 常见错误与规避方法

Students lose marks on trigonometry for predictable reasons. Read these carefully:

学生在三角函数题中失分的原因往往高度相似。请仔细阅读以下注意事项：

Using degrees instead of radians, or vice versa. IGCSE Mathematics uses degrees unless the question explicitly says radians. Always check the calculator mode.
混淆度数与弧度。除非题目明确说明使用弧度，IGCSE 数学统一使用度数。计算前务必检查计算器的角度模式。
Applying sine rule when cosine rule is needed. Use sine rule for AAS/ASA/SSA; use cosine rule for SAS/SSS.
该用余弦定理时误用正弦定理。AAS/ASA/SSA 用正弦定理；SAS/SSS 用余弦定理。
Forgetting the ambiguous case in sine rule. Check whether 180° − θ also produces a valid triangle.
忽略正弦定理中的两解情况。务必检验 180° − θ 是否也能构成有效三角形。
Sign errors in cosine rule rearrangement. When finding angle A, write cos A = (b² + c² − a²) / (2bc) and substitute carefully.
余弦定理变形时符号错误。求角 A 时，公式应写为 cos A = (b² + c² − a²) / (2bc)，代入时需仔细。
Not giving answers to the required degree of accuracy. IGCSE usually asks for 1 decimal place or 3 significant figures unless stated otherwise.
未按题目要求的精度作答。IGCSE 通常要求保留一位小数或三位有效数字，除非另有说明。
Poor diagram labelling. Always draw and label the triangle with given values before starting any calculation.
示意图标注不全。解题前务必画出三角形并标出所有已知量。

Prepare a checklist before each exam: calculator in degrees, diagram drawn, correct rule identified, and final answer checked for reasonableness.

每次考试前请自查清单：计算器处于度数模式、示意图已画好、已选择正确的定理、答案经合理性检验。

11. Worked Example: Mixed Problem | 综合例题解析

Let us solve a complete IGCSE-style problem that combines the cosine rule, sine rule and area formula.

下面我们完整解答一道 IGCSE 风格的综合题，融合余弦定理、正弦定理与面积公式。

Question: In triangle PQR, PQ = 7 cm, PR = 9 cm and angle P = 52°.

题目：在三角形 PQR 中，PQ = 7 厘米，PR = 9 厘米，∠P = 52°。

(a) Find the length of QR. (b) Find angle R. (c) Find the area of triangle PQR.

(a) 求 QR 的长度。(b) 求角 R。(c) 求三角形 PQR 的面积。

Solution (a): We know two sides (7 and 9) and the included angle 52°. Use the cosine rule.

解答 (a)：已知两边（7 和 9）及其夹角 52°，使用余弦定理。

QR² = 7² + 9² − 2 × 7 × 9 × cos 52° = 49 + 81 − 126 × 0.616 = 130 − 77.6 = 52.4

QR = √52.4 ≈ 7.24 cm

Solution (b): Now use the sine rule to find angle R, which is opposite side PQ = 7 cm.

解答 (b)：利用正弦定理求角 R，它对应的边为 PQ = 7 厘米。

7 / sin R = 7.24 / sin 52° → sin R = 7 × sin 52° / 7.24 = 7 × 0.788 / 7.24 = 0.762

R = sin⁻¹(0.762) ≈ 49.7°

Check angle sum: P + R = 52° + 49.7° = 101.7°, so Q = 78.3°, which is valid. The ambiguous case does not apply here because 180° − 49.7° = 130.3° would make the angle sum exceed 180°.

检验内角和：P + R = 52° + 49.7° = 101.7°，因此 Q = 78.3°，成立。此处不存在两解问题，因为 180° − 49.7° = 130.3° 会使内角超过 180°。

Solution (c): Use the area formula with two sides and the included angle.

解答 (c)：使用两边及其夹角计算面积。

Area = ½ × 7 × 9 × sin 52° = 31.5 × 0.788 ≈ 24.8 cm²

This problem demonstrates how to move fluidly between the three triangle formulas. Mastering this process earns full marks in multi-part questions.

本题展示了如何灵活切换三种三角公式。掌握这一整串解题流程，你就能在多问答题中拿到满分。

12. Summary and Final Revision Strategy | 总结与最终复习策略

Here is what you must know for IGCSE trigonometry, condensed into one checklist:

以下是 IGCSE 三角函数必须掌握的精华清单：

SOH CAH TOA for right-angled triangles: sin θ = O/H, cos θ = A/H, tan θ = O/A.
SOH CAH TOA 用于直角三角形：sin θ = 对/斜，cos θ = 邻/斜，tan θ = 对/邻。
Exact values for 0°, 30°, 45°, 60°, 90° — memorise the pattern, not just the table.
特殊角精确值 0°、30°、45°、60°、90° —— 记住规律，而非死背表格。
Sine rule a/sin A = b/sin B = c/sin C for AAS/ASA/SSA; mind the ambiguous case.
正弦定理 a/sin A = b/sin B = c/sin C 适用于 AAS/ASA/SSA；注意两解情况。
Cosine rule a² = b² + c² − 2bc cos A for SAS/SSS; rearrange to find angles.
余弦定理 a² = b² + c² − 2bc cos A 适用于 SAS/SSS；变形后可求角度。
Area = ½ ab sin C for any triangle with two sides and included angle.
面积公式 = ½ ab sin C 适用于已知两边及夹角的任意三角形。
Graphs of sin, cos and tan: know amplitude, period, zeros, and vertical asymptote for tan.
三角函数图像：掌握 sin、cos、tan 的振幅、周期、零点及 tan 的垂直渐近线。
Transformations y = a sin(bx) + c: amplitude |a|, period 360°/b, midline y = c.
图像变换 y = a sin(bx) + c：振幅 |a|，周期 360°/b，中线 y = c。
Equations: isolate the ratio, find reference angle, then use CAST for all solutions.
解方程：分离三角函数，求参考角，再用 CAST 图确定全部解。

Practice at least one past-paper question from each category every week. Time yourself, mark your work strictly, and revisit any mistakes until you can solve the full question without looking at notes.

每周至少练习每类题目中的一道真题。计时作答、严格批改，并反复回做错题，直到能在不看笔记的情况下完整解出。

Trigonometry rewards consistent practice more than raw talent. With the rules, exact values and graph shapes firmly in memory, you will approach every trig question with confidence on exam day. Good luck!

三角函数更青睐持续练习，而非天赋。只要熟练记忆公式、特殊角精确值与图像特征，考场上你便能自信应对每一道三角函数题。祝考试顺利！

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IGCSE Trigonometry Masterclass: Sine Rule, Cosine Rule & Graphs | IGCSE 三角函数精讲：正弦定理、余弦定理与图像