📚 Chapter Review 6: Trigonometric Ratios | 第六章复习：三角比
This chapter review gives a complete, exam-focused summary of the trigonometric ratios studied in Edexcel A Level Pure Mathematics Year 1 Chapter 6. You will revisit sine, cosine and tangent for right-angled triangles and for angles of any magnitude, memorise exact values for key angles, sketch and transform trigonometric graphs, apply the two fundamental identities, solve trigonometric equations in a given interval, and use the sine rule, cosine rule and area formula for non-right-angled triangles. Mastering these skills is essential for success in the AS and A Level examinations.
本章复习全面总结了 Edexcel A Level 纯数第一册第六章的三角比内容。你将重温直角三角形和任意角的三角函数值，熟记特殊角的精确值，绘制并变换三角函数图像，运用两个基本三角恒等式，在给定区间内解三角方程，并对非直角三角形使用正弦定理、余弦定理与面积公式。掌握这些技能是 AS 和 A Level 考试成功的关键。
1. Trigonometric Ratios in Right-Angled Triangles | 直角三角形中的三角比
e, cos θ = adjacent ÷ hypotenuse and tan θ = opposite ÷ adjacent. The hypotenuse is always the longest side, opposite the right angle; the opposite side faces θ; the adjacent side touches θ. These ratios link an angle to the side lengths and are the foundation of every right-triangle problem.
对于含锐角 θ 的直角三角形，基本三角比定义为：sin θ = 对边 ÷ 斜边，cos θ = 邻边 ÷ 斜边，tan θ = 对边 ÷ 邻边。斜边永远是最长边，直角所对；对边与 θ 相对；邻边与 θ 相邻。这些比值把角与边长联系起来，是解决直角三角形问题的基础。
θ = O ÷ H cos θ = A ÷ H tan θ = O ÷ A (SOH CAH TOA)
To find an unknown side, substitute the known angle and side into the correct ratio and rearrange; multiplying by the hypotenuse or adjacent side is usually the quickest move. To find an unknown angle, apply the inverse functions sin⁻¹, cos⁻¹ and tan⁻¹, and always confirm that your calculator is in degree mode.
求未知边时，把已知角和已知边代入正确的比值再整理；通常直接乘以斜边或邻边最快。求未知角时，要使用反函数 sin⁻¹、cos⁻¹、tan⁻¹，并务必确认计算器处于度数模式。
e side is 5 cm and the adjacent side is 9 cm, then θ = tan⁻¹(5 ÷ 9) ≈ 29.1°. Note that Pythagoras' theorem combines with these ratios to give the identity sin²θ + cos²θ = 1.
例题：直角三角形斜边 12 cm，一锐角为 35°，则对边 = 12 × sin 35° ≈ 6.88 cm。若对边 5 cm、邻边 9 cm，则 θ = tan⁻¹(5 ÷ 9) ≈ 29.1°。注意勾股定理与这些比值结合，即可得到恒等式 sin²θ + cos²θ = 1。
2. Exact Values for Key Angles | 特殊角的精确值
The specification requires you to know the exact values of sine, cosine and tangent for 0°, 30°, 45°, 60° and 90°. They come from two special triangles: the isosceles right-angled triangle with sides 1, 1, √2, and the half-equilateral triangle with sides 1, √3, 2. In the exam you must be able to write them down without a calculator.
考纲要求你熟记 0°、30°、45°、60° 和 90° 的正弦、余弦和正切精确值。它们来自两个特殊三角形：边长 1、1、√2 的等腰直角三角形，以及边长 1、√3、2 的等边三角形的一半。考试中你必须能不用计算器直接写出。
| Angle θ | sin θ | cos θ | tan θ |
|---|---|---|---|
| 0° | 0 | 1 | 0 |
| 30° | ½ | √3 ÷ 2 | 1 ÷ √3 = √3 ÷ 3 |
| 45° | 1 ÷ √2 = √2 ÷ 2 | 1 ÷ √2 = √2 ÷ 2 | 1 |
| 60° | √3 ÷ 2 | ½ | √3 |
| 90° | 1 | 0 | undefined |
cute angles of a right-angled triangle are complementary. Also remember tan 90° is undefined since it would require division by zero, and that rationalising denominators is expected in final answers.
注意其对称性：sin 30° = cos 60°，sin 60° = cos 30°，因为直角三角形两锐角互余。还要记住 tan 90° 无定义，因为会出现除以零；最终答案中分母一般要求有理化。
These exact values are also used in radian form: sin(π ÷ 6) = ½, cos(π ÷ 4) = √2 ÷ 2 and tan(π ÷ 3) = √3. Recognising both forms saves time when a question mixes degrees and radians.
这些精确值也常用于弧度形式：sin(π ÷ 6) = ½，cos(π ÷ 4) = √2 ÷ 2，tan(π ÷ 3) = √3。两种形式都熟悉，可以在题目混用角度与弧度时节省时间。
3. The Unit Circle and Angles of Any Magnitude | 单位圆与任意角
minal side meets the circle of radius 1 has coordinates (cos θ, sin θ). Thus cos θ is the x-coordinate and sin θ is the y-coordinate, while tan θ = sin θ ÷ cos θ is the gradient of the radius.
要把三角比推广到 90° 以外，需要借助单位圆。角 θ 从 x 轴正方向逆时针量起，终边与半径为 1 的圆交点的坐标即为 (cos θ, sin θ)。因此 cos θ 是横坐标，sin θ 是纵坐标，而 tan θ = sin θ ÷ cos θ 就是该半径的斜率。
The signs of the ratios in each quadrant are summarised by the CAST diagram: in quadrant 1 all three are positive; in quadrant 2 only sine (and cosecant) is positive; in quadrant 3 only tangent is positive; in quadrant 4 only cosine is positive. A quick alternative is the phrase "All Students Take Calculus".
各象限三角比的符号可用 CAST 图记忆：第一象限三者皆正；第二象限只有正弦为正；第三象限只有正切为正；第四象限只有余弦为正。也可以用口诀 "All Students Take Calculus" 快速判断。
Related-angle results follow directly from the symmetry of the circle, and they are the fastest way to find a second solution:
借助圆的对称性可直接得到相关角公式，它们是寻找第二个解最快的方法：
- sin(180° − θ) = sin θ, cos(180° − θ) = −cos θ, tan(180° − θ) = −tan θ
sin(180° − θ) = sin θ，cos(180° − θ) = −cos θ，tan(180° − θ) = −tan θ - sin(180° + θ) = −sin θ, cos(180° + θ) = −cos θ, tan(180° + θ) = tan θ
sin(180° + θ) = −sin θ，cos(180° + θ) = −cos θ，tan(180° + θ) = tan θ - sin(360° − θ) = −sin θ, cos(360° − θ) = cos θ, tan(360° − θ) = −tan θ
sin(360° − θ) = −sin θ，cos(360° − θ) = cos θ，tan(360° − θ) = −tan θ
For example, sin 150° = sin 30° = ½, cos 210° = −cos 30° = −√3 . Negative angles are measured clockwise, so sin(−40°) = −sin 40° and cos(−40°) = cos 40°.
例如，sin 150° = sin 30° = ½，cos 210° = −cos 30° = −√3 ÷ 2，tan 315° = −tan 45° = −1。负角按顺时针度量，所以 sin(−40°) = −sin 40°，cos(−40°) = cos 40°。
4. Graphs of Sine, Cosine and Tangent | 正弦、余弦与正切图像
with period 360° and range −1 ≤ y ≤ 1, while the tangent curve repeats every 180° and has vertical asymptotes where cosine is zero.
你必须能准确画出三条曲线，标出截距、最高点、最低点和渐近线。正弦与余弦曲线是周期 360°、值域为 −1 ≤ y ≤ 1 的波浪线；正切曲线每 180° 重复一次，并在余弦为零处有垂直渐近线。
| Function | Period | Range | Key features |
|---|---|---|---|
| y = sin θ | 360° | −1 ≤ y ≤ 1 | passes through origin, max at 90°, min at 270° |
| y = cos θ | 360° | −1 ≤ y ≤ 1 | max at 0°, min at 180°, sin curve shifted 90° left |
| y = tan θ | 180° | all real y | asymptotes at 90°, 270°, −90°; intercepts at 0°, 180° |
−θ) = −f(θ)), while the cosine graph is symmetric about the y-axis (an even function, so f(−θ) = f(θ)). These explain why cos(−θ) = cos θ but sin(−θ) = −sin θ.
两个对称性质值得牢记：正弦与正切图像关于原点中心对称（奇函数，f(−θ) = −f(θ)），余弦图像关于 y 轴对称（偶函数，f(−θ) = f(θ)）。这解释了为什么 cos(−θ) = cos θ，而 sin(−θ) = −sin θ。
or any integer n. This is exactly the tool you need when a question asks for all solutions rather than those in a stated interval.
周期性可以写成：对任意整数 n，sin(θ ± 360°n) = sin θ，cos(θ ± 360°n) = cos θ，tan(θ ± 180°n) = tan θ。当题目要求"所有解"而非某区间内的解时，这正是所需的工具。
5. Transformations of Trigonometric Graphs | 三角函数图像的变换
s y = a sin(bθ + c) + d, and each constant has a clear graphical meaning. The amplitude is |a| ertical translation; and the term c ÷ b produces a horizontal translation (a phase shift).
一般的变换形式可写为 y = a sin(bθ + c) + d，每个常数都有明确的图像含义：振幅为 |a|，最大值 a + d，最小值 −a + d；周期为 360° ÷ b；d 是上下平移；c ÷ b 则产生左右平移（相位移动）。
y = a sin(bθ + c) + d amplitude = |a| period = 360° ÷ b
tude 3, period 180°, maximum 2 and minimum −4. A stretch parallel to the y-axis is described as a stretch of scale factor a in the y-direction, whereas the change caused by b is a stretch of scale factor 1 ÷ b in the x-direction.
例如 y = 3 cos(2θ) − 1 的振幅为 3，周期为 180°，最大值 2，最小值 −4。沿 y 轴的拉伸称为 y 方向比例因子 a 的拉伸，而 b 带来的变化是 x 方向比例因子 1 ÷ b 的拉伸。
When sketching, mark the key points of one full period first — intercepts, maxima, minima and asymptotes — then repeat the pattern across the required domain. Showing the correct number of complete cycles within the domain is often worth a mark on its own.
作图时，先标出一个完整周期的关键点——截距、最高点、最低点和渐近线——再在给定区间内重复该形状。在区间内画出正确数量的完整周期，本身往往就是一个得分点。
6. The Two Fundamental Identities | 两个基本三角恒等式
Two identities underpin almost every trigonometric proof and equation in the course. The first comes from Pythagoras' theorem applied to the unit circle, and the second from the definitions of sine, cosine and tangent.
课程中几乎所有三角恒等式证明与解方程都建立在两个恒等式之上。第一个来自单位圆上的勾股定理，第二个来自正弦、余弦、正切的定义。
sin²θ + cos²θ = 1 tan θ = sin θ ÷ cos θ (cos θ ≠ 0)
sin²θ, which is exactly the substitution needed when an equation mixes squares with a single term. The tangent identity converts a tangent into a ratio of sine and cosine, allowing factorisation and cancellation.
整理后，勾股恒等式给出 sin²θ = 1 − cos²θ 与 cos²θ = 1 − sin²θ，正是方程中同时出现平方项与一次项时所需的代换。正切恒等式把正切化为正弦与余弦之比，便于因式分解与约分。
ine fractions over a common denominator, and use sin²θ + cos²θ = 1 to remove pairs of squared terms. Never move terms across the equals sign as though the identity were an equation to be solved.
证明恒等式时，从较复杂的一边出发，逐步化简到与另一边相同。把全部函数化为正弦和余弦，用通分合并分式，并用 sin²θ + cos²θ = 1 消去平方项的组合。切勿把它当成方程，把项移过等号。
7. Solving Trigonometric Equations in a Given Interval | 在给定区间内解三角方程
lows a fixed routine. First find the principal value α using the inverse function on your calculator. Then use the symmetry of the relevant graph, or the CAST diagram, to obtain every solution in the required interval. Finally check each answer lies inside the stated range.
解 sin θ = k、cos θ = k 或 tan θ = k 有一套固定步骤。先用计算器反函数求出主值 α，再借助相应图像的对称性或 CAST 图，得到区间内的全部解，最后逐一检验答案是否落在给定范围内。
The general solutions, valid for any integer n, are:
对任意整数 n，通解公式为：
- sin θ = k → θ = α + 360°n or θ = 180° − α + 360°n
sin θ = k → θ = α + 360°n 或 θ = 180° − α + 360°n - cos θ = k → θ = α + 360°n or θ = −α + 360°n
cos θ = k → θ = α + 360°n 或 θ = −α + 360°n - tan θ = k → θ = α + 180°n
tan θ = k → θ = α + 180°n
Remember the ranges of the inverse functions: sin⁻¹ and tan⁻¹ return values between −90° and 90°, while cos⁻¹ returns values between 0° and 180°. This is why a positive cosine gives a principal value in the first quadrant but a negative cosine gives one in the second.
记住反函数的值域：sin⁻¹ 与 tan⁻¹ 返回 −90° 到 90° 之间的值，cos⁻¹ 返回 0° 到 180° 之间的值。这解释了为什么余弦为正时主值在第一象限，而余弦为负时主值落在第二象限。
question had asked for −360° ≤ θ ≤ 360°, we would also include 30° − 360° = −330° and 150° − 360° = −210°.
例：在 0° ≤ θ ≤ 360° 内解 sin θ = 0.5。主值为 30°，由对称性得第二个解 180° − 30° = 150°，故 θ = 30° 或 150°。若区间改为 −360° ≤ θ ≤ 360°，还要加上 30° − 360° = −330° 与 150° − 360° = −210°。
8. Equations with Double Angles and Quadratics | 倍角与二次型三角方程
ssion appears), solve for u within the correspondingly enlarged interval, then convert back to θ. Solving for θ directly usually loses solutions.
当未知量以 2θ、3θ 或 θ ± 30° 的形式出现时，最稳妥的方法是换元。令 u = 2θ（或题目中出现的表达式），在相应放大的区间内解出 u，再换回 θ。直接对 θ 求解往往会丢解。
2sin θ + 1)(sin θ − 1) = 0, so sin θ = −½ or sin θ = 1. Hence θ = 210°, 330°, or θ = 90°. The solution set is θ = 90°, 210° and 330°.
若方程是关于 sin θ 或 cos θ 的二次式，就把该三角比当作变量进行因式分解。例如在 0° ≤ θ ≤ 360° 内解 2sin²θ − sin θ − 1 = 0。分解得 (2sin θ + 1)(sin θ − 1) = 0，故 sin θ = −½ 或 sin θ = 1，于是 θ = 210°、330° 或 90°。解集为 θ = 90°、210°、330°。
If an equation contains both sine and cosine squared, replace one of them using sin
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