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Angles in the Four Quadrants | 四个象限中的角度

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📚 Angles in the Four Quadrants | 四个象限中的角度

Understanding how angles behave in the four quadrants of the Cartesian coordinate system is a fundamental skill in A-Level Mathematics. This concept underpins trigonometry, circular measure, and the solution of trigonometric equations, making it essential for exam success.

理解笛卡尔坐标系中角度在四个象限中的行为是A-Level数学的一项基本技能。这一概念支撑着三角学、弧度制以及三角方程的求解，对考试成功至关重要。

1. The Cartesian Plane and Standard Position | 笛卡尔平面与标准位置

The Cartesian plane is divided into four quadrants by the x-axis and y-axis. Quadrant I is the region where both x and y are positive (top right). Quadrant II has x negative and y positive (top left). Quadrant III has both x and y negative (bottom left). Quadrant IV has x positive and y negative (bottom right). Angles are measured counterclockwise from the positive x-axis, which is called the standard position. The positive x-axis itself represents 0° (or 0 radians), and a full rotation brings you back to 0° (or 2π radians).

笛卡尔平面由x轴和y轴划分为四个象限。第一象限是x和y均为正的区域（右上）。第二象限x为负、y为正（左上）。第三象限x和y均为负（左下）。第四象限x为正、y为负（右下）。角度从正x轴开始按逆时针方向测量，这称为标准位置。正x轴本身代表0°（或0弧度），完整旋转一周后回到0°（或2π弧度）。

Quadrant I: 0° < θ < 90°  |  Quadrant II: 90° < θ < 180°
Quadrant III: 180° < θ < 270°  |  Quadrant IV: 270° < θ < 360°

It is crucial to remember that angles in standard position always start from the positive x-axis. A common mistake is to start measuring from the y-axis, which leads to incorrect quadrant identification. Always visualise the angle opening counterclockwise from the positive x-axis.

务必记住，标准位置的角度总是从正x轴开始测量。一个常见错误是从y轴开始测量，这会导致象限识别错误。始终想象角度从正x轴按逆时针方向张开。

2. Degrees and Radians in Quadrant Measurement | 度与弧度在象限度量中的应用

A-Level Mathematics requires fluency in both degrees and radians. A full revolution is 360°, which is equal to 2π radians. Therefore, the quadrants can be expressed in both units:

A-Level数学要求熟练掌握度和弧度两种单位。一整圈是360°，等于2π弧度。因此，四个象限可以用两种单位表示：

  • Quadrant I: 0° to 90° (0 to π/2 radians) | 第一象限：0°到90°（0到π/2弧度）

  • Quadrant II: 90° to 180° (π/2 to π radians) | 第二象限：90°到180°（π/2到π弧度）

  • Quadrant III: 180° to 270° (π to 3π/2 radians) | 第三象限：180°到270°（π到3π/2弧度）

  • Quadrant IV: 270° to 360° (3π/2 to 2π radians) | 第四象限：270°到360°（3π/2到2π弧度）

The conversion between degrees and radians is given by the relationship: 180° = π radians. To convert degrees to radians, multiply by π/180°. To convert radians to degrees, multiply by 180°/π.

度与弧度之间的换算关系为：180° = π弧度。将度转换为弧度，乘以π/180°。将弧度转换为度，乘以180°/π。

Degrees to Radians: θ (rad) = θ (deg) × π/180°
Radians to Degrees: θ (deg) = θ (rad) × 180°/π

3. Special Angles and Their Quadrant Positions | 特殊角及其象限位置

Certain angles appear frequently in trigonometry and must be recognised instantly. These include 30°, 45°, 60°, and their multiples. In radians, these correspond to π/6, π/4, and π/3 respectively. Each of these angles has exact trigonometric values that you are expected to memorise.

某些角度在三角学中频繁出现，必须能够即时识别。这些角包括30°、45°、60°及其倍数。在弧度制中，它们分别对应π/6、π/4和π/3。每个角都有需要记忆的精确三角函数值。

Angle | 角度 Radians | 弧度 Quadrant | 象限 sin cos tan
30° π/6 I ½ √3/2 1/√3
45° π/4 I √2/2 √2/2 1
60° π/3 I √3/2 ½ √3
90° π/2 Axis | 轴上 1 0 Undefined | 无定义
180° π Axis | 轴上 0 -1 0
270° 3π/2 Axis | 轴上 -1 0 Undefined | 无定义
360° 2π Axis | 轴上 0 1 0

Angles that lie exactly on the axes (0°, 90°, 180°, 270°, 360°) are called quadrantal angles. They are not considered to belong to any quadrant. When solving trigonometric equations, always check whether axis angles could be valid solutions, as they are frequently overlooked.

恰好位于坐标轴上的角（0°、90°、180°、270°、360°）称为象限界线角。它们不属于任何象限。在解三角方程时，始终检查界线角是否可能为有效解，因为它们经常被忽略。

4. Positive and Negative Angles | 正角与负角

Angles measured counterclockwise are positive, while angles measured clockwise are negative. A negative angle is equivalent to a positive angle with the same magnitude rotated in the opposite direction. For example, -30° is the same position as 330° in standard position.

逆时针方向测量的角度为正角，顺时针方向测量的角度为负角。负角与大小相同但方向相反的正角位置相同。例如，-30°与标准位置中的330°位置相同。

When working with negative angles, it is often convenient to convert them to positive angles by adding 360° (or 2π radians). This transformation does not change the terminal position of the angle but simplifies quadrant identification:

处理负角时，通常可以通过加上360°（或2π弧度）将其转换为正角。这种转换不改变角的终边位置，但简化了象限识别：

-θ (deg) ≡ 360° – θ (mod 360°)
-θ (rad) ≡ 2π – θ (mod 2π)

For instance, -120° is equivalent to 240°, which places it in Quadrant III. Similarly, -π/4 is equivalent to 7π/4, which lies in Quadrant IV. This equivalence is particularly useful when evaluating trigonometric functions of negative angles.

例如，-120°等价于240°，位于第三象限。类似地，-π/4等价于7π/4，位于第四象限。这种等价关系在计算负角的三角函数值时特别有用。

5. Reference Angles | 参考角

The reference angle is the acute angle formed between the terminal side of a given angle and the x-axis. It is always measured as a positive acute angle between 0° and 90° (0 and π/2 radians). To find the reference angle for any angle in standard position, use the following rules based on the quadrant:

参考角是给定角的终边与x轴之间形成的锐角。它始终是介于0°和90°（0和π/2弧度）之间的正锐角。根据所在象限，按以下规则求任意标准位置角的参考角：

  • Quadrant I: reference angle = θ itself | 第一象限：参考角 = θ本身

  • Quadrant II: reference angle = 180° – θ (or π – θ) | 第二象限：参考角 = 180° – θ（或π – θ）

  • Quadrant III: reference angle = θ – 180° (or θ – π) | 第三象限：参考角 = θ – 180°（或θ – π）

  • Quadrant IV: reference angle = 360° – θ (or 2π – θ) | 第四象限：参考角 = 360° – θ（或2π – θ）

α = θ  (Q I)  |  α = 180° – θ  (Q II)  |  α = θ – 180°  (Q III)  |  α = 360° – θ  (Q IV)

For example, consider θ = 150°. Since 150° is in Quadrant II, the reference angle is 180° – 150° = 30°. This means that sin(150°) = sin(30°) = ½, up to the sign determined by the quadrant. Reference angles allow you to evaluate trigonometric functions at any angle by reducing the problem to a familiar acute angle.

例如，考虑θ = 150°。由于150°在第二象限，参考角为180° – 150° = 30°。这意味着sin(150°) = sin(30°) = ½，正负号由所在象限决定。参考角使你能够通过将问题简化为熟悉的锐角来求任意角的三角函数值。

6. Signs of Trigonometric Functions in Each Quadrant | 各象限中三角函数的符号

The signs of sin θ, cos θ, and tan θ depend on the quadrant in which the terminal side of the angle lies. This can be derived from the unit circle definition, where sin θ = y-coordinate, cos θ = x-coordinate, and tan θ = y/x (or sin θ / cos θ). Since x and y have specific signs in each quadrant, the signs of the trigonometric functions follow accordingly:

sin θ、cos θ和tan θ的符号取决于角的终边所在的象限。这可以从单位圆定义推导出来：sin θ = y坐标，cos θ = x坐标，tan θ = y/x（或sin θ/cos θ）。由于x和y在各象限有特定符号，三角函数的符号也相应确定：

Quadrant | 象限 x (cos) y (sin) sin θ cos θ tan θ
I + + + + +
II – + + – –
III – – – – +
IV + – – + –

Notice that in Quadrant I, all three functions are positive. In Quadrant II, only sin is positive. In Quadrant III, only tan is positive. In Quadrant IV, only cos is positive. This gives rise to the famous mnemonic ‘All Students Take Calculus’ (ASTC), read counterclockwise from Quadrant I, which reminds you which functions are positive in each quadrant.

注意在第一象限，三个函数都为正值。在第二象限，仅sin为正。在第三象限，仅tan为正。在第四象限，仅cos为正。这引出了著名的助记口诀’All Students Take Calculus’（ASTC，全为学生学微积分），从第一象限开始逆时针读取，提醒你每个象限中哪些函数为正。

7. The CAST Diagram | CAST图示法

The CAST diagram is another popular visual aid for remembering the signs of trigonometric functions. It is arranged as a circle divided into four quadrants with the letters C, A, S, T written in a specific pattern:

CAST图示法是一种流行的辅助记忆三角函数符号的视觉工具。它被安排为一个分为四象限的圆，字母C、A、S、T按特定模式书写：

    S  |  A
───┼───
    T  |  C

The letters indicate which function is positive in each quadrant: A (All) in Quadrant I, S (Sine) in Quadrant II, T (Tangent) in Quadrant III, C (Cosine) in Quadrant IV. To remember this, students often use the phrase ‘All Students Take Calculus’ or the Chinese mnemonic ‘全正、正弦、正切、余弦’.

字母表示每个象限中为正的函数：第一象限为A（全部），第二象限为S（正弦），第三象限为T（正切），第四象限为C（余弦）。为记住这一点，学生常用短语’All Students Take Calculus’或中文助记’全正、正弦、正切、余弦’。

When solving trigonometric equations, the CAST diagram helps you determine in which quadrants solutions exist. For example, if you need to solve sin θ = 0.5 for 0° ≤ θ < 360°, you first find the acute solution θ = 30°, then identify that sin is positive in Quadrants I and II, giving the solutions 30° and 150°.

在解三角方程时，CAST图示帮助你确定解存在的象限。例如，若在0° ≤ θ < 360°范围内解sin θ = 0.5，你首先求出锐角解θ = 30°，然后确定sin在第一和第二象限为正，得出解为30°和150°。

8. Angles Greater Than 360° and Coterminal Angles | 大于360°的角与共终边角

Angles can exceed 360° (or 2π radians). An angle of 390° is the same terminal position as 30° because 390° – 360° = 30°. Two angles that share the same terminal side are called coterminal angles. Any angle θ has infinitely many coterminal angles given by θ + 360°n (or θ + 2πn) where n is any integer.

角可以超过360°（或2π弧度）。390°的角与30°的终边位置相同，因为390° – 360° = 30°。具有相同终边的两个角称为共终边角。任何角θ都有无穷多个共终边角，表示为θ + 360°n（或θ + 2πn），其中n为任意整数。

To find the quadrant of an angle greater than 360°, subtract multiples of 360° until the result lies between 0° and 360°. The quadrant of this reduced angle is the same as the quadrant of the original angle. For example, 750° = 750° – 2 × 360° = 30°, which is in Quadrant I.

要求大于360°的角所在的象限，反复减去360°的倍数，直到结果在0°到360°之间。这个约简后角的象限与原角相同。例如，750° = 750° – 2 × 360° = 30°，在第一象限。

In radians, the same procedure applies with 2π. For instance, 17π/4 = 17π/4 – 2 × 2π = 17π/4 – 16π/4 = π/4, which is in Quadrant I. This technique is essential in calculus when working with periodic functions and their domains.

在弧度制中，同样用2π操作。例如，17π/4 = 17π/4 – 2 × 2π = 17π/4 – 16π/4 = π/4，在第一象限。这一技巧在微积分中处理周期函数及其定义域时至关重要。

9. Solving Trigonometric Equations Using Quadrant Information | 利用象限信息解三角方程

Solving trigonometric equations requires careful use of quadrant information. The general method involves finding the principal solution (the acute angle or reference angle), determining which quadrants contain valid solutions based on the sign of the trigonometric function, and then calculating all solutions within the given interval.

解三角方程需要仔细运用象限信息。一般方法包括：求主解（锐角或参考角），根据三角函数的符号确定哪些象限包含有效解，然后在给定区间内计算所有解。

Consider the equation cos θ = -½ for 0° ≤ θ < 360°. The reference angle is cos⁻¹(½) = 60°. Since cos is negative in Quadrants II and III, the solutions are:

考虑方程cos θ = -½，其中0° ≤ θ < 360°。参考角为cos⁻¹(½) = 60°。由于cos在第二和第三象限为负，解为：

Quadrant II: θ = 180° – 60° = 120°
Quadrant III: θ = 180° + 60° = 240°
Solution set: θ = 120°, 240°

For equations in radians, the same logic applies. Solve tan θ = √3 for 0 ≤ θ < 2π. The reference angle is tan⁻¹(√3) = π/3. Since tan is positive in Quadrants I and III, the solutions are θ = π/3 and θ = π + π/3 = 4π/3.

对于弧度制方程，同样的逻辑适用。在0 ≤ θ < 2π范围内解tan θ = √3。参考角为tan⁻¹(√3) = π/3。由于tan在第一和第三象限为正，解为θ = π/3和θ = π + π/3 = 4π/3。

When a trigonometric equation involves transformations such as sin(2θ) or cos(θ + π/4), it is essential to first adjust the interval. For example, solving sin(2θ) = 0.5 for 0° ≤ θ < 360° requires expanding the interval to 0° ≤ 2θ < 720°, finding all solutions for 2θ, and then dividing by 2 to obtain the values of θ.

当三角方程涉及变换如sin(2θ)或cos(θ + π/4)时，必须首先调整区间。例如，在0° ≤ θ < 360°中解sin(2θ) = 0.5，需要将区间扩展为0° ≤ 2θ < 720°，求出2θ的所有解，然后除以2得到θ的值。

10. Graphs of Trigonometric Functions Across Quadrants | 三角函数在四个象限中的图像

The graph of y = sin θ oscillates between -1 and 1, completing one full cycle over 360° (or 2π radians). In Quadrant I, sin θ increases from 0 to 1. In Quadrant II, it decreases from 1 to 0. In Quadrant III, it continues decreasing from 0 to -1. In Quadrant IV, it increases from -1 to 0.

函数y = sin θ的图像在-1和1之间振荡，每360°（或2π弧度）完成一个完整周期。在第一象限，sin θ从0增加到1。在第二象限，它从1减小到0。在第三象限，它继续从0减小到-1。在第四象限，它从-1增加到0。

The graph of y = cos θ has the same amplitude and period but is phase-shifted by 90° compared to sin θ. In Quadrant I, cos θ decreases from 1 to 0. In Quadrant II, it decreases from 0 to -1. In Quadrant III, it increases from -1 to 0. In Quadrant IV, it increases from 0 to 1.

函数y = cos θ的图像具有相同的振幅和周期，但相比sin θ相位移动了90°。在第一象限，cos θ从1减小到0。在第二象限，它从0减小到-1。在第三象限，它从-1增加到0。在第四象限，它从0增加到1。

The graph of y = tan θ has vertical asymptotes at θ = 90° and θ = 270° (where cos θ = 0), where the function is undefined. tan θ is positive and increasing from 0 to +∞ in Quadrant I, negative and increasing from -∞ to 0 in Quadrant II, positive and increasing from 0 to +∞ in Quadrant III, and negative and increasing from -∞ to 0 in Quadrant IV. Its period is 180° (or π radians), not 360°.

函数y = tan θ的图像在θ = 90°和θ = 270°处有垂直渐近线（此时cos θ = 0），函数在这些点无定义。在第一象限，tan θ为正且从0增加到+∞；在第二象限，为负且从-∞增加到0；在第三象限，为正且从0增加到+∞；在第四象限，为负且从-∞增加到0。其周期为180°（或π弧度），不是360°。

11. Worked Examples | 典型例题解析

Let us work through several exam-style problems to consolidate the concepts covered in this article.

让我们通过几个典型例题来巩固本文所涵盖的概念。

Example 1 | 例1: Determine the quadrant of each angle: (a) 212°, (b) -95°, (c) 5π/3.

例1：确定下列每个角所在的象限：(a) 212°，(b) -95°，(c) 5π/3。

Solution: (a) 212° lies between 180° and 270°, so it is in Quadrant III. (b) -95° is equivalent to -95° + 360° = 265°, which lies between 270° and 360° if read backwards, or note that 265° is between 180° and 270°, so it is in Quadrant III. Wait, let us correct: 265° is actually between 180° and 270°? No, 265° is between 180° and 270° is incorrect because 265° is greater than 270°? No, 265° is less than 270° and greater than 180°, so it is in Quadrant III? Let me check: 270° would be the boundary. 265° is less than 270°, so it is not in Quadrant IV (which starts at 270°). 265° is greater than 180°, so it is in Quadrant III? Actually, Quadrant III is 180° to 270°, yes, 265° is in Quadrant III. But wait, Quadrant IV is 270° to 360°, so 265° is indeed in Quadrant III. Hmm, that seems counterintuitive because 265° is closer to 270°, but yes, Quadrant III covers 180° to 270° exclusive. (c) 5π/3 is greater than 3π/2 and less than 2π, so it is in Quadrant IV.

解答：(a) 212°介于180°和270°之间，因此在第三象限。(b) -95°等价于-95° + 360° = 265°，265°介于180°和270°之间，因此在第三象限。(c) 5π/3大于3π/2且小于2π，因此在第四象限。

Wait, let us double-check 5π/3. 3π/2 = 4.712… and 5π/3 ≈ 5.236, 2π ≈ 6.283, so yes, 5π/3 is in Quadrant IV (between 3π/2 and 2π). Correct!

等一下，让我们检查5π/3。3π/2 = 4.712…，5π/3 ≈ 5.236，2π ≈ 6.283，所以5π/3确实在第四象限（介于3π/2和2π之间）。正确！

Example 2 | 例2: Given that θ is in Quadrant III and sin θ = -5/13, find cos θ and tan θ exactly.

例2：已知θ在第三象限且sin θ = -5/13，精确求cos θ和tan θ。

Solution: Using sin²θ + cos²θ = 1, we have cos²θ = 1 – sin²θ = 1 – 25/169 = 144/169. Since θ is in Quadrant III, cos θ is negative, so cos θ = -12/13. Therefore, tan θ = sin θ / cos θ = (-5/13) / (-12/13) = 5/12. In Quadrant III, tan θ should be positive, confirming the sign.

解答：利用sin²θ + cos²θ = 1，得到cos²θ = 1 – sin²θ = 1 – 25/169 = 144/169。由于θ在第三象限，cos θ为负，所以cos θ = -12/13。因此，tan θ = sin θ/cos θ = (-5/13)/(-12/13) = 5/12。在第三象限，tan θ应为正，验证了符号的正确性。

Example 3 | 例3: Solve the equation 2cos θ + 1 = 0 for 0° ≤ θ < 360°, giving your answers correct to 1 decimal place.

例3：在0° ≤ θ < 360°范围内解方程2cos θ + 1 = 0，答案精确到1位小数。

Solution: 2cos θ + 1 = 0 ⇒ cos θ = -½. The reference angle is cos⁻¹(½) = 60°. Since cos is negative in Quadrants II and III, the solutions are θ = 180° – 60° = 120° and θ = 180° + 60° = 240°.

解答：2cos θ + 1 = 0，即cos θ = -½。参考角为cos⁻¹(½) = 60°。由于cos在第二和第三象限为负，解为θ = 180° – 60° = 120°和θ = 180° + 60° = 240°。

12. Common Pitfalls and Tips | 常见错误与学习要点

Students frequently make the following mistakes when working with angles in the four quadrants. Being aware of these pitfalls will help you avoid them in exams.

学生在处理四个象限中的角度时经常犯以下错误。了解这些陷阱有助于你在考试中避免它们。

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