📚 F - Multiples of π/6 and π/4 | π/6 和 π/4 的倍数
Understanding angles that are integer multiples of π/6 and π/4 is essential for mastering trigonometric functions without a calculator. These angles generate the famous 'special angles' on the unit circle, making it possible to evaluate sine, cosine, and tangent exactly using surds and simple fractions. This topic appears throughout IB Mathematics, from solving equations to modelling periodic phenomena. In this article, we will explore how to identify these angles, find their coordinates on the unit circle, and determine their trigonometric values accurately.
理解 π/6 和 π/4 整数倍的角度，是无需计算器就能掌握三角函数的关键。这些角在单位圆上产生著名的“特殊角”，使得我们可以用根式和简单分数精确求出正弦、余弦和正切值。从解方程到周期现象建模，这一内容贯穿 IB 数学始终。本文将探讨如何识别这些角、找出它们在单位圆上的坐标，并准确确定其三角函数值。
1. The Unit Circle and Radian Measure | 单位圆与弧度制
2) and kπ/4 (k = 0, 1, ..., 8) lie on the unit circle and have exact trigonometric values.
单位圆是以原点为圆心、半径为 1 的圆。在弧度制下，π 对应 180°，因此 π/6 = 30°，π/4 = 45°。当我们考虑这些基本角的整数倍时，所有特殊角便会出现：所有形如 kπ/6（k=0,1,2,...,12）和 kπ/4（k=0,1,...,8）的角都位于单位圆上，并具有精确的三角函数值。
Working in radians is vital for calculus and trigonometric analysis. The circumference of the unit circle is 2π, so moving by multiples of π/6 or π/4 sweeps out arcs that are easy to compare with the axes and quadrants.
使用弧度对于微积分和三角分析至关重要。单位圆的周长为 2π，因此以 π/6 或 π/4 的倍数移动所扫过的弧，可以方便地与坐标轴和象限比较。
2. Angles as Multiples of π/6 | π/6 的倍数角
The set of angles kπ/6 with integer k from 0 to 12 gives the following radian measures: 0, π/6, π/3, π/2, 2π/3, 5π/6, π, 7π/6, 4π/3, 3π/2, 5π/3, 11π/6, 2π. These correspond to 0°, 30°, 60°, 90°, 120°, 150°, 180°, 210°, 240°, 270°, 300°, 330°, 360°. Every 30° step on the unit circle is covered.
k 为 0 到 12 的整数时，kπ/6 这组角度的弧度值包括：0、π/6、π/3、π/2、2π/3、5π/6、π、7π/6、4π/3、3π/2、5π/3、11π/6、2π。它们对应 0°、30°、60°、90°、120°、150°、180°、210°、240°、270°、300°、330°、360°，覆盖了单位圆上每 30° 一步的所有角。
Notice how some of these angles reduce to fractions with denominator 6: π/6, 5π/6, 7π/6, 11π/6; others have denominator 3: π/3, 2π/3, 4π/3, 5π/3; and others are quadrantal angles: 0, π/2, π, 3π/2, 2π. This grouping helps when recalling exact values.
注意有些角度可化简为分母为 6 的分数：π/6、5π/6、7π/6、11π/6；有些是分母为 3：π/3、2π/3、4π/3、5π/3；还有轴角：0、π/2、π、3π/2、2π。这样的分组有助于回忆准确值。
3. Angles as Multiples of π/4 | π/4 的倍数角
π/4, 3π/2, 7π/4, 2π. These correspond to 0°, 45°, 90°, 135°, 180°, 225°, 270°, 315°, 360°. This set provides a 45° step around the circle and includes quadrantal angles at every multiple of π/2.
对于 π/4 的倍数，有 kπ/4（k=0,1,...,8）：0、π/4、π/2、3π/4、π、5π/4、3π/2、7π/4、2π。它们对应 0°、45°、90°、135°、180°、225°、270°、315°、360°。这组角度在单位圆上以 45° 步进，并在每个 π/2 倍数处包含轴角。
are not quadrantal are π/4, 3π/4, 5π/4, and 7π/4. These all lie on the lines y = x or y = –x, which simplifies finding coordinates and trigonometric values.
分母为 4 而非轴角的角是 π/4、3π/4、5π/4 和 7π/4。它们都位于直线 y=x 或 y=–x 上，这简化了坐标和三角函数值的求解。
4. Special Right Triangles | 特殊直角三角形
Two right triangles underpin all exact trigonometric values for these angles. The first is the isosceles right triangle with angles 45°, 45°, 90° (π/4, π/4, π/2); its sides are in the ratio 1 : 1 : √2. The second is the 30°-60°-90° triangle (π/6, π/3, π/2); its sides are in the ratio 1 : √3 : 2, where the shortest side is opposite the 30° angle.
所有特殊角的精确三角函数值都源于两个直角三角形。第一个是内角为 45°、45°、90°（π/4、π/4、π/2）的等腰直角三角形，边长比为 1 : 1 : √2。第二个是 30°-60°-90° 三角形（π/6、π/3、π/2），边长比为 1 : √3 : 2，最短边对着 30° 角。
By placing these triangles inside the unit circle so that the hypotenuse becomes the radius (length 1), we can scale the sides. For the 45° triangle, the legs become 1/√2 = √2/2. For the 30°-60° triangle, the legs become 1/2 (opposite 30°) and √3/2 (opposite 60°).
把这些三角形放在单位圆内，使斜边成为半径（长度为 1），就可以缩放边长。对于 45° 三角形，直角边变成 1/√2 = √2/2。对于 30°-60° 三角形，直角边变成 1/2（对 30° 角）和 √3/2（对 60° 角）。
5. Coordinates on the Unit Circle for π/6 Multiples | π/6 倍数在单位圆上的坐标
of π/6 in the first quadrant are derived from the 30°-60° triangle. At θ = π/6, cos(π/6) = √3/2, sin(π/6) = 1/2. At θ = π/3 (which is 2π/6), cos(π/3) = 1/2, sin(π/3) = √3/2.
第一象限中 π/6 倍数角的坐标 (cosθ, sinθ) 由 30°-60° 三角形导出。当 θ=π/6 时，cos(π/6)=√3/2，sin(π/6)=1/2。当 θ=π/3（即 2π/6）时，cos(π/3)=1/2，sin(π/3)=√3/2。
o the quadrant. For instance, 5π/6 is in Quadrant II with reference angle π/6: cos(5π/6) = –√3/2, sin(5π/6) = 1/2. Similarly, 7π/6 gives cos = –√3/2, sin = –1/2; 11π/6 gives cos = √3/2, sin = –1/2.
在第二、三、四象限，坐标的绝对值不变，符号根据象限改变。例如，5π/6 在第二象限，参考角为 π/6：cos(5π/6)=–√3/2，sin(5π/6)=1/2。类似地，7π/6 的 cos = –√3/2，sin = –1/2；11π/6 的 cos = √3/2，sin = –1/2。
For the π/3 family (reference angle π/3), the coordinates swap: 2π/3 (QII) gives cos = –1/2, sin = √3/2; 4π/3 (QIII) gives (–1/2, –√3/2); 5π/3 (QIV) gives (1/2, –√3/2).
对于 π/3 家族（参考角 π/3），坐标互换：2π/3（第二象限）给出 cos = –1/2，sin = √3/2；4π/3（第三象限）为 (–1/2, –√3/2)；5π/3（第四象限）为 (1/2, –√3/2)。
6. Coordinates on the Unit Circle for π/4 Multiples | π/4 倍数在单位圆上的坐标
For angles that are odd multiples of π/4, the coordinates involve √2/2. In the first quadrant, at π/4, cos = sin = √2/2. In the second quadrant, 3π/4 has cos = –√2/2, sin = √2/2. In the third quadrant, 5π/4: cos = –√2/2, sin = –√2/2. In the fourth quadrant, 7π/4: cos = √2/2, sin = –√2/2.
对于 π/4 的奇数倍角，坐标都涉及 √2/2。第一象限 π/4 处，cos = sin = √2/2。第二象限 3π/4 的 cos = –√2/2，sin = √2/2。第三象限 5π/4：cos = –√2/2，sin = –√2/2。第四象限 7π/4：cos = √2/2，sin = –√2/2。
Because all these points lie on the lines y = x or y = – , and the tangent values are simply ±1. This symmetry makes mental calculation very quick.
由于这些点都在直线 y=x 或 y=–x 上，坐标具有对称性，正切值为 ±1。这种对称性使得心算非常快。
7. Sine, Cosine and Tangent Values | 正弦、余弦和正切值
given by sin/cos. For angles where cos = 0 (π/2, 3π/2), tangent is undefined. Key values:
利用坐标，我们可以列出正弦、余弦和正切的精确值。正切由 sin/cos 给出。对于 cos=0 的角（π/2, 3π/2），正切无定义。关键值如下表：
| θ (rad) | sin θ | cos θ | tan θ |
|---|---|---|---|
| π/6 | 1/2 | √3/2 | 1/√3 = √3/3 |
| π/4 | √2/2 | √2/2 | 1 |
| π/3 | √3/2 | 1/2 | √3 |
These three first-quadrant angles are the building blocks. All other special angles can be obtained by reflecting these values into other quadrants and adjusting signs according to the mnemonic 'All Students Take Calculus' or CAST.
这三个第一象限角是基础构件。所有其他特殊角都可以通过将这些值反射到其他象限，并根据“All Students Take Calculus”（CAST）助记规则调整符号得到。
8. Using the CAST Diagram | 使用 CAST 图
The CAST diagram reminds us which trigonometric functions are positive in each quadrant. In Quadrant I (0 to π/2): All positive. Quadrant II (π/2 to π): Sin positive. Quadrant III (π to 3π/2): Tan positive. Quadrant IV (3π/2 to 2π): Cos positive. This determines the signs of coordinates and of tan.
CAST 图帮助我们记忆每个象限中哪些三角函数为正。第一象限（0 至 π/2）：全部为正。第二象限（π/2 至 π）：正弦为正。第三象限（π 至 3π/2）：正切为正。第四象限（3π/2 至 2π）：余弦为正。这决定了坐标和正切的符号。
For example, at 5π/6 (QII), sine is positive, cosine negative, so tan is negative. At 7π/4 (QIV), cosine positive, sine negative, so tan negative. Applying CAST systematically prevents sign errors.
例如，在 5π/6（第二象限）处，正弦为正，余弦为负，因此正切为负。在 7π/4（第四象限）处，余弦为正，正弦为负，因此正切为负。系统运用 CAST 可以避免符号错误。
9. Worked Example: Finding sin(11π/6) and cos(11π/6) | 实例：求 sin(11π/6) 和 cos(11π/6)
The angle 11π/6 is a multiple of π/6 and lies in Quadrant IV. Its reference angle is 2π – = π/6. From the unit circle, the absolute values match those of π/6: sine = 1/2, cosine = √3/2. In QIV, sine is negative, cosine positive. Hence sin(11π/6) = –1/2, cos(11π/6) = √3/2.
角 11π/6 是 π/6 的倍数，位于第四象限。它的参考角为 2π – 11π/6 = π/6。由单位圆可知，绝对值与 π/6 相同：正弦为 1/2，余弦为 √3/2。在第四象限，正弦为负，余弦为正。因此 sin(11π/6) = –1/2，cos(11π/6) = √3/2。
We can verify: tan(11π/6) = sin/cos = (–1/2) / (√3/2) = –1/√3, rationalised to –√3/3. This is consistent with the reference angle tan(π/6) but negative.
可验证：tan(11π/6) = sin/cos = (–1/2) / (√3/2) = –1/√3，有理化为 –√3/3。这与参考角 tan(π/6) 的值一致，但为负。
10. Worked Example: tan(5π/4) and Reciprocal Functions | 实例：tan(5π/4) 与倒数函数
The angle 5π/4 is a multiple of π/4, located in Quadrant III. Its reference angle is 5π/4 – π = π/4. The coordinates for π/4 are (√2/2, √2/2), so for QIII both are negative: cos = –√2/2, sin = –√2/2. Therefore tan(5π/4) = (sin)/(cos) = (–√2/2)/(–√2/2) = 1. In QIII, tangent is positive, which matches CAST.
角 5π/4 是 π/4 的倍数，位于第三象限。它的参考角为 5π/4 – π = π/4。π/4 的坐标为 (√2/2, √2/2)，所以在第三象限两者均为负：cos = –√2/2，sin = –√2/2。因此 tan(5π/4) = (sin)/(cos) = (–√2/2)/(–√2/2) = 1。第三象限正切为正，与 CAST 相符。
Reciprocal functions are also exact: sec(5π/4) = 1/cos = –2/√2 = –√2; csc(5π/4) = 1/sin = –√2; cot(5π/4) = 1/tan = 1. These values are often required in IB problems.
倒数函数同样精确：sec(5π/4) = 1/cos = –2/√2 = –√2；csc(5π/4) = 1/sin = –√2；cot(5π/4) = 1/tan = 1。这些值在 IB 题目中经常出现。
11. Solving Trigonometric Equations with π/6 and π/4 Multiples | 运用 π/6 与 π/4 倍数解三角方程
+ 2kπ or x = 5π/6 + 2kπ for any integer k. This method relies on knowing multiples of π/6.
解方程如 sin x = 1/2 时，我们首先识别出参考角为 π/6。由于正弦为正，解位于第一和第二象限。因此 x = π/6 + 2kπ 或 x = 5π/6 + 2kπ，其中 k 为整数。这种方法依赖于熟悉 π/6 的倍数。
If the equation is cos x = –√2 ress the final answer using the smallest positive angle first if the domain is 0 ≤ x < 2π.
若方程为 cos x = –√2/2，参考角为 π/4，且余弦在第二、三象限为负。故 x = 3π/4 + 2kπ 或 x = 5π/4 + 2kπ。若定义域为 0 ≤ x < 2π，答案通常先给出最小正角。
12. Summary and Memory Tips | 总结与记忆技巧
The multiples of π/6 and π/4 give all the special angles you need to evaluate trigonometric functions exactly. Build a strong mental map of the unit circle with coordinates at every 30° and 45° step. Use the special triangles to recall the surd values: √2/2 for π/4, and the pair (1/2, √3/2) for π/6 and π/3.
π/6 和 π/4 的倍数提供了精确计算三角函数所需的所有特殊角。在脑海中构建一个清晰的单位圆图，标出每 30° 和 45° 步的坐标。利用特殊三角形记忆根式值：π/4 为 √2/2，π/6 和 π/3 为 (1/2, √3/2)。
cosine, reverse the order. This reinforces the pattern: sin(0)=0, sin(30°)=1/2, sin(45°)=√2/2, sin(60°)=√32, sin(90°)=1. Practising with both π/6 and π/4 families will make you fast and confident.
一个手指记忆法：伸出左手，掌心朝向自己。将手指依次对应 0°、30°、45°、60°、90°（小指到拇指）。对于正弦，取下方手指根号再除以 2；对于余弦，则反转顺序。这强化了规律：sin(0)=0，sin(30°)=1/2，sin(45°)=√2/2，sin(60°)=√3/2，sin(90°)=1。将 π/6 和 π/4 系列反复练习，可使你解题又快又有信心。
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