📚 Modelling with Trigonometric Functions | 利用三角函数建模
riodic phenomena. From tidal patterns and seasonal temperature variations to the motion of a Ferris wheel, trigonometric models help us describe and predict behaviour over time. In the Edexcel A-Level Mathematics course, modelling with trigonometric functions requires understanding how to construct equations of the form y = a sin(bx + c) + d or y = a cos(bx + c) + d and interpret the parameters in context.
正弦和余弦等三角函数不仅是抽象的数学概念，它们还是建模现实世界周期现象的强大工具。从潮汐模式和季节性温度变化到摩天轮的运动，三角函数模型帮助我们描述和预测随时间变化的行为。在爱德思A-Level数学课程中，利用三角函数建模需要理解如何构造形如 y = a sin(bx + c) + d 或 y = a cos(bx + c) + d 的方程，并能在具体情境中解释参数的含义。
1. Introduction to Trigonometric Modelling | 三角函数建模导论
Many natural and engineered systems exhibit repetitive behaviour over equal time intervals. In mathematics, such behaviour is called periodic, and the simplest periodic functions are sine and cosine. By adjusting a few parameters, we can stretch, shift, and reposition these basic waves to match real data. The process of modelling essentially involves finding the right values for amplitude, period, phase shift, and vertical shift so that the trigonometric curve fits observed measurements.
许多自然和工程系统在相等的时间间隔内表现出重复的行为。在数学中，这种行为称为周期性的，最简单的周期函数就是正弦和余弦。通过调整几个参数，我们可以拉伸、平移和重新定位这些基本波形，使之与实际数据匹配。建模的过程本质上就是找到合适的振幅、周期、相移和垂直位移的值，使三角函数曲线与观测到的测量值相吻合。
) + d. Although the sine and cosine curves are simply horizontal translations of each other, the choice between them often depends on the starting point of the cycle in the given context.
你会遇到的一般形式是 y = a sin(bx + c) + d 和 y = a cos(bx + c) + d。虽然正弦和余弦曲线只是彼此的左右平移，但选择哪一个常常取决于给定情境中周期的起始点。
In the Edexcel specification, x is typically time or an angle measure, and y is the variable being modelled, such as height, temperature, or depth. The independent variable x is often in radians, but degrees may also appear in contextual problems – always pay attention to the unit specified.
在爱德思考试大纲中，x 通常是时间或角度度量，y 是被建模的变量，如高度、温度或深度。自变量 x 常用弧度制，但在情境问题中也可能出现角度制——务必留意题中指定的单位。
2. The Standard Trigonometric Forms | 标准三角函数形式
argument as c. The second is y = a sin(b(x + c)) + d, which explicitly shows the horizontal translation by – the form y = a sin(bx + c) + d, so you need to be comfortable converting between the two.
表达变形后的正弦或余弦函数有两种等价方式。第一种是 y = a sin(bx + c) + d，此时相移以 c 的形式内置在自变量中。第二种是 y = a sin(b(x + c)) + d，它明确地显示出水平平移了 –c。爱德思的题目常使用 y = a sin(bx + c) + d 的形式，因此你需要能熟练地在两者之间转换。
d (period = 2π/b or 360°/b if using degrees), c influences the phase shift (actual shift = – he vertical shift, which is also the equilibrium line about which the function oscillates.
参数 a 控制振幅，b 决定周期（周期 = 2π/b 或使用度数时为 360°/b），c 影响相移（实际平移量 = –c/b），而 d 给出垂直位移，它也是函数振荡所围绕的平衡线。
sin(bx + c) + d. Understanding these interpretations is essential for setting up a model from a real-world description.
下表总结了模型 y = a sin(bx + c) + d 中每个参数的含义。理解这些含义对于根据实际描述建立模型至关重要。
| Parameter | Role | How it affects the graph |
|---|---|---|
| a | Amplitude | Vertical stretch; peak deviation from the equilibrium line is |a| |
| b | Angular frequency | Period = 2π/b (radians) or 360°/b (degrees) |
| c | Phase parameter | Horizontal translation = –c/b |
| d | Vertical shift | Moves the central line up or down; equilibrium value |
Once these four parameters are pinned down, the entire periodic behaviour is captured by the equation. The art of modelling lies in extracting these numbers from data like maximum and minimum values, times of high and low points, and the length of one full cycle.
一旦这四个参数确定下来，整个周期性行为就被方程捕捉到了。建模的技巧在于从数据（如最大值和最小值、高低点出现的时间以及一个完整周期的长度）中提取出这些数字。
3. Amplitude – The Height of Oscillation | 振幅 – 振荡的高度
central equilibrium value. In the equation y = a sin(bx + c) + d, the maximum value is d + |a| and the minimum value is d – |a| he amplitude is half the difference between the maximum and minimum observations: a = (max – min) / 2.
振幅 a 衡量输出值在中心平衡值上下波动的距离。在方程 y = a sin(bx + c) + d 中，最大值为 d + |a|，最小值为 d – |a|。因此，振幅就是观察到的最大值与最小值之差的一半：a = (最大值 – 最小值) / 2。
In many modelling problems, you will be given the highest and lowest points of a cyclical phenomenon, such as the depth of water in a harbour at high and low tide. The amplitude is always taken as a positive number, but you may write the function with a negative a to reflect a reflection in the horizontal axis. In Edexcel exams, it is common to use positive a and adjust the phase shift if the graph starts by going downwards.
在许多建模问题中，你会得到周期性现象的最高点和最低点，例如港口在涨潮和落潮时的水深。振幅总是取为正数，但你也可以通过取负 a 来反映沿水平轴的翻转。在爱德思考试中，通常使用正的 a，如果图形一开始是向下的，则通过调节相移来体现。
For example, if the water depth varies between 2 m and 10 m, then amplitude a = (10 – 6 m. The order of calculation is important: find a first from the range, then determine d.
例如，若水深在 2 米和 10 米之间变化，则振幅 a = (10 – 2) / 2 = 4 米。中心线 d 则为 (10 + 2) / 2 = 6 米。计算的顺序很重要：先通过范围求出 a，再确定 d。
4. Period and Angular Frequency b | 周期与角频率 b
ransformed to y = sin(bx), the period becomes 2π / b (in radians) or 360° / b (in degrees). The coefficient b is sometimes called the angular frequency because it indicates how many cycles fit into an interval of length 2π.
三角函数的周期是一个完整循环的长度。对于 y = sin x 和 y = cos x，周期为 2π 弧度或 360°。当图像变为 y = sin(bx) 时，周期变为 2π / b（弧度制）或 360° / b（角度制）。系数 b 有时被称为角频率，因为它表明在 2π 长度的区间内包含多少个循环。
To find b from a real-world problem, identify the time taken for one full repetition – independent variable is time in hours, days, or months, as this affects the numerical value.
要从实际问题中求出 b，需要确定一个完整重复所需的时间——这就是周期 T。若 x 为弧度，则利用关系式 b = 2π / T。如果题目指定使用度数，那么 b = 360° / T。务必检查自变量是以小时、天还是月为单位的时间，因为这会影响数值。
, b = 2π / 30 = π/15. If the model uses degrees, b = 360° / 30 = 12. Selecting the correct angular measure is crucial for a valid model.
假设一座摩天轮每 30 分钟完成一圈旋转。若 x 以分钟计量，则周期 T = 30 分钟。如果用弧度建模，b = 2π / 30 = π/15。如果使用度数，b = 360° / 30 = 12。选择正确的角度度量对于建立一个有效的模型至关重要。
5. Phase Shift and Horizontal Translation | 相移与水平平移
sin(bx + c) + d determines the horizontal shift. The graph is translated left or right by – he left; if c is negative, the shift is to the right. In contexts, this shift aligns the mathematical curve with the timing of the real event, such as placing the high tide at a specific hour.
表达式 y = a sin(bx + c) + d 中的参数 c 决定水平平移量。图像向左或向右移动 –c/b。如果 c 为正，向左平移；如果 c 为负，向右平移。在情境中，这一平移将数学曲线与实际事件的发生时间对齐，例如把高潮时间放在某个具体的小时上。
When using the alternative form y = a sin(b(x + c)) + d, the shift is simply – o express the phase shift consistently within the brackets when constructing the equation.
当使用另一种形式 y = a sin(b(x + c)) + d 时，平移量就是 –c，许多学生觉得这更直观。但爱德思的评分方案两种正确形式都接受。只要在构建方程时括号内的相移表达一致，注意即可。
ine model that starts at the equilibrium line going up, the shift is found by solving bx + c = 0 at that time. For a cosine model starting at a maximum, solve bx + c = 0.
要根据数据确定平移量，你需要一个参考点，比如第一个最大值、最小值或平衡线交叉点出现的时间。对于从平衡线向上启动的正弦模型，通过在该时刻求解 bx + c = 0 来求出平移。对于从最大值启动的余弦模型，则求解 bx + c = 0。
6. Vertical Shift – The Equilibrium Line | 垂直位移 – 平衡线
ity over one cycle. Mathematically, d = (maximum + minimum) / 2. In many modelling situations, d is the mean sea level, average daily temperature, or the centre height of a ride.
参数 d 是垂直平移，它确定了振荡的中心轴位置。它代表了一个周期内周期性量的平均值。数学上，d = (最大值 + 最小值) / 2。在许多建模场景中，d 就是平均海平面、日均温度或游乐设施的中心高度。
When reading off a graph, d is simply the y-coordinate halfway between the peaks and troughs. Never confuse d with the amplitude – the amplitude is the vertical distance from d to a peak, while d itself is the baseline. A common mistake is to set d equal to the minimum value when the graph starts low; instead, always compute it from max and min.
读取图形时，d 就是波峰和波谷中间位置的 y 坐标。切勿将 d 与振幅混淆——振幅是从 d 到波峰的垂直距离，而 d 本身是基线。一个常见的错误是当图形从低位开始时将 d 设为最小值；应该始终通过最大值和最小值来计算。
from 5°C in winter to 25°C in summer. Then d = (25 + 5) / 2 = 15°C, and amplitude a = (25 – sin(bt + c) + 15, with b and c to be determined from the period and timing of the seasons.
例如，某地的年气温模型在冬季为 5°C，夏季为 25°C。那么 d = (25 + 5) / 2 = 15°C，振幅 a = (25 – 5) / 2 = 10°C。结合起来就得到模型 T = 10 sin(bt + c) + 15，其中 b 和 c 可以通过周期和季节时间点来确定。
7. Worked Example: Modelling Tide Heights | 实例：潮汐高度建模
A harbour’s water depth varies between a maximum of 12 m at 4:00 a.m. and a minimum of 4 m at 10:00 a.m. on a particular day. Model the depth D (in metres) as a function of time t (in hours after midnight) using a cosine function.
某个港口的某天水深在凌晨 4 点达到最高 12 米，上午 10 点降到最低 4 米。请以时间 t（午夜后的小时数）为自变量，用余弦函数建立水深 D（米）的模型。
First, amplitude a: a = (12 – r (not given but typical semi-diurnal tide), the period T = 12 hours. Assuming radians, b = 2π / 12 = π/6.
首先，振幅 a：a = (12 – 4) / 2 = 4 米。垂直位移 d：d = (12 + 4) / 2 = 8 米。周期是相邻两次高潮之间的时间。从 4:00 的高潮到 12 小时之后的下一个高潮（题目未给出，但典型的半日潮是如此），周期 T = 12 小时。假设使用弧度，b = 2π / 12 = π/6。
A cosine model starting at a maximum has the form D = a cos(b(t – t₀)) + d, where t₀ is the time of the first maximum. Here t₀ = 4, so D = 4 cos(π/6 (t – 4)) + 8. Expanding gives D = 4 cos((π/6)t – 2π/3) + 8, which matches y = a cos(bx + c) + d with c = –2π/3.
从最大值开始的余弦模型具有形式 D = a cos(b(t – t₀)) + d，其中 t₀ 是第一个最大值发生的时间。这里 t₀ = 4，因此 D = 4 cos(π/6 (t – 4)) + 8。展开得到 D = 4 cos((π/6)t – 2π/3) + 8，与 y = a cos(bx + c) + d 一致，其中 c = –2π/3。
ow be used to predict the depth at any time, e.g., at 7:00 a.m. (t = 7): D = 4 cos(π/6 × 7 – 2π/3) + 8 = 4 cos(7π/6 – 8 = 8 m. The calculation illustrates how a well-constructed model gives quick predictions.
这个模型现在可用于预测任意时刻的水深，例如早上 7 点（t = 7）：D = 4 cos(π/6 × 7 – 2π/3) + 8 = 4 cos(7π/6 – 4π/6) + 8 = 4 cos(π/2) + 8 = 4 × 0 + 8 = 8 米。这一计算展示了一个良好构建的模型如何快速给出预测。
8. Solving Trigonometric Equations in Context | 情境中解三角方程
Once a trigonometric model is established, exam questions frequently ask you to find the times at which the output reaches a specific value, or when it exceeds a certain threshold. This requires solving a trigonometric equation over a given domain, often involving the general solution or the use of the CAST diagram.
一旦建立了三角函数模型，考试题目常常要求你找出输出达到特定数值的时刻，或者何时超过某个阈值。这就需要在一个给定的定义域内求解三角方程，往往会涉及通解或 CAST 图的使用。
For example, using the tide model D = 4 cos(π/6 t – 2π/3) + 8, to find when the depth is exactly 10 m, set 4 cos(π/6 t – 2π/3) + 8 = 10. Then cos(π/6 t – and the general solutions in 0 ≤ t < 24 can be found by considering θ = π/3 + 2nπ and θ = –π/3 + 2nπ.
例如，利用潮汐模型 D = 4 cos(π/6 t – 2π/3) + 8，要找出水深恰好为 10 米的时刻，令 4 cos(π/6 t – 2π/3) + 8 = 10，得到 cos(π/6 t – 2π/3) = 0.5。cos(θ) = 0.5 的主值是 θ = π/3，通过考虑 θ = π/3 + 2nπ 和 θ = –π/3 + 2nπ，可以求出 0 ≤ t < 24 范围内的全部解。
in at 10:00 a.m. This matches the given data that low tide is at 10:00 a.m. with depth 4 m? alls from 12 m at 4:00 to 4 m at 10:00, so passing 10 m at t = 6 and t = 10 is consistent.
求解得到 t = 6 和 t = 10，意味着水深在早上 6 点和上午 10 点达到 10 米。这与给定数据一致：10 点是最低潮 4 米，但模型显示水深从 4 点的 12 米下降到 10 点的 4 米过程中，在 6 点和 10 点经过 10 米是合理的。
Always remember to check that your solutions lie within the stated domain and to interpret them in the real-world context, discarding any extraneous times that fall outside the range of interest.
务必记住检查所求得的解是否位于指定的定义域内，并在实际背景中解释它们，舍去任何落在关注范围之外的额外时间。
9. Choosing Between Sine and Cosine Models | 正弦与余弦模型的选择
ses. The choice between them in modelling depends on what is most convenient for the starting condition described in the problem.
基本的正弦曲线 y = sin x 从平衡线（y = 0）启动并上升；余弦曲线 y = cos x 则从最大值（y = 1）启动并下降。在建模时选择哪个，取决于问题中描述的初始条件用哪种形式更方便。
al shift is natural: y = a sin(bx) + d. If it starts at a maximum (e.g., depth is highest at midnight), a cosine function fits best: y = a cos(bx) + d.
如果现象从平均值开始随后上升（例如在 t = 0 时水位为平均海平面且正在上涨），那么不带水平平移的正弦函数就很自然：y = a sin(bx) + d。如果从最大值开始（例如午夜时水深最高），余弦函数最为合适：y = a cos(bx) + d。
lways convert a sine model to an equivalent cosine model using the identity sin(θ) = cos(θ – ). This means that a sine model y = a sin(bx + c) + d can be rewritten as y = a cos(bx + c – will award full marks for either valid form, provided it accurately models the given data.
你随时可以利用恒等式 sin(θ) = cos(θ – π/2) 将正弦模型转换为等价的余弦模型。这意味着正弦模型 y = a sin(bx + c) + d 可以重写为 y = a cos(bx + c – π/2) + d。只要准确模拟了给定的数据，阅卷人会对两种正确形式中的任一种都给予满分。
It is wise to choose the form that requires the least complicated phase shift. This often reduces algebraic errors and makes the subsequent solving of equations clearer.
明智的做法是选择所需相移最简洁的形式。这常常能减少代数错误，并使后续方程的求解更加清晰。
10. Common Pitfalls and Exam Tips | 常见陷阱与考试技巧
Many marks are lost in trigonometric modelling questions due to simple avoidable mistakes. The most frequent errors include: confusing amplitude and vertical shift, using degrees when radians are required (or vice versa), and misidentifying the period from a graph because of incorrect reading of the time axis.
在三角函数建模题中，很多失分是由于一些简单的可避免错误造成的。最常见的错误包括：混淆振幅和垂直位移、在需要弧度时使用了度数（或反之），以及由于误读时间轴而从图形中错误识别周期。
lf. Always remember: period T = 2π / b for radians, so b = 2π / T. Carefully note whether the x-axis is labelled in hours, days, or minutes, and ensure b and c are consistent with the unit of x.
另一个常见错误是忘记 b 通过除法与周期关联，而不是乘法。学生有时会直接把 b 等同于周期。务必记住：对于弧度，周期 T = 2π / b，因此 b = 2π / T。认真
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