Trigonometric Modeling in Real-World Applications | 三角函数建模的实际应用问题

📚 Trigonometric Modeling in Real-World Applications | 三角函数建模的实际应用问题

Trigonometric functions are powerful tools for describing any phenomenon that repeats periodically. From ocean tides and Ferris wheels to alternating current and seasonal temperatures, sine and cosine functions help us model the world mathematically, allowing us to make accurate predictions. This article explores the key principles of trigonometric modeling and walks through a range of typical exam-style application problems.

三角函数是描述一切周期现象的强大概力工具。从海洋潮汐、摩天轮到交流电和季节性气温，正弦函数与余弦函数帮助我们以数学方式建立世界模型，从而做出准确预测。本文将探讨三角函数建模的核心原理，并带你系统梳理解答各类典型考试应用问题的方法。

1. The General Sine Function | 正弦函数的一般形式

Every trigonometric model can be expressed in the form:

y = A sin(B(x − C)) + D

Here, |A| represents the amplitude — half the vertical distance between the maximum and minimum values. The period, or the time for one complete cycle, is 2π / |B|. The value C is the horizontal shift (also called the phase shift), while D is the vertical shift, which positions the midline of the wave. For a cosine model, we simply replace sin with cos.

其中，|A| 代表振幅——即最大值与最小值之间垂直距离的一半。周期的时长为 2π / |B|。C 是水平位移（也称相位移动），而 D 是垂直位移，决定波浪中线的位置。若使用余弦模型，只需将 sin 换成 cos 即可。

The midline equation is y = D, and the maximum and minimum values are D + A and D − A respectively. Understanding how each parameter affects the graph is the first step toward solving any modeling problem.

中线方程为 y = D，最大值和最小值分别为 D + A 和 D − A。理解每个参数对图形的影响，是解答一切建模问题的第一步。

2. Identifying a Periodic Context | 识别周期情境

How can you tell whether a real-world situation should be modeled by a trigonometric function? Look for three essential features: repetition over time, a stable range between maximum and minimum values, and a smooth, wave-like transition between extremes. Examples include daily tides, the rotation of a wheel, the motion of a piston, and the cycle of daylight hours.

如何判断一个实际问题是否适合用三角函数建模？你需要寻找三个关键特征：随着时间重复出现、在最大值与最小值之间存在稳定的范围，以及在极端值之间有平滑的波浪形过渡。典型例子包括每日潮汐、车轮旋转、活塞运动和日照时长变化。

In contrast, linear or exponential models are appropriate for steady growth or decay, while quadratic models suit symmetrical projectile paths. If the problem contains keywords such as “cycles”, “periodic”, “rotates”, “oscillates”, or “repeats”, a trigonometric model is likely required.

相比之下，线性或指数模型适合描述稳定增长或衰减，二次函数模型适合对称的抛射路径。如果题目中出现 “cycles”（周期）、”periodic”（周期性）、”rotates”（旋转）、”oscillates”（摆动）或 “repeats”（重复）等关键词，那很可能需要使用三角函数模型。

3. Ferris Wheel Problems | 摩天轮问题

The Ferris wheel is a classic modeling scenario. Suppose a Ferris wheel has a diameter of 40 m, with its lowest point 2 m above the ground. It completes one full revolution every 60 seconds. We begin timing when the rider is at the lowest point. Model the height h(t) as a function of time t.

摩天轮是经典的建模场景。假设一个摩天轮的直径为 40 米，最低点距地面 2 米，每 60 秒完成一整圈。我们从乘客位于最低点时开始计时。请以时间 t 的函数 h(t) 来建立高度的模型。

Since we start at the lowest point, a negative cosine function is the natural choice. The amplitude A equals the radius, 20 m. The midline D is the height of the centre: 2 + 20 = 22 m. The period is 60 s, so B = 2π / 60 = π / 30. Hence:

由于我们从最低点开始计时，自然选择负余弦函数。振幅 A 等于半径，即 20 米。中线 D 是圆心的离地高度：2 + 20 = 22 米。周期为 60 秒，因此 B = 2π / 60 = π / 30。于是：

h(t) = −20 cos(πt / 30) + 22

To find the height after 15 seconds, evaluate h(15) = −20 cos(π/2) + 22 = 22 m, which corresponds to the level of the centre. After 45 seconds, h(45) = −20 cos(3π/2) + 22 = 22 m, again at the centre height but on the descending side. This simple check confirms the model behaves correctly.

要求 15 秒后的高度，代入计算 h(15) = −20 cos(π/2) + 22 = 22 米，对应圆心高度。45 秒后，h(45) = −20 cos(3π/2) + 22 = 22 米，同样位于圆心高度但处于下降阶段。这一简单验证表明模型行为正确。

4. Tides and Oceanography | 潮汐与海洋学

Coastal towns experience two high tides and two low tides each day, making tide levels a natural candidate for sine modeling. Suppose the water depth at a harbour is modelled by:

沿海城镇每天经历两次涨潮和两次退潮，使潮汐水位成为正弦建模的理想对象。假设某港口的水深可以用以下模型描述：

d(t) = 6 + 4 sin(πt / 6)

where d is measured in metres and t in hours after midnight. The amplitude is 4 m, so the depth oscillates between 2 m and 10 m. The period is 2π ÷ (π/6) = 12 hours, meaning two complete cycles occur in a 24-hour day, which is consistent with real tidal behaviour.

其中 d 以米为单位，t 为自午夜起的小时数。振幅为 4 米，因此水深在 2 米与 10 米之间波动。周期为 2π ÷ (π/6) = 12 小时，即在 24 小时内有两次完整循环，这符合实际潮汐规律。

If a ship requires at least 8 m of water to enter the harbour, we solve 6 + 4 sin(πt / 6) ≥ 8, which simplifies to sin(πt / 6) ≥ 0.5. The solution within one cycle is πt/6 ∈ [π/6, 5π/6], giving t ∈ [1, 5]. Thus the ship can enter between 1:00 and 5:00, and again between 13:00 and 17:00. This demonstrates how inequalities with trigonometric functions translate directly into scheduling decisions.

如果一艘船需要至少 8 米水深才能进港，我们解不等式 6 + 4 sin(πt / 6) ≥ 8，化简得 sin(πt / 6) ≥ 0.5。在一个周期内的解为 πt/6 ∈ [π/6, 5π/6]，即 t ∈ [1, 5]。因此船只可在 1:00 至 5:00 之间进港，以及在 13:00 至 17:00 之间再次进港。这展示了三角不等式如何直接转化为实际调度决策。

5. Seasonal Temperature Models | 季节性温度模型

Average monthly temperatures follow a yearly cycle and are often modelled with a cosine function. Suppose the hottest month in a city is July (average 28°C) and the coldest month is January (average 4°C). Let t = 0 represent January 1st.

城市月平均气温按年度循环变化，通常用余弦函数建模。假设某城市最热月份为七月（平均 28°C），最冷月份为一月（平均 4°C）。令 t = 0 表示 1 月 1 日。

The midline is (28 + 4) / 2 = 16°C. The amplitude is (28 − 4) / 2 = 12°C. The period is 12 months, so B = 2π / 12 = π / 6. Since the maximum occurs at t = 6 (July), we write the model as:

中线为 (28 + 4) / 2 = 16°C。振幅为 (28 − 4) / 2 = 12°C。周期为 12 个月，因此 B = 2π / 12 = π / 6。由于最大值出现在 t = 6（七月），模型可写成：

T(t) = 12 cos(π(t − 6) / 6) + 16

To predict the temperature in April (t = 3), compute T(3) = 12 cos(−π/2) + 16 = 16°C. In October (t = 9), we also get 16°C. These symmetric results make sense: April and October lie halfway between the extremes on the cosine curve.

要预测四月（t = 3）的温度，计算 T(3) = 12 cos(−π/2) + 16 = 16°C。十月（t = 9）同样得到 16°C。这组对称结果是合理的：四月和十月在余弦曲线上正好位于两个极值之间。

6. Alternating Current (AC) | 交流电

Electrical engineering relies heavily on sine functions. Standard household electricity in many countries has a frequency of 50 Hz, meaning the current completes 50 cycles per second. The voltage can be modelled as:

电气工程高度依赖正弦函数。许多国家的家用电源频率为 50 Hz，即电流每秒完成 50 个周期。电压可以建模为：

V(t) = V_max sin(2π f t)

where V_max is the peak voltage and f is the frequency in hertz. For a 230 V (RMS) supply, the peak voltage is approximately 230 × √2 ≈ 325 V, so V(t) = 325 sin(100πt).

其中 V_max 是峰值电压，f 是以赫兹为单位的频率。对于 230 V（有效值）的电源，峰值电压约为 230 × √2 ≈ 325 V，因此 V(t) = 325 sin(100πt)。

The period is 1/50 = 0.02 s, or 20 ms. The time between consecutive voltage zeros is half a period, 10 ms. Understanding this sinusoidal nature is essential for rectifier design, transformer analysis, and power transmission calculations.

周期为 1/50 = 0.02 秒，即 20 毫秒。连续两个电压零点之间的时间为半个周期，即 10 毫秒。理解这一正弦特性对于整流器设计、变压器分析和电力传输计算至关重要。

7. Light and Sound Waves | 光波与声波

Both light and sound propagate as waves, and their intensity or pressure can be modelled using trigonometric functions. A pure musical tone, for example, produces a sinusoidal pressure variation. The pitch of the sound is determined by its frequency, while loudness relates to amplitude.

光和声都以波的形式传播，其强度或压力可以用三角函数建模。例如，一个纯音会产生正弦式的气压变化。音调由频率决定，响度则与振幅相关。

A sound wave with frequency 440 Hz (the musical note A₄) can be expressed as P(t) = 0.5 sin(2π × 440 × t), where P is the pressure variation in pascals. The period is 1/440 ≈ 0.00227 s. In exam problems, you may be asked to find the frequency from a graph, determine the amplitude, or write the equation of a wave given its period and range.

频率为 440 Hz 的声波（即音乐中的 A₄ 音）可表示为 P(t) = 0.5 sin(2π × 440 × t)，其中 P 是以帕斯卡为单位的压力变化。周期为 1/440 ≈ 0.00227 秒。在考试中，你可能会被要求从图像求频率、确定振幅，或根据周期和值域写出波动方程。

8. Pendulums and Springs | 单摆与弹簧振动

In mechanics, simple harmonic motion is the textbook example of trigonometric modeling. A mass attached to a spring oscillates back and forth with a period determined by the mass and the spring constant. If a spring has a period of 4 seconds, the angular frequency is B = 2π / 4 = π / 2.

在力学中，简谐运动是三角函数建模的教科书级例证。连接在弹簧上的物体会来回振动，其周期由质量和弹簧常数共同决定。若弹簧的周期为 4 秒，则角频率为 B = 2π / 4 = π / 2。

Suppose the mass starts at its maximum displacement of 6 cm above equilibrium, moving downward. The displacement is:

假设物体从平衡位置上方最大位移 6 厘米处开始向下运动，其位移为：

x(t) = 6 cos(πt / 2)

At t = 1 s, x(1) = 6 cos(π/2) = 0, meaning the mass passes through equilibrium. At t = 2 s, x(2) = 6 cos(π) = −6, meaning it reaches its lowest point. The cycle completes at t = 4 s. These problems often combine trig modeling with differentiation to find velocity and acceleration.

在 t = 1 秒时，x(1) = 6 cos(π/2) = 0，即物体经过平衡位置。在 t = 2 秒时，x(2) = 6 cos(π) = −6，即到达最低点。整个周期在 t = 4 秒时完成。这类问题常将三角建模与导数结合，以求解速度和加速度。

9. A Step-by-Step Method for Modeling Problems | 建模问题的分步解法

To solve any trigonometric modeling problem reliably, follow these five steps:

要稳妥地解答任何三角建模问题，请遵循以下五个步骤：

Step 1 — Identify the cycle: Find the period from the problem or graph. Convert words like “every 8 hours” into B = 2π / 8 = π / 4.

第一步——确定周期：从题干或图像中找出周期。像”每 8 小时”这样的表述可换算为 B = 2π / 8 = π / 4。
Step 2 — Find the midline: Calculate D as (maximum + minimum) ÷ 2.

第二步——求出中线：计算 D =（最大值 + 最小值）÷ 2。
Step 3 — Find the amplitude: Calculate A as (maximum − minimum) ÷ 2.

第三步——求出振幅：计算 A =（最大值 − 最小值）÷ 2。
Step 4 — Determine the phase shift: Observe where the first maximum or first minimum occurs, then choose sin or cos and solve for C.

第四步——确定相位移动：观察第一个最大值或最小值出现在何处，然后选择 sin 或 cos 并解出 C。
Step 5 — Verify: Substitute a known value of t to check whether the function gives the stated result.

第五步——验证：代入一个已知的 t 值，检查函数是否给出题目所述的结果。

This structured approach minimises errors and is particularly valuable under exam pressure.

这种结构化方法能最大限度地减少错误，在考试压力下尤其有价值。

10. Common Pitfalls and How to Avoid Them | 常见陷阱与规避方法

Students frequently lose marks on modeling questions due to a handful of recurring mistakes.

学生在建模题上失分，往往源于一些反复出现的错误。

Using degrees instead of radians: In calculus-level modeling, B is always derived using radians. Remember that 2π corresponds to 360°.

使用度数而非弧度：在涉及微积分的建模中，B 始终基于弧度推导。请记住 2π 对应 360°。
Confusing amplitude with range: A student may write A = 40 when the range is 40; the actual amplitude is 40 ÷ 2 = 20.

混淆振幅与值域：若值域为 40，学生可能直接写 A = 40，而实际振幅应为 40 ÷ 2 = 20。
Incorrect initial condition: Starting at the maximum requires cos, while starting at the midline rising requires sin. Starting at the minimum requires −cos.

初始条件判断错误：从最大值出发需要用 cos，从中线上升出发需要用 sin，从最小值出发需要用 −cos。
Setting the midline at zero: If data oscillates between 2 and 10, the midline is 6, not 0. The vertical shift D = 6 must be included.

将中线设为零：若数据在 2 与 10 之间波动，中线的位置是 6 而非 0。必须加上垂直位移 D = 6。

Condition at t = 0	Starting function	t = 0 时的状态
Maximum value	y = A cos(Bt) + D	处于最大值
Minimum value	y = −A cos(Bt) + D	处于最小值
Midline, rising	y = A sin(Bt) + D	中线处，上升中
Midline, falling	y = −A sin(Bt) + D	中线处，下降中

11. Worked Exam-Style Example | 考点例题精解

Let us apply the full method to a typical examination question.

下面我们运用完整的方法解析一道典型的考试题目。

Problem: The depth of water in a harbour, y metres, at time t hours after midnight, is modelled by y = 5 + 3 sin(πt / 6). Find (a) the maximum depth and the time of the first maximum; (b) the times between midnight and midday when the depth is exactly 5 m.

题目：某港口水深 y 米与午夜后 t 小时的关系为 y = 5 + 3 sin(πt / 6)。求 (a) 最大水深及第一次达到最大值的时间；(b) 从午夜到正午之间水深恰好为 5 米的时刻。

Solution (a): The maximum value of sin is 1, so y_max = 5 + 3 = 8 m. The first maximum occurs when sin(πt / 6) = 1, that is, πt / 6 = π/2, giving t = 3. So the deepest water occurs at 3:00 a.m.

解答 (a)：sin 的最大值为 1，因此最大水深 y_max = 5 + 3 = 8 米。第一次达到最大值时 sin(πt / 6) = 1，即 πt / 6 = π/2，解得 t = 3。因此最深水位出现在凌晨 3:00。

Solution (b): Set y = 5, giving 5 + 3 sin(πt / 6) = 5, so sin(πt / 6) = 0. Within the interval 0 ≤ t ≤ 12, this occurs when πt/6 = 0, π, 2π, giving t = 0, 6, 12. Hence the depth is exactly 5 m at midnight, 6:00 a.m., and midday.

解答 (b)：令 y = 5，得到 5 + 3 sin(πt / 6) = 5，即 sin(πt / 6) = 0。在 0 ≤ t ≤ 12 区间内，这发生在 πt/6 = 0、π、2π，即 t = 0、6、12。因此水深恰好为 5 米的时刻是午夜、早上 6:00 和正午。

This example illustrates the importance of understanding the unit circle: the sine function equals zero at integer multiples of π, and equals ±1 at odd multiples of π/2.

这个例子说明了理解单位圆的重要性：正弦函数在 π 的整数倍处取值为 0，在 π/2 的奇数倍处取值为 ±1。

12. Connecting Graphs to Real-World Meaning | 图像与实际意义的相互转化

Examiners often provide a sinusoidal graph and ask you to interpret its features in context. For example, the distance between two consecutive maxima on the t-axis represents the period. The vertical distance from a peak to the midline gives the amplitude. The horizontal position of the first peak gives the phase shift.

考官常常给出一幅正弦图像，要求你结合实际情境解释其特征。例如，t 轴上相邻两个最大值之间的距离代表周期。从峰值到中线的垂直距离给出了振幅。第一个峰值的位置则给出了相位移动。

It is equally important to translate algebra into meaning. When asked “when does the object first return to its starting position?”, you are being asked to find the period. When asked “how many times does the tide reach its maximum in one day?”, you are being asked to count cycles, which equals the elapsed time divided by the period.

同样重要的是将代数转化为意义。当题目问”物体何时第一次回到初始位置？”时，其实是在求周期。当题目问”一天内潮汐达到最大值几次？”时，其实是在数循环次数，即用经过的总时间除以周期。

Mastering these translations between graphs, equations, tables, and words is the ultimate skill in trigonometric modeling, and it reliably appears in both AS and A-Level examinations across major boards.

掌握图像、方程、表格和文字之间的相互转化，是三角函数建模的最终技能，也是各大考试局 AS 与 A-Level 考试中的高频考查内容。

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Trigonometric Modeling in Real-World Applications | 三角函数建模的实际应用问题