IB Mathematics: Real-World Applications of Trigonometric Models | IB数学：三角模型的实际应用

📚 IB Mathematics: Real-World Applications of Trigonometric Models | IB数学：三角模型的实际应用

Trigonometric functions are powerful tools for describing phenomena that repeat or oscillate over time. In this article, we explore how the general sine and cosine model y = A sin(B(x − C)) + D can be applied to tides, temperatures, daylight hours, alternating current, sound waves, seasonal demand, pendulum motion, and more. We will also examine how to interpret each parameter and how to construct and refine a trigonometric model from given data.

三角函数是描述随时间重复或振荡现象的強大工具。在本文中，我们将探讨一般正弦与余弦模型 y = A sin(B(x − C)) + D 如何应用于潮汐、气温、日照时长、交流电、声波、季节性需求、单摆运动等领域。我们还将讨论如何解读每个参数的含义，以及如何根据已知数据构建和优化三角模型。

1. The General Trigonometric Model | 一般三角模型

The standard model is y = A sin(B(x − C)) + D, where |A| is the amplitude, B controls the period (period = 2π/|B| for sine and cosine in radians), C is the horizontal shift (phase shift), and D is the vertical shift (midline). The midline y = D represents the average value around which the function oscillates.

标准模型为 y = A sin(B(x − C)) + D，其中 |A| 为振幅，B 控制周期（正弦和余弦在弧度制下周期为 2π/|B|），C 为水平平移（相移），D 为垂直平移（中线）。中线 y = D 表示函数围绕其振荡的平均值。

For a maximum M and minimum m, we have:

D = (M + m)/2, A = (M − m)/2

This is often the quickest way to estimate A and D from data.

对于最大值 M 和最小值 m，我们有：

D = (M + m)/2, A = (M − m)/2

这往往是从数据中快速估算 A 和 D 的方法。

2. Tidal Modelling | 潮汐建模

Ocean tides rise and fall approximately twice a day, making them a classic example of sinusoidal behaviour. Suppose the water depth at a harbour is 10 m at high tide and 2 m at low tide, and the period is 12.5 hours. If high tide occurs at t = 0, a cosine model is natural:

海洋潮汐每天大约涨落两次，是正弦行为的经典实例。假设某港口在水位高潮时深度为 10 米，低潮时为 2 米，周期为 12.5 小时。若高潮发生在 t = 0，则自然使用余弦模型：

y = 4 cos(2πt/12.5) + 6

Here D = (10 + 2)/2 = 6, A = (10 − 2)/2 = 4, and B = 2π/12.5. The model predicts the depth at any time t hours after high tide.

这里 D = (10 + 2)/2 = 6，A = (10 − 2)/2 = 4，B = 2π/12.5。该模型可以预测高潮后 t 小时任意时刻的水深。

To determine when a ship requiring 7 m depth can enter, solve 4 cos(2πt/12.5) + 6 > 7. This yields intervals for safe navigation.

若要判断吃水深度为 7 米的船只何时可以进港，可解不等式 4 cos(2πt/12.5) + 6 > 7，从而得到安全通航的时间区间。

3. Temperature Cycles | 气温变化周期

Daily or annual temperature fluctuations can be modelled by sine or cosine curves. For an annual model, let x be the day of the year. The coldest day is around day 15 in the northern hemisphere, and the warmest is around day 200. If the average annual temperature is 15 °C and the seasonal variation is ±10 °C, we can write:

每日或每年的气温波动可以用正弦或余弦曲线建模。以年周期模型为例，设 x 为一年的第几天。在北半球，最冷的日子大约在第 15 天左右，最热的日子大约在第 200 天左右。若年平均气温为 15 °C，季节性变化幅度为 ±10 °C，则可写为：

T(x) = 10 sin(2π(x − 105)/365) + 15

Why x − 105? Because a sine curve reaches its maximum at 1/4 of the period, i.e. day 91.25 plus 105 gives the phase that aligns the maximum with day 200. Adjusting the phase shift C is essential for matching observed data.

为什么是 x − 105？因为正弦曲线在周期的 1/4 处达到最大值，即第 91.25 天，加上 105 可使最大值对应到第 200 天附近。调整相移 C 对匹配实际数据至关重要。

4. Daylight Hours | 日照时长

The number of daylight hours in a day also follows a periodic pattern. In Oslo, the longest day has about 18.5 hours of daylight and the shortest about 6 hours. The period is 365 days. Using a sine model, first compute:

每天的日照时长也呈现周期性变化。在奥斯陆，最长日照约为 18.5 小时，最短约为 6 小时，周期为 365 天。使用正弦模型时，先计算：

D = (18.5 + 6)/2 = 12.25, A = (18.5 − 6)/2 = 6.25

If the maximum occurs on June 21 (day 172), a cosine model is convenient:

L(t) = 6.25 cos(2π(t − 172)/365) + 12.25

This model allows us to estimate daylight on any date and to find dates with a specified number of daylight hours.

若最大值出现在 6 月 21 日（第 172 天），则使用余弦模型更为方便：

L(t) = 6.25 cos(2π(t − 172)/365) + 12.25

该模型可用于估计任意日期的日照时数，并找出具有特定日照时数的日期。

5. Alternating Current | 交流电

Electrical voltage in a mains supply is a sinusoid. In many countries, the voltage has an amplitude of about 325 V and a frequency of 50 Hz. Since frequency f is related to the angular frequency by B = 2πf, the voltage is:

市电电压是正弦波。在许多国家，电压振幅约为 325 伏，频率为 50 赫兹。由于频率 f 与角频率的关系为 B = 2πf，因此电压为：

V(t) = 325 sin(100πt)

The root mean square (RMS) voltage is V_RMS = A/√2 ≈ 230 V. The period is 1/50 = 0.02 s. This model is essential for understanding AC circuits, rectification, and electrical safety.

均方根电压为 V_RMS = A/√2 ≈ 230 伏。周期为 1/50 = 0.02 秒。该模型对于理解交流电路、整流以及用电安全非常重要。

6. Sound Waves | 声波

A pure tone can be represented by a sinusoidal pressure wave. If a tuning fork produces a sound with frequency 440 Hz and amplitude 0.5 Pa, the pressure variation is:

纯音可以用正弦压力波表示。若音叉发出频率为 440 Hz、振幅为 0.5 帕斯卡的声音，则压力变化为：

P(t) = 0.5 sin(2π × 440 t) = 0.5 sin(880πt)

Human hearing perceives frequency as pitch and amplitude as loudness. Adding multiple sine waves of different frequencies creates more complex sounds, which is the basis of Fourier analysis.

人耳将频率感知为音调，将振幅感知为响度。不同频率的正弦波叠加会产生更复杂的声音，这也是傅里叶分析的基础。

7. Ferris Wheel Motion | 摩天轮运动

Consider a Ferris wheel of radius 20 m, with its centre 22 m above the ground. It completes one revolution every 40 seconds. A passenger starts at the lowest point. The height h(t) above the ground after t seconds can be modelled by a cosine function with a phase shift:

设摩天轮半径为 20 米，轮心离地 22 米，旋转周期为 40 秒。乘客从最低点开始，则 t 秒后离地高度 h(t) 可用余弦函数并带有相移来建模：

h(t) = 20[1 − cos(2πt/40)] + 2? No.

Let us derive correctly. The centre is 22 m above ground. At the bottom, height = 2 m; at the top, height = 42 m. The vertical position relative to the centre is −20 at t = 0 if starting at bottom. Using cosine:

让我们正确推导。轮心离地 22 米。在最低点时高度为 2 米，最高点为 42 米。若从最低点开始，相对轮心的竖直位移为 −20。使用余弦函数：

h(t) = 22 − 20 cos(2πt/40)

Check: at t = 0, h = 22 − 20 = 2 m. At t = 20 (half revolution), h = 22 − 20 cos(π) = 42 m. This is correct.

验证：t = 0 时，h = 22 − 20 = 2 米；t = 20（半圈）时，h = 22 − 20 cos(π) = 42 米。这样是正确的。

8. Seasonal Sales and Marketing | 季节性销售与营销

Retail sales often show annual periodicity. An ice cream shop has monthly sales that are highest in July (around 1200 units) and lowest in January (around 200 units). Using months x = 1 to 12, with the maximum at x = 7, the sales model is:

零售销售往往呈年度周期性。某冰淇淋店的月销量在 7 月最高（约 1200 件），在 1 月最低（约 200 件）。以月份 x = 1 到 12 为变量，最大值出现在 x = 7，则销售模型为：

S(x) = 500 cos(2π(x − 7)/12) + 700

Check: S(1) = 500 cos(2π(−6)/12) + 700 = 500 cos(−π) + 700 = 200; S(7) = 1200. The model helps managers plan inventory and staffing.

验证：S(1) = 500 cos(2π(−6)/12) + 700 = 500 cos(−π) + 700 = 200；S(7) = 1200。该模型有助于管理者规划库存和人员安排。

9. Pendulum and Simple Harmonic Motion | 单摆与简谐运动

A pendulum swinging with small amplitude undergoes simple harmonic motion. The angular displacement θ from the vertical can be written as:

小角度摆动的单摆作简谐运动。其偏离竖直方向的角位移 θ 可以写为：

θ(t) = θ_max sin(2πt/T)

where T is the period. For small angles, T ≈ 2π√(L/g), where L is the pendulum length and g ≈ 9.81 m/s². This model is also applied to springs, vibrating strings, and building oscillations during earthquakes.

其中 T 为周期。在小角度条件下，T ≈ 2π√(L/g)，L 为摆长，g ≈ 9.81 m/s²。该模型同样适用于弹簧、琴弦振动以及地震时建筑物的振荡。

10. Constructing a Model from Data | 从数据构建模型

To fit a trigonometric model to real data, follow these steps:

要将三角模型拟合到真实数据，请遵循以下步骤：

Identify the period: Look for repeating highs and lows; determine the time between two maxima or two minima.
Find the midline and amplitude: Use D = (M + m)/2 and A = (M − m)/2.
Choose sine or cosine and locate the phase shift: If the first maximum is at x = a, use y = A cos(B(x − a)) + D, or use y = A sin(B(x − c)) + D with c adjusted appropriately.
Validate: Substitute data points to check whether the model matches the observed values.

确定周期：观察数据中重复出现的高点和低点，求出两个最大值或两个最小值之间的时间间隔。

求中线和振幅：使用 D = (M + m)/2 和 A = (M − m)/2。

选择正弦或余弦并确定相移：若第一个最大值出现在 x = a，可使用 y = A cos(B(x − a)) + D，或使用 y = A sin(B(x − c)) + D 并适当调整 c。

验证：将数据点代入模型，检查是否与观测值匹配。

11. Example Problem | 例题解析

The water depth at a dock is recorded over 24 hours. High tide at midnight: 5 m. Low tide at 6 a.m.: 1 m. Find a trigonometric model for depth d(t) in hours after midnight.

某码头记录 24 小时内的水深变化。午夜高潮时水深 5 米，早晨 6 点低潮时水深 1 米。求以午夜后小时数 t 为变量的水深 d(t) 三角模型。

Here the period is 12 hours (from high to low is 6 hours, so half period = 6 h). Midline D = (5 + 1)/2 = 3, amplitude A = (5 − 1)/2 = 2. Since high tide occurs at t = 0, use cosine:

此处理论上周期为 12 小时（从高潮到低潮为 6 小时，即半个周期为 6 小时）。中线 D = (5 + 1)/2 = 3，振幅 A = (5 − 1)/2 = 2。由于高潮发生在 t = 0，使用余弦函数：

d(t) = 2 cos(2πt/12) + 3 = 2 cos(πt/6) + 3

At t = 6, d = 2 cos(π) + 3 = 1, correct. To find when depth exceeds 4 m, solve 2 cos(πt/6) + 3 > 4, giving cos(πt/6) > 1/2. This yields 0 ≤ t < 2 or 10 < t < 14 within a 12-hour cycle.

t = 6 时，d = 2 cos(π) + 3 = 1，正确。若求水深超过 4 米的时间，可解 2 cos(πt/6) + 3 > 4，即 cos(πt/6) > 1/2。在一个 12 小时周期内解得 0 ≤ t < 2 或 10 < t < 14。

12. Common Pitfalls | 常见易错点

When working with trigonometric models, students often forget to check the units of B. In IB mathematics, unless otherwise specified, angles are in radians. Therefore period = 2π/B, not 360/B. Another common error is confusing horizontal shift with phase shift when using the sine model: in y = A sin(B(x − C)), the shift is C, not B·C.

在处理三角模型时，学生经常忘记检查 B 的单位。在 IB 数学中，除非另有说明，角度采用弧度制。因此周期 = 2π/B，而不是 360/B。另一个常见错误是在使用正弦模型时混淆水平平移与相移：在 y = A sin(B(x − C)) 中，平移量为 C，而不是 B·C。

Also, be careful with the sign of A. A negative amplitude reflects the graph across the midline, which is equivalent to a phase shift of π. Always check that your model matches at least three data points: maximum, minimum, and midline crossing.

此外要注意 A 的符号。负振幅会使图像关于中线上下翻转，相当于相移 π。务必验证模型至少匹配三个数据点：最大值、最小值和中线交点。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

IB Mathematics: Real-World Applications of Trigonometric Models | IB数学：三角模型的实际应用