Trigonometry and Modelling | 三角学与建模

📚 Trigonometry and Modelling | 三角学与建模

In Edexcel A-Level Mathematics, “Trigonometry and modelling” brings together trigonometric identities, harmonic forms such as R sin(x ± α) and R cos(x ± α), and real-world periodic behaviour. This topic tests your ability to rewrite expressions, solve trigonometric equations, and interpret maximum and minimum values in practical contexts.

在 Edexcel A-Level 数学中，“三角学与建模”将三角恒等式、R sin(x ± α) 与 R cos(x ± α) 等调和形式以及现实世界的周期性行为结合起来。本主题考查你将表达式改写、解三角方程，以及在实际情境中解释最大值和最小值的能力。

1. Core Trigonometric Identities | 核心三角恒等式

Before you can work fluently with harmonic form and modelling, you must be confident with the basic trigonometric identities. The Pythagorean identity sin² x + cos² x = 1 is used constantly to simplify expressions or to change between sine and cosine.

在你能够熟练处理调和形式和建模之前，必须先掌握基本三角恒等式。毕达哥拉斯恒等式 sin² x + cos² x = 1 经常用于化简表达式，或在正弦和余弦之间进行转换。

sin² x + cos² x = 1
sin 2x = 2 sin x cos x
cos 2x = cos² x − sin² x = 2 cos² x − 1 = 1 − 2 sin² x

These identities allow you to reduce powers, rewrite products, and solve equations that look different but are actually quadratic in sin x or cos x. In modelling questions, they help you convert a given expression into the standard harmonic form.

这些恒等式让你能够降幂、改写乘积，并求解那些看似不同、但实际上是关于 sin x 或 cos x 的二次方程的题目。在建模题中，它们帮助你将其它给定表达式转换为标准调和形式。

2. Deriving the Harmonic Form R sin(x + α) | 推导调和形式 R sin(x + α)

Any expression of the type a sin x + b cos x can be written as R sin(x ± α) or R cos(x ± α). This is possible because the compound angle formula gives sin(x + α) = sin x cos α + cos x sin α.

任何形如 a sin x + b cos x 的表达式都可以写成 R sin(x ± α) 或 R cos(x ± α)。这是可行的，因为复合角公式给出 sin(x + α) = sin x cos α + cos x sin α。

To express a sin x + b cos x as R sin(x + α), compare coefficients after expanding:

要将 a sin x + b cos x 写成 R sin(x + α)，展开后比较系数：

R sin(x + α) = R sin x cos α + R cos x sin α

Therefore we require R cos α = a and R sin α = b. Squaring and adding these two equations gives R² = a² + b², so R = √(a² + b²). Dividing gives tan α = b/a.

因此我们需要 R cos α = a 且 R sin α = b。将两式平方后相加，可得 R² = a² + b²，所以 R = √(a² + b²)。两式相除则给出 tan α = b/a。

The angle α must be chosen in the quadrant that matches the signs of a and b. For example, if a is negative and b is positive, then cos α is negative and sin α is positive, so α lies in the second quadrant.

角度 α 必须选择在与 a 和 b 的符号相匹配的象限。例如，如果 a 为负、b 为正，那么 cos α 为负且 sin α 为正，所以 α 位于第二象限。

3. The R cos(x ± α) Form and Choosing a Form | R cos(x ± α) 形式与形式选择

Sometimes a question or a modelling situation is easier to handle using R cos(x ± α). The expansion R cos(x − α) = R cos x cos α + R sin x sin α shows that a sin x + b cos x can also be written with cosine as the main function.

有时题目或建模情境使用 R cos(x ± α) 更容易处理。展开式 R cos(x − α) = R cos x cos α + R sin x sin α 表明，a sin x + b cos x 也可以用余弦作为主函数来表示。

For the form R cos(x − α), comparing coefficients gives R cos α = b and R sin α = a. The amplitude R is still √(a² + b²), but the tangent ratio becomes tan α = a/b.

对于形式 R cos(x − α)，比较系数得到 R cos α = b 且 R sin α = a。振幅 R 仍然是 √(a² + b²)，但正切比变为 tan α = a/b。

The choice between R sin and R cos is often guided by the structure of the problem. If you need to maximise a sin x + b cos x, either form will give the same amplitude, but the location of the maximum will be expressed differently because the two forms are phase shifts of each other.

在 R sin 和 R cos 之间的选择通常由题目结构决定。如果你需要求 a sin x + b cos x 的最大值，两种形式给出的振幅相同，但最大值出现的位置会表达不同，因为这两种形式只是彼此的相位移动。

4. Worked Example: Finding R and α | 例题：求 R 和 α

Express 3 sin x + 4 cos x in the form R sin(x + α), where R > 0 and 0 < α < π/2. First find the amplitude R:

将 3 sin x + 4 cos x 写成 R sin(x + α) 的形式，其中 R > 0 且 0 < α < π/2。首先求振幅 R：

R = √(3² + 4²) = √25 = 5

Then find α using tan α = b/a = 4/3. Since both coefficients are positive, α is in the first quadrant, so α = arctan(4/3) ≈ 0.927 rad.

然后使用 tan α = b/a = 4/3 求 α。由于两个系数都是正的，α 位于第一象限，所以 α = arctan(4/3) ≈ 0.927 rad。

Therefore 3 sin x + 4 cos x = 5 sin(x + 0.927). If the coefficients had been 5 sin x − 12 cos x, then R would be 13 and tan α = 12/5, but the sign of b would place α in the fourth quadrant, giving a negative α or an equivalent positive angle.

因此 3 sin x + 4 cos x = 5 sin(x + 0.927)。如果系数是 5 sin x − 12 cos x，那么 R 会是 13，tan α = 12/5，但 b 的符号会使 α 位于第四象限，从而得到负 α 或等效的正角度。

5. Solving Equations Using Harmonic Form | 使用调和形式解方程

Equations such as 3 sin x + 4 cos x = 2 are difficult to solve in their original form. Rewriting the left-hand side as R sin(x + α) produces a single sine equation, which can then be solved using inverse sine and symmetry.

像 3 sin x + 4 cos x = 2 这样的方程在原始形式下很难求解。将左边改写为 R sin(x + α) 后，得到单一正弦方程，然后可以利用反正弦和对称性求解。

Using 3 sin x + 4 cos x = 5 sin(x + 0.927), the equation becomes 5 sin(x + 0.927) = 2, so sin(x + 0.927) = 0.4.

利用 3 sin x + 4 cos x = 5 sin(x + 0.927)，方程变为 5 sin(x + 0.927) = 2，所以 sin(x + 0.927) = 0.4。

For solutions in 0 ≤ x ≤ 2π, write x + 0.927 = arcsin 0.4 + 2nπ or x + 0.927 = π − arcsin 0.4 + 2nπ. Subtract 0.927 and choose the values of n that keep x inside the required interval.

对于 0 ≤ x ≤ 2π 内的解，写出 x + 0.927 = arcsin 0.4 + 2nπ 或 x + 0.927 = π − arcsin 0.4 + 2nπ。减去 0.927，并选择使 x 保持在所求区间内的 n 值。

This method is extremely common in Edexcel papers. You should be comfortable working with inverse trigonometric values in radians and using arcsin, arccos, or arctan notation.

这种方法在 Edexcel 试卷中非常常见。你应该熟练使用弧度制下的反三角函数值，并习惯使用 arcsin、arccos 或 arctan 记号。

6. Maximum and Minimum Values | 最大值与最小值

The harmonic form makes maximum and minimum values immediately visible. Since the sine and cosine functions oscillate between −1 and 1, the expression R sin(x + α) has maximum value R and minimum value −R.

调和形式使最大值和最小值一目了然。由于正弦和余弦函数在 −1 与 1 之间振荡，表达式 R sin(x + α) 的最大值为 R，最小值为 −R。

For 3 sin x + 4 cos x = 5 sin(x + 0.927), the maximum value is 5 and the minimum value is −5. To locate the maximum, set x + 0.927 = π/2 + 2nπ, giving x = π/2 − 0.927 + 2nπ.

对于 3 sin x + 4 cos x = 5 sin(x + 0.927)，最大值为 5，最小值为 −5。要确定最大值出现的位置，令 x + 0.927 = π/2 + 2nπ，得到 x = π/2 − 0.927 + 2nπ。

In modelling, the amplitude R is often the most important feature. It gives the size of the oscillation above or below the central line, such as the height of a wave, the variation in temperature, or the depth of a tide.

在建模中，振幅 R 通常是最重要的特征。它给出了围绕中心线上下振荡的大小，例如波高、温度变化或潮汐深度。

7. Building Trigonometric Models | 构建三角模型

Trigonometric functions model any phenomenon that repeats at regular intervals. A general model can be written as H(t) = A sin(ωt + φ) + B, where A is amplitude, B is vertical shift, ω is angular frequency, and φ is phase shift.

三角函数可以模拟任何以固定间隔重复的现象。一般模型可写为 H(t) = A sin(ωt + φ) + B，其中 A 是振幅，B 是垂直位移，ω 是角频率，φ 是相位移动。

The period is given by T = 2π/ω when t is measured in radians, or T = 360°/ω when degrees are used. The vertical shift B represents the central value or mean level around which the function oscillates.

当 t 以弧度为单位时，周期由 T = 2π/ω 给出；当使用度时，周期为 T = 360°/ω。垂直位移 B 表示函数围绕振荡的中心值或平均水平。

For example, the tide model h(t) = 5 sin(πt/6) + 8 has amplitude 5 metres, mean depth 8 metres, and period T = 2π ÷ (π/6) = 12 hours.

例如，潮汐模型 h(t) = 5 sin(πt/6) + 8 的振幅为 5 米，平均深度为 8 米，周期为 T = 2π ÷ (π/6) = 12 小时。

The argument πt/6 controls the period. After every 12 hours, the sine function completes one full cycle, which matches the natural tidal rhythm in many coastal regions.

自变量中的 πt/6 控制着周期。每经过 12 小时，正弦函数完成一个完整循环，这与许多沿海地区自然的潮汐节律相吻合。

8. Interpreting Graphs and Phase Shifts | 图解读与相位移动

When a graph is given, you can construct a trigonometric model by identifying the maximum, minimum, period, and horizontal shift. The amplitude is (maximum − minimum)/2, and the vertical shift is (maximum + minimum)/2.

当给出图像时，你可以通过确定最大值、最小值、周期和水平位移来构建三角模型。振幅为 (最大值 − 最小值)/2，垂直位移为 (最大值 + 最小值)/2。

The horizontal or phase shift can be found by locating a key point, such as the first maximum or a point where the curve crosses the central line moving upwards. This shift corresponds to φ/ω in the model A sin(ωt + φ) + B.

水平位移或相位移动可以通过确定一个关键点来找到，例如第一个最大值点，或曲线向上穿过中心线的点。这个位移对应于模型 A sin(ωt + φ) + B 中的 φ/ω。

If a graph has a maximum of 10 and a minimum of 2, then the amplitude is (10 − 2)/2 = 4 and the vertical shift is (10 + 2)/2 = 6. If one full cycle takes 12 hours, then ω = 2π/12 = π/6, giving a model of the form y = 4 sin(πt/6 + φ) + 6.

如果图像最大值为 10、最小值为 2，则振幅为 (10 − 2)/2 = 4，垂直位移为 (10

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Trigonometry and Modelling | 三角学与建模