Simplifying a cos x ± b sin x | 化简 a cos x ± b sin x

📚 Simplifying a cos x ± b sin x | 化简 a cos x ± b sin x

The expression a cos x ± b sin x appears constantly in A-Level Mathematics, Physics and Engineering. It is a linear combination of a cosine wave and a sine wave with the same period, and it can always be rewritten as a single sinusoidal function of the form R cos(x ∓ α) or R sin(x ± α). This technique, known as the harmonic form or the R-α method, turns messy combinations into a single transformed trig curve, making it far easier to analyse amplitude, maximum and minimum values, sketch graphs, and solve equations.

表达式 a cos x ± b sin x 在 A-Level 数学、物理和工程学中频繁出现。它是同周期余弦波与正弦波的线性组合，总可以被改写为 R cos(x ∓ α) 或 R sin(x ± α) 形式的单一正弦型函数。这种技巧称为谐波形式（harmonic form）或 R-α 方法，能将复杂的组合转化为单一变换后的三角函数曲线，从而极大地方便我们分析振幅、最大值与最小值、绘制图像以及解方程。

1. The Compound Angle Foundation | 复合角公式基础

Before we can simplify a cos x ± b sin x, we must recall the four compound angle identities for cosine and sine. These are given in the Edexcel formula booklet, but you must know them fluently in both directions.

在化简 a cos x ± b sin x 之前，我们必须回顾 cos 与 sin 的四个复合角恒等式。这些公式在 Edexcel 公式册中会给出，但你必须能够熟练地正用和反用它们。

cos(A − B) = cos A cos B + sin A sin B

cos(A + B) = cos A cos B − sin A sin B

sin(A + B) = sin A cos B + cos A sin B

sin(A − B) = sin A cos B − cos A sin B

The key idea of the R-α method is to read these identities in reverse: starting from an expanded expression like R cos x cos α + R sin x sin α, we compress it back into R cos(x − α).

R-α 方法的核心思想就是反向使用这些公式：从一个展开式（如 R cos x cos α + R sin x sin α）出发，把它压缩回 R cos(x − α)。

2. The Harmonic Form | 谐波形式（R-α 形式）

The harmonic form states that any expression of the type a cos x + b sin x, with a and b positive constants, can be written as a single cosine or sine function. The most commonly tested form in Edexcel is R cos(x − α), but three other forms are equally valid and appear in exam questions.

谐波形式指出：任何形如 a cos x + b sin x 的表达式（其中 a、b 为正常数）都可以写成单一的余弦或正弦函数。Edexcel 考试中最常考察的形式是 R cos(x − α)，但其他三种形式同样有效，也常出现在考题中。

For a cos x + b sin x = R cos(x − α), the expansion is:

对于 a cos x + b sin x = R cos(x − α)，其展开为：

a cos x + b sin x = R cos x cos α + R sin x sin α

Comparing coefficients of cos x and sin x gives two matching equations. This comparison is the heart of the method, and every derivation follows from it.

比较 cos x 与 sin x 的系数，可以得到两个对应方程。这种系数比较是整个方法的核心，所有推导都由此出发。

R cos α = a and R sin α = b
R cos α = a 且 R sin α = b

3. Deriving R and α | 推导 R 与 α

From the two matching equations we can find R by squaring and adding. This eliminates α in one step.

由这两个对应方程，我们可以通过平方再相加来求 R。这一步能直接消去 α。

(R cos α)² + (R sin α)² = a² + b²

Since cos²α + sin²α = 1, the left-hand side simplifies to R². Therefore the amplitude is always the square root of the sum of the squares of the two coefficients.

由于 cos²α + sin²α = 1，左边化简为 R²。因此振幅始终等于两个系数平方之和再开根号。

R = √(a² + b²)

Dividing the two matching equations gives tan α = b/a, so α is found using the inverse tangent function. When a and b are both positive, α lies in the first quadrant and is acute.

将两个对应方程相除得到 tan α = b/a，因此用反正切函数求 α。当 a、b 均为正数时，α 位于第一象限，是一个锐角。

α = tan⁻¹(b/a), with 0 < α < 90°

α = tan⁻¹(b/a)，其中 0 < α < 90°

It is essential to remember that this derivation assumes a and b are positive. If one of them is negative, the sign of tan α changes and α must be placed in the correct quadrant. We will return to this in the pitfalls section.

必须记住，上述推导假设 a、b 均为正数。若其中有负数，tan α 的符号会改变，α 必须放在正确的象限中。我们将在“常见易错点”一节中详细讨论。

4. The Four Standard Forms | 四种标准形式

Depending on which single function you choose as the target, there are four equivalent rewrites. They all share the same value of R, and the same acute reference angle α = tan⁻¹(b/a). The only difference is the sign inside the bracket and whether the leading function is sine or cosine.

根据你选择的目标单一函数不同，一共有四种等价的改写形式。它们共享相同的 R 值和相同的锐角参考角 α = tan⁻¹(b/a)，唯一的区别是括号内的符号以及前导函数是正弦还是余弦。

Original form	Single form	R	α
a cos x + b sin x	R cos(x − α)	√(a² + b²)	tan α = b/a
a cos x − b sin x	R cos(x + α)	√(a² + b²)	tan α = b/a
a sin x + b cos x	R sin(x + α)	√(a² + b²)	tan α = b/a
a sin x − b cos x	R sin(x − α)	√(a² + b²)	tan α = b/a

Notice the beautiful symmetry: a negative sign in front of the sine term corresponds to a plus sign inside the cosine form, and a negative sign in front of the cosine term corresponds to a minus sign inside the sine form.

注意这里优美的对称性：sin 项前是负号，对应余弦形式括号内取加号；cos 项前是负号，对应正弦形式括号内取减号。

5. Worked Conversion | 转换实例

Express 3 cos x + 4 sin x in the form R cos(x − α), where R > 0 and 0 < α < 90°. Give α correct to 2 decimal places.

将 3 cos x + 4 sin x 表示为 R cos(x − α) 的形式，其中 R > 0，0 < α < 90°，α 精确到小数点后两位。

Step 1: Compare with the compound angle form.

第一步：与复合角形式比较。

3 cos x + 4 sin x = R cos(x − α) = R cos x cos α + R sin x sin α

Step 2: Match coefficients. R cos α = 3 and R sin α = 4.

第二步：比较系数。R cos α = 3 且 R sin α = 4。

Step 3: Square and add. R² = 3² + 4² = 25, so R = 5.

第三步：平方相加。R² = 3² + 4² = 25，因此 R = 5。

Step 4: Divide. tan α = 4/3, so α = tan⁻¹(4/3) ≈ 53.13°.

第四步：相除。tan α = 4/3，所以 α = tan⁻¹(4/3) ≈ 53.13°。

3 cos x + 4 sin x = 5 cos(x − 53.13°)

You should always check your answer by expanding the result back: 5 cos(x − 53.13°) should reproduce 3 cos x + 4 sin x exactly.

你应当始终通过展开来检验答案：5 cos(x − 53.13°) 应能精确还原 3 cos x + 4 sin x。

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

Once an expression is written as R cos(x − α), finding its maximum and minimum becomes trivial. Since the cosine function always lies between −1 and 1, the whole expression must lie between −R and R.

一旦表达式写成 R cos(x − α)，求最大值和最小值就变得非常简单。由于余弦函数始终在 −1 与 1 之间，整个表达式的取值范围必然在 −R 与 R 之间。

Maximum value = R, occurring when cos(x − α) = 1, i.e. x = α + 360n°
最大值 = R，当 cos(x − α) = 1 时取得，即 x = α + 360n°
Minimum value = −R, occurring when cos(x − α) = −1, i.e. x = α + 180° + 360n°
最小值 = −R，当 cos(x − α) = −1 时取得，即 x = α + 180° + 360n°

For the example 3 cos x + 4 sin x = 5 cos(x − 53.13°), the maximum is 5, occurring at x = 53.13°. The minimum is −5, occurring at x = 233.13°. This immediately tells you the range of the function: −5 ≤ y ≤ 5.

对于例 3 cos x + 4 sin x = 5 cos(x − 53.13°)，最大值为 5，在 x = 53.13° 处取得；最小值为 −5，在 x = 233.13° 处取得。这立刻告诉我们函数的值域：−5 ≤ y ≤ 5。

If the original expression had been 3 cos x − 4 sin x, the single form would be 5 cos(x + 53.13°). The maximum still equals 5, but it now occurs at x = −53.13°, or equivalently x = 306.87° for the smallest positive angle.

若原表达式为 3 cos x − 4 sin x，则单一形式为 5 cos(x + 53.13°)。最大值仍为 5，但此时在 x = −53.13° 处取得，等价于最小正角 x = 306.87°。

7. Solving Trigonometric Equations | 解三角方程

Equations of the form a cos x + b sin x = c are impossible to solve directly by basic inverse functions, because the two terms cannot be separated easily. The R-α method converts the left-hand side into a single cosine, after which the equation becomes standard.

形如 a cos x + b sin x = c 的方程无法直接用基本反函数求解，因为两个项很难分离。R-α 方法将左边化为单一余弦，方程就变成了标准形式。

Solve 3 cos x + 4 sin x = 2 for 0° ≤ x < 360°.

解方程 3 cos x + 4 sin x = 2，其中 0° ≤ x < 360°。

Using the result above, rewrite the equation as 5 cos(x − 53.13°) = 2.

利用上面的结果，将方程改写为 5 cos(x − 53.13°) = 2。

cos(x − 53.13°) = 2/5 = 0.4

Let θ = x − 53.13°. Then cos θ = 0.4. The principal value is θ = 66.42°. Since cosine is positive in the first and fourth quadrants, the solutions in the range 0° ≤ θ < 360° are:

令 θ = x − 53.13°，则 cos θ = 0.4。主值为 θ = 66.42°。由于余弦在第一、第四象限为正，在 0° ≤ θ < 360° 范围内的解为：

θ = 66.42° or θ = 360° − 66.42° = 293.58°

θ = 66.42° 或 θ = 360° − 66.42° = 293.58°

Finally convert back: x = θ + 53.13°, giving x = 119.55° or x = 346.71°. Both lie in the required range, so the solution set is complete.

最后换算回去：x = θ + 53.13°，得到 x = 119.55° 或 x = 346.71°。两者都在要求范围内，因此解集完整。

8. Sketching Graphs | 绘制函数图像

The single-form rewrite reveals everything needed to sketch the graph without plotting many points. For y = R cos(x − α), the graph is the standard cosine curve with amplitude R, period 360° (or 2π radians), shifted to the right by α.

单一形式改写揭示了绘制图像所需的全部信息，无需逐点描图。对于 y = R cos(x − α)，图像是标准余弦曲线，振幅为 R，周期为 360°（或 2π

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Simplifying a cos x ± b sin x | 化简 a cos x ± b sin x