📚 Simplifying α cos x ± b sin x | 简化 α cos x ± b sin x 形式
et expressions like α cos x + b sin x or α cos x - b sin x. These are linear combinations of sine and cosine. By using compound angle identities, we can rewrite them as a single trigonometric function, typically in the form R cos(x ∓ . This is a key skill for equation solving, sketching graphs, and finding extremum values.
在A-Level数学中，你经常会遇到 α cos x + b sin x 或 α cos x - b sin x 这样的表达式。它们属于正弦和余弦的线性组合。利用复合角恒等式，我们可以将它们重写成单个三角函数，通常形如 R cos(x ∓ θ) 或 R sin(x ± θ)。这对于解方程、画函数图像和求极值来说是一项关键技能。
1. Why Simplify These Expressions? | 为什么要简化这些表达式？
ine or cosine term reveals the amplitude R and phase shift θ directly. This unifies a sum into one wave, making it far easier to solve trigonometric equations, differentiate, integrate, or analyse the maximum and minimum values of the function.
将 α cos x ± b sin x 表示为单个正弦或余弦项可以直接显示出振幅 R 和相位差 θ。这样一来，多个波的叠加就统一成了一个波，从而让解三角方程、求导、积分或者分析函数的最大值和最小值都变得简单得多。
2. The Compound Angle Formulas | 复合角公式
Everything relies on the expansions of cos(A ± B) and sin(A ± B). Recall:
cos(A ± B) = cos A cos B ∓ sin A sin B
sin(A ± B) = sin A cos B ± cos A sin B
qual to, say, R cos(x - θ). By expanding the right-hand side and comparing coefficients of cos x and sin x, we can determine R and θ.
一切都基于 cos(A ± B) 和 sin(A ± B) 的展开式。回忆：
cos(A ± B) = cos A cos B ∓ sin A sin B
sin(A ± B) = sin A cos B ± cos A sin B
当我们令 α cos x ± b sin x 等于例如 R cos(x - θ) 时，这些恒等式让我们可以通过比较 cos x 和 sin x 的系数来求出 R 和 θ。
3. Choosing the Right Form | 选择正确的形式
We can express α cos x ± b sin x as R cos(x ∓ ds on the signs and on whether you prefer cosine or sine. The most common standard form is R cos(x - θ) for α cos x + b sin x and R cos(x + θ) for α cos x - b sin x. However, all four combinations are valid and interchangeable.
我们可以将 α cos x ± b sin x 表示为 R cos(x ∓ θ) 或 R sin(x ± θ)。选择哪种形式通常取决于符号以及你更习惯用余弦还是正弦。最常见的形式是：对于 α cos x + b sin x 用 R cos(x - θ)，对于 α cos x - b sin x 用 R cos(x + θ)。不过，四种组合都是有效的，并且可以互相转化。
| Expression | Rewrite as | Conditions |
| α cos x + b sin x | R cos(x - θ) | α = R cos θ, b = R sin θ |
| α cos x - b sin x | R cos(x + θ) | α = R cos θ, b = R sin θ |
| α cos x + b sin x | R sin(x + θ) | α = R sin θ, b = R cos θ |
| α cos x - b sin x | R sin(x - θ) | α = R sin θ, b = R cos θ |
For simplicity, exam questions often favour the cosine forms because they avoid swapping the roles of α and b in the tan formula. However, you should be comfortable with all variants.
简单来说，考试题偏爱余弦形式，因为这样可以避免在 tan 公式中交换 α 和 b 的角色。不过，你应该熟练掌握所有变体。
4. Expressing as R cos(x - θ) | 表示为 R cos(x - θ)
rt with α cos x + b sin x. Assume α cos x + b sin x = R cos(x - θ). Expand the right side:
R cos(x - θ) = R cos x cos θ + R sin x sin θ
Compare coefficients of cos x and sin x: α = R cos θ and b = R sin θ. Hence:
R = √α² + b²
tan θ = b / α
, b positive → 2nd quadrant; etc.). Always give θ in radians unless degrees are specified.
从 α cos x + b sin x 开始。设 α cos x + b sin x = R cos(x - θ)。展开右边：
R cos(x - θ) = R cos x cos θ + R sin x sin θ
比较 cos x 和 sin x 的系数得到：α = R cos θ 且 b = R sin θ。于是有：
R = √α² + b²
tan θ = b / α
θ 的象限由 α 和 b 的符号决定（两者为正 → 第一象限；α 负，b 正 → 第二象限；等等）。除非题目指明度数，否则 θ 一律用弧度表示。
5. Expressing as R cos(x + θ) | 表示为 R cos(x + θ)
When we have α cos x - b sin x, it is natural to use R cos(x + θ). Expand:
R cos(x + θ) = R cos x cos θ - R sin x sin θ
Matching terms gives α = R cos θ and b = R sin θ — exactly the same relationships as before! Therefore R = √α² + b² de is +θ. This means the expression is written as R cos(x + θ), with θ found in the same way.
当表达式为 α cos x - b sin x 时，很自然地使用 R cos(x + θ) 形式。展开得：
R cos(x + θ) = R cos x cos θ - R sin x sin θ
匹配项得到 α = R cos θ 且 b = R sin θ —— 与前面完全相同的关系！因此 R = √α² + b²，tan θ = b / α，但要注意角度里面是 +θ。这意味着表达式被写成 R cos(x + θ)，而 θ 的求法完全一样。
6. Expressing as R sin(x + θ) | 表示为 R sin(x + θ)
Alternatively, α cos x + b sin x can be written as R sin(x + θ). Expand:
R sin(x + θ) = R sin x cos θ + R cos x sin θ
Now the coefficients swap roles: α = R sin θ and b = R cos θ. Therefore:
R = √α² + b²
tan θ = α / b
This form is useful when the coefficient of cos x is larger and you want an angle addition in sine. Pay careful attention to which coefficient gives tan θ — it is α/b, not b/α.
另一种方法，α cos x + b sin x 可以写成 R sin(x + θ)。展开：
R sin(x + θ) = R sin x cos θ + R cos x sin θ
此时系数角色互换：α = R sin θ 且 b = R cos θ。因此：
R = √α² + b²
tan θ = α / b
当 cos x 的系数较大，且你想用正弦的加法角时，这种形式很有用。要格外注意 tan θ 的表达式 —— 它是 α/b，而不是 b/α。
7. Expressing as R sin(x - θ) | 表示为 R sin(x - θ)
For α cos x - b sin x, the sine form with subtraction is R sin(x - θ). Expand:
R sin(x - θ) = R sin x cos θ - R cos x sin θ
Matching: b = R cos θ and α = R sin θ? Wait — cient is R cos θ which must equal b; for cos x, coefficient is -R sin θ which must equal α? cos θ - R cos x sin θ, then the term with sin x is R cos θ sin x, so we need R cos θ = -b? ess the question specifically asks for it. Usually we prefer the cosine forms for clarity.
对于 α cos x - b sin x，相应的正弦减法形式是 R sin(x - θ)。展开：
R sin(x - θ) = R sin x cos θ - R cos x sin θ
匹配时需谨慎：sin x 的系数是 R cos θ，应等于 -b；cos x 的系数是 -R sin θ，应等于 α。这容易引起符号错误。除非题目明确要求，我们通常更倾向于使用余弦形式以避免混淆。
8. Worked Example 1: 3 cos x + 4 sin x | 例题1：3 cos x + 4 sin x
Write 3 cos x + 4 sin x in the form R cos(x - θ), where R > 0 and 0 < θ < π/2.
将 3 cos x + 4 sin x 写成 R cos(x - θ) 的形式，其中 R > 0，0 < θ < π/2。
Solution:
os x + 4 sin x = R cos(x - θ). Expand: R cos x cos θ + R sin x sin θ. Equate coefficients:
R cos θ = 3, R sin θ = 4
Square and add: R²(cos²θ + sin²θ) = 3² + 4² = 25 ⇒ R = √25 = 5.
first quadrant. In radians, θ ≈ 0.9273 rad, but exact answer can be given as arctan(4/3).
Thus, 3 cos x + 4 sin x = 5 cos(x - arctan(4/3)).
解：
设 3 cos x + 4 sin x = R cos(x - θ)。展开得：R cos x cos θ + R sin x sin θ。比较系数：
R cos θ = 3， R sin θ = 4
平方相加：R²(cos²θ + sin²θ) = 3² + 4² = 25 ⇒ R = √25 = 5。
相除：tan θ = (R sin θ)/(R cos θ) = 4/3。所以 θ = arctan(4/3)。由于 3 和 4 均为正，θ 在第一象限。用弧度表示为 θ ≈ 0.9273 rad，精确值可写作 arctan(4/3)。
因此，3 cos x + 4 sin x = 5 cos(x - arctan(4/3))。
9. Worked Example 2: 5 cos x - 12 sin x | 例题2：5 cos x - 12 sin x
Express 5 cos x - 12 sin x in the form R cos(x + θ), where R > 0 and 0 < θ < π.
将 5 cos x - 12 sin x 写成 R cos(x + θ) 的形式，其中 R > 0，0 < θ < π。
Let 5 cos x - 12 sin x = R cos(x + θ). Expand: R cos x cos θ - R sin x sin θ. Equating:
R cos θ = 5, R sin θ = 12
Then R = √5² + 12² = √25 + 144 = √169 = 13.
le is arctan(12/5). However, we must check the signs: R cos θ = 5 > 0 and R sin θ = 12 > 0 ⇒ rst quadrant, 0 < θ < π/2. The question allows θ up to π, so θ = arctan(12/5) ≈ 1.176 rad.
Therefore, 5 cos x - 12 sin x = 13 cos(x + arctan(12/5)).
设 5 cos x - 12 sin x = R cos(x + θ)。展开得：R cos x cos θ - R sin x sin θ。比较：
R cos θ = 5， R sin θ = 12
于是 R = √5² + 12² = √25 + 144 = √169 = 13。
tan θ = 12/5。由于 R cos θ = 5 > 0 且 R sin θ = 12 > 0，θ 在第一象限，0 < θ < π/2。题目要求 θ 在 (0, π) 内，因此 θ = arctan(12/5) ≈ 1.176 rad。
因此，5 cos x - 12 sin x = 13 cos(x + arctan(12/5))。
10. Solving Equations Using the Simplified Form | 利用简化形式解方程
in x = c becomes straightforward. For example, solve 3 cos x + 4 sin x = 2 for 0 ≤ x ≤ 2π.
一旦你重写了表达式，求解形如 α cos x ± b sin x = c 的方程就变得很简单。例如，在 0 ≤ x ≤ 2π 范围内解 3 cos x + 4 sin x = 2。
le 1, 3 cos x + 4 sin x = 5 cos(x - 0.9273). So the equation becomes 5 cos(x - 0.9273) = 2 ⇒ cos(x - 0.9273) = 0.4.
arccos(0.4) ≈ 1.1593 rad. The second solution in [0, 2π] is u = 2π - 1.1593 ≈ 5.1239 rad.
12 - 2π ≈ 6.0512 - 6.2832 = -0.232 → no, we need to keep within 0 to 2π. Actually 5.1239 + 0.9273 = 6.0512, which is less than 2π (6.2832), so it's valid. But careful: 6.0512 is within 0 to 2π? tions are approximately x = 2.09 rad and x = 6.05 rad. Always verify by substituting back.
从例 1 知，3 cos x + 4 sin x = 5 cos(x - 0.9273)。所以方程变为 5 cos(x - 0.9273) = 2 ⇒ cos(x - 0.9273) = 0.4。
令 u = x - 0.9273。则 cos u = 0.4。主值：u = arccos(0.4) ≈ 1.1593 rad。在 [0, 2π] 中的第二个解为 u = 2π - 1.1593 ≈ 5.1239 rad。
将每个解加回 0.9273：x = 1.1593 + 0.9273 = 2.0866 rad，以及 x = 5.1239 + 0.9273 = 6.0512 rad（6.0512 < 6.2832，有效）。因此解约为 x = 2.09 rad 和 x = 6.05 rad。务必代回验证。
11. Maximum and Minimum Values | 最大值和最小值
Since R cos(x ∓ nd the minimum value is -R. The values of x at which they occur are found by solving cos(x ∓ θ) = ±1.
因为 R cos(x ∓ θ) 或 R sin(x ± θ) 在 -R 和 R 之间振荡，我们可以立刻说出 α cos x ± b sin x 的最大值和最小值。最大值是 R，最小值是 -R。它们发生时的 x 值可通过解 cos(x ∓ θ) = ±1 求得。
1.176). Hence the maximum is 13, occurring when cos(x + 1.176) = 1, i.e., x + 1.176 = 2kπ ⇒ - 1.176. The minimum is -13, occurring when cos(x + 1.176) = -1, i.e., x + 1.176 = (2k+1)π ⇒ x = (2k+1)π - 1.176.
例如，f(x) = 5 cos x - 12 sin x。我们求得 f(x) = 13 cos(x + 1.176)。因此最大值为 13，发生在 cos(x + 1.176) = 1 时，即 x + 1.176 = 2kπ ⇒ x = 2kπ - 1.176。最小值为 -13，发生在 cos(x + 1.176) = -1 时，即 x + 1.176 = (2k+1)π ⇒ x = (2k+1)π - 1.176。
12. Common Mistakes and Key Tips | 常见错误与关键提示
s sometimes slip up when simplifying α cos x ± b sin x. Watch out for these common errors:
- Forgetting to square and add to find R: always R = √α² + b², not α + b.
- Using the wrong tan formula: for R cos(x ∓ θ), tan θ = b/α; for R sin(x + θ), tan θ = α/b. Read the form carefully.
- Ignoring the quadrant when determining θ. Always check the signs of the coefficients to place θ in the correct quadrant.
- Confusing radians and degrees. If the interval is in radians, your θ must be in radians.
- When solving equations, remember to add the phase shift correctly and check that all solutions lie within the required domain by adding or subtracting 2π as needed.
即使能力强的学生在简化 α cos x ± b sin x 时有时也会出错。警惕以下常见错误：
- 忘记平方相加求 R：一定是 R = √α² + b²，而不是 α + b。
- 用错 tan 公式：对于 R cos(x ∓ θ)，tan θ = b/α；对于 R sin(x + θ)，tan θ = α/b。要仔细辨别形式。
- 确定 θ 时忽略象限。务必根据系数的符号将 θ 置于正确的象限。 Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
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