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

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

The expression a cos x ± b sin x appears frequently in trigonometry, calculus, and physics. It can be rewritten as a single sine or cosine function with a phase shift, which makes solving equations, finding maxima and minima, and sketching graphs much simpler.

表达式 a cos x ± b sin x 在三角学、微积分和物理中经常出现。它可以改写为带有相位移动的单一正弦或余弦函数，从而使解方程、求最大值最小值以及绘制图形变得更加简单。

---

1. The Basic Principle | 基本恒等式

The key idea is to express a cos x + b sin x in the form R cos (x - α) or R sin (x + α), where R > 0 and α is a constant angle.

关键思想是将 a cos x + b sin x 表示成 R cos (x - α) 或 R sin (x + α) 的形式，其中 R > 0，α 为常数角。

• R is the amplitude (always positive).
• α is the phase shift, usually measured in radians.
• The exact form chosen depends on the problem: sine form or cosine form.

• R 是振幅（始终为正）。
• α 是相位移，通常以弧度为单位。
• 选择哪种形式取决于问题：正弦形式或余弦形式。

a cos x + b sin x = R cos (x - α)

---

2. Finding R | 求 R 的值

Using the compound angle formula for cosine:

利用余弦的复合角公式：

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

Therefore:

因此：

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

Comparing with a cos x + b sin x, we obtain:

与 a cos x + b sin x 比较，得到：

R cos α = a, R sin α = b

Squaring and adding these two equations gives:

将两式平方后相加，得到：

R² (cos² α + sin² α) = a² + b²

Since cos² α + sin² α = 1, we have:

由于 cos² α + sin² α = 1，因此：

R = √a² + b²

---

3. Finding α | 求 α 的值

From the same comparison, we have:

由同样的比较，可得：

tan α = ba

However, we must be careful about the quadrant of α. The signs of a and b determine which quadrant α lies in.

然而，我们必须注意 α 所在的象限。a 和 b 的符号决定了 α 落在哪个象限。

Sign of a	Sign of b	Quadrant of α
+	+	First (0 to π/2)
−	+	Second (π/2 to π)
−	−	Third (π to 3π/2)
+	−	Fourth (3π/2 to 2π)

In many exam questions, a > 0 and b > 0, so α = tan^-1(b/a) directly. But for other signs, use the atan2 function or adjust by adding π.

在许多考试题目中，a > 0 且 b > 0，因此可直接使用 α = tan^-1(b/a)。但对于其他符号，请使用 atan2 函数或通过加 π 进行调整。

---

4. Alternative Forms | 不同形式

The same expression can be written in several equivalent forms. Each has its own use.

同一个表达式可以写成几种等价形式，每种形式都有其用途。

For a cos x + b sin x:

对于 a cos x + b sin x：

R cos (x - α)

R sin (x + β), where β = π2 - α

For a cos x - b sin x:

对于 a cos x - b sin x：

R cos (x + α)

R sin (x - β)

The choice between sine and cosine form is often determined by what you need: cosine form is convenient for maxima/minima, sine form is convenient for solving equations involving sin x.

选择正弦形式还是余弦形式通常取决于需求：余弦形式适合求最大值和最小值，正弦形式适合解包含 sin x 的方程。

---

5. Worked Example 1 | 例题 1

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

将 3 cos x + 4 sin x 表示为 R cos (x - α) 的形式，其中 R > 0，0 < α < π/2。

Here a = 3, b = 4.

这里 a = 3，b = 4。

R = √3² + 4² = √9 + 16 = √25 = 5

Now tan α = 43. Since a > 0, b > 0, α is in the first quadrant.

现在 tan α = 43。因为 a > 0, b > 0，所以 α 在第一象限。

α = tan^-1(43) ≈ 0.927 radians

Therefore:

因此：

3 cos x + 4 sin x = 5 cos (x - 0.927)

---

6. Worked Example 2 (Negative Coefficient) | 例题 2（负系数）

Express 5 cos x - 12 sin x in the form R cos (x + α), where R > 0 and 0 < α < π/2.

将 5 cos x - 12 sin x 表示为 R cos (x + α) 的形式，其中 R > 0，0 < α < π/2。

We use the identity:

我们使用恒等式：

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

Thus:

因此：

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

Comparing with 5 cos x - 12 sin x:

与 5 cos x - 12 sin x 比较：

R cos α = 5, R sin α = 12

So R = √5² + 12² = 13 and tan α = 125. Since both are positive, α = tan^-1(12/5) ≈ 1.176 radians.

所以 R = √5² + 12² = 13，且 tan α = 125。由于两者均为正，α = tan^-1(12/5) ≈ 1.176 弧度。

Hence:

因此：

5 cos x - 12 sin x = 13 cos (x + 1.176)

---

广告 / Sponsored

7. Solving Equations | 解方程

When solving an equation of the form a cos x + b sin x = c, the single-function form reduces it to a standard trigonometric equation.

解形如 a cos x + b sin x = c 的方程时，单一函数形式将其化为标准的三角方程。

Example: Solve 3 cos x + 4 sin x = 2 for 0 ≤ x < 2π, giving answers to 3 significant figures.

例：解 3 cos x + 4 sin x = 2，其中 0 ≤ x < 2π，答案保留 3 位有效数字。

From Example 1, we have 3 cos x + 4 sin x = 5 cos (x - 0.927).

由例 1，我们有 3 cos x + 4 sin x = 5 cos (x - 0.927)。

So the equation becomes:

因此方程变为：

5 cos (x - 0.927) = 2

cos (x - 0.927) = 0.4

Let θ = x - 0.927. Then cos θ = 0.4.

令 θ = x - 0.927，则 cos θ = 0.4。

The principal value is θ = cos^-1(0.4) ≈ 1.159. The general solutions are θ = 2π n ± 1.159.

主值为 θ = cos^-1(0.4) ≈ 1.159。通解为 θ = 2π n ± 1.159。

• θ = 1.159 ⇒ x = 1.159 + 0.927 = 2.086
• θ = 2π - 1.159 = 5.124 ⇒ x = 5.124 + 0.927 = 6.051

• θ = 1.159 ⇒ x = 1.159 + 0.927 = 2.086
• θ = 2π - 1.159 = 5.124 ⇒ x = 5.124 + 0.927 = 6.051

Both solutions lie in the interval 0 ≤ x < 2π. So x ≈ 2.09 and x ≈ 6.05 (3 s.f.).

两个解都在区间 0 ≤ x < 2π 内。所以 x ≈ 2.09 和 x ≈ 6.05（3 位有效数字）。

---

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

Since a single sine or cosine function always lies between −1 and 1, the maximum and minimum values of a cos x + b sin x are easy to find.

由于单一正弦或余弦函数总是在 −1 和 1 之间，因此 a cos x + b sin x 的最大值和最小值很容易求出。

For example, 3 cos x + 4 sin x = 5 cos (x - 0.927).

例如，3 cos x + 4 sin x = 5 cos (x - 0.927)。

• Maximum value = R = 5, occurring when cos (x - 0.927) = 1, i.e. x = 0.927 + 2π n.
• Minimum value = -R = -5, occurring when cos (x - 0.927) = -1, i.e. x = 0.927 + π + 2π n.

• 最大值 = R = 5，当 cos (x - 0.927) = 1 时取得，即 x = 0.927 + 2π n。
• 最小值 = -R = -5，当 cos (x - 0.927) = -1 时取得，即 x = 0.927 + π + 2π n。

In general, the maximum value of a cos x + b sin x is √a² + b² and the minimum value is -√a² + b².

一般地，a cos x + b sin x 的最大值为 √a² + b²，最小值为 -√a² + b²。

---

9. Sketching the Graph | 绘制图像

Rewriting the expression as a single cosine function immediately gives the amplitude, period, and phase shift for sketching.

将表达式改写为单一余弦函数后，可以立即得到用于绘图的振幅、周期和相位移。

For y = 3 cos x + 4 sin x = 5 cos (x - 0.927):

对于 y = 3 cos x + 4 sin x = 5 cos (x - 0.927)：

• Amplitude = 5
• Period = 2π
• Phase shift: the graph is shifted right by 0.927 radians relative to y = 5 cos x.

• 振幅 = 5
• 周期 = 2π
• 相位移：图像相对于 y = 5 cos x 向右平移 0.927 弧度。

This is much easier than plotting the original sum of two functions point by point.

这比逐点绘制两个函数之和的图像要容易得多。

---

10. Common Mistakes | 常见错误

Students often make errors when determining α, especially when a or b is negative.

学生在确定 α 时常犯错误，尤其是当 a 或 b 为负数时。

• Using tan^-1(b/a) without checking the quadrant. For example, if a = -3 and b = 4, the angle is in the second quadrant, not the first.
• Forgetting that R is always positive.
• Using degrees when the problem expects radians (or vice versa).
• Confusing the signs in the compound angle formulas.

• 不检查象限就直接使用 tan^-1(b/a)。例如，若 a = -3，b = 4，角度在第二象限，而不是第一象限。
• 忘记 R 始终为正。
• 题目要求弧度却使用角度（或反之）。
• 混淆复合角公式中的符号。

---

11. Practice Problems | 练习题

Try these problems to check your understanding.

尝试以下练习以检查你的理解。

1. Express 2 cos x + 5 sin x in the form R cos (x - α).
2. Express 6 cos x - 8 sin x in the form R sin (x - β).
3. Solve 4 cos x + 3 sin x = 1 for 0 ≤ x < 2π.
4. Find the maximum and minimum values of 12 cos x + 5 sin x.

1. 将 2 cos x + 5 sin x 表示为 R cos (x - α) 的形式。
2. 将 6 cos x - 8 sin x 表示为 R sin (x - β) 的形式。
3. 解 4 cos x + 3 sin x = 1，其中 0 ≤ x < 2π。
4. 求 12 cos x + 5 sin x 的最大值和最小值。

Answers:

答案：

1. √29 cos (x - 1.19)
2. 10 sin (x - 0.644) (since 6cos x - 8sin x = 10 sin (x - 0.644), check using compound angle formula)
3. x ≈ 1.29, 4.99
4. Max = 13, Min = −13

1. √29 cos (x - 1.19)
2. 10 sin (x - 0.644)（因为 6cos x - 8sin x = 10 sin (x - 0.644)，请用复合角公式验证）
3. x ≈ 1.29, 4.99
4. 最大值 = 13，最小值 = −13

---

12. Summary | 总结

To simplify a cos x ± b sin x:

化简 a cos x ± b sin x 的方法：

1. Identify a and b from the given expression.
2. Calculate R = √a² + b².
3. Determine α from tan α = b/a, respecting the quadrant.
4. Choose the appropriate form: R cos (x ∓ α) or R sin (x ± β).

1. 从给定表达式中识别 a 和 b。
2. 计算 R = √a² + b²。
3. 根据 tan α = b/a 确定 α，注意象限。
4. 选择合适的形式：R cos (x ∓ α) 或 R sin (x ± β)。

Mastering this technique is essential for A-Level Mathematics. It appears in trigonometry, differentiation, integration, and even mechanics problems. Practice it until the process becomes automatic.

掌握这一技巧对 A-Level 数学至关重要。它在三角学、微分、积分甚至力学问题中都会出现。请多加练习，直到该过程变得熟练自如。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）