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 \(a \cos x \pm b \sin x\) 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 \(a \cos x \pm b \sin x\) 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 \(R \cos (x – \alpha)\) or \(R \sin (x + \alpha)\), where \(R > 0\) and \(\alpha\) is a constant angle.

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

\(R\) is the amplitude (always positive).
\(\alpha\) 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）。
\(\alpha\) 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。

\(a \cos x + b \sin x = R \cos (x – \alpha)\)

2. Finding R | 求 R 的值

Using the compound angle formula for cosine:

利用余弦的复合角公式：

\(\cos (x – \alpha) = \cos x \cos \alpha + \sin x \sin \alpha\)

Therefore:

因此：

\(R \cos (x – \alpha) = R \cos \alpha \cos x + R \sin \alpha \sin x\)

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

与 \(a \cos x + b \sin x\) 比较，得到：

\(R \cos \alpha = a\), \(R \sin \alpha = b\)

Squaring and adding these two equations gives:

将两式平方后相加，得到：

\(R^2 (\cos^2 \alpha + \sin^2 \alpha) = a^2 + b^2\)

Since \(\cos^2 \alpha + \sin^2 \alpha = 1\), we have:

由于 \(\cos^2 \alpha + \sin^2 \alpha = 1\)，因此：

\(R = \sqrt{a^2 + b^2}\)

3. Finding α | 求 α 的值

From the same comparison, we have:

由同样的比较，可得：

\(\tan \alpha = \frac{b}{a}\)

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

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

Sign of \(a\)	Sign of \(b\)	Quadrant of \(\alpha\)
+	+	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 \(\alpha = \tan^{-1}(b/a)\) 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 \(\alpha = \tan^{-1}(b/a)\)。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 \(a \cos x + b \sin x\)：

\(R \cos (x – \alpha)\)

\(R \sin (x + \beta)\), where \(\beta = \frac{\pi}{2} – \alpha\)

For \(a \cos x – b \sin x\):

对于 \(a \cos x – b \sin x\)：

\(R \cos (x + \alpha)\)

\(R \sin (x – \beta)\)

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 \(R \cos (x – \alpha)\), where \(R > 0\) and \(0 < \alpha < \pi/2\).

将 \(3 \cos x + 4 \sin x\) 表示为 \(R \cos (x – \alpha)\) 的形式，其中 \(R > 0\)，\(0 < \alpha < \pi/2\)。

Here \(a = 3\), \(b = 4\).

这里 \(a = 3\)，\(b = 4\)。

\(R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

Now \(\tan \alpha = \frac{4}{3}\). Since \(a > 0, b > 0\), \(\alpha\) is in the first quadrant.

现在 \(\tan \alpha = \frac{4}{3}\)。因为 \(a > 0, b > 0\)，所以 \(\alpha\) 在第一象限。

\(\alpha = \tan^{-1}\left(\frac{4}{3}\right) \approx 0.927\) 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 \(R \cos (x + \alpha)\), where \(R > 0\) and \(0 < \alpha < \pi/2\).

将 \(5 \cos x – 12 \sin x\) 表示为 \(R \cos (x + \alpha)\) 的形式，其中 \(R > 0\)，\(0 < \alpha < \pi/2\)。

We use the identity:

我们使用恒等式：

\(\cos (x + \alpha) = \cos x \cos \alpha – \sin x \sin \alpha\)

Thus:

因此：

\(R \cos (x + \alpha) = R \cos \alpha \cos x – R \sin \alpha \sin x\)

Comparing with \(5 \cos x – 12 \sin x\):

与 \(5 \cos x – 12 \sin x\) 比较：

\(R \cos \alpha = 5\), \(R \sin \alpha = 12\)

So \(R = \sqrt{5^2 + 12^2} = 13\) and \(\tan \alpha = \frac{12}{5}\). Since both are positive, \(\alpha = \tan^{-1}(12/5) \approx 1.176\) radians.

所以 \(R = \sqrt{5^2 + 12^2} = 13\)，且 \(\tan \alpha = \frac{12}{5}\)。由于两者均为正，\(\alpha = \tan^{-1}(12/5) \approx 1.176\) 弧度。

Hence:

因此：

\(5 \cos x – 12 \sin x = 13 \cos (x + 1.176)\)

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 \(0 \le x < 2\pi\), giving answers to 3 significant figures.

例：解 \(3 \cos x + 4 \sin x = 2\)，其中 \(0 \le x < 2\pi\)，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 \(\theta = x – 0.927\). Then \(\cos \theta = 0.4\).

令 \(\theta = x – 0.927\)，则 \(\cos \theta = 0.4\)。

The principal value is \(\theta = \cos^{-1}(0.4) \approx 1.159\). The general solutions are \(\theta = 2\pi n \pm 1.159\).

主值为 \(\theta = \cos^{-1}(0.4) \approx 1.159\)。通解为 \(\theta = 2\pi n \pm 1.159\)。

\(\theta = 1.159\) ⇒ \(x = 1.159 + 0.927 = 2.086\)
\(\theta = 2\pi – 1.159 = 5.124\) ⇒ \(x = 5.124 + 0.927 = 6.051\)

\(\theta = 1.159\) ⇒ \(x = 1.159 + 0.927 = 2.086\)
\(\theta = 2\pi – 1.159 = 5.124\) ⇒ \(x = 5.124 + 0.927 = 6.051\)

Both solutions lie in the interval \(0 \le x < 2\pi\). So \(x \approx 2.09\) and \(x \approx 6.05\) (3 s.f.).

两个解都在区间 \(0 \le x < 2\pi\) 内。所以 \(x \approx 2.09\) 和 \(x \approx 6.05\)（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. \(x = 0.927 + 2\pi n\).
Minimum value = \(-R = -5\), occurring when \(\cos (x – 0.927) = -1\), i.e. \(x = 0.927 + \pi + 2\pi n\).

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

In general, the maximum value of \(a \cos x + b \sin x\) is \(\sqrt{a^2 + b^2}\) and the minimum value is \(-\sqrt{a^2 + b^2}\).

一般地，\(a \cos x + b \sin x\) 的最大值为 \(\sqrt{a^2 + b^2}\)，最小值为 \(-\sqrt{a^2 + b^2}\)。

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 = \(2\pi\)
Phase shift: the graph is shifted right by 0.927 radians relative to \(y = 5 \cos x\).

振幅 = 5
周期 = \(2\pi\)
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 \(\alpha\), especially when \(a\) or \(b\) is negative.

学生在确定 \(\alpha\) 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。

Express \(2 \cos x + 5 \sin x\) in the form \(R \cos (x – \alpha)\).
Express \(6 \cos x – 8 \sin x\) in the form \(R \sin (x – \beta)\).
Solve \(4 \cos x + 3 \sin x = 1\) for \(0 \le x < 2\pi\).
Find the maximum and minimum values of \(12 \cos x + 5 \sin x\).

将 \(2 \cos x + 5 \sin x\) 表示为 \(R \cos (x – \alpha)\) 的形式。
将 \(6 \cos x – 8 \sin x\) 表示为 \(R \sin (x – \beta)\) 的形式。
解 \(4 \cos x + 3 \sin x = 1\)，其中 \(0 \le x < 2\pi\)。
求 \(12 \cos x + 5 \sin x\) 的最大值和最小值。

Answers:

答案：

\(\sqrt{29} \cos (x – 1.19)\)
\(10 \sin (x – 0.644)\) (since \(6\cos x – 8\sin x = 10 \sin (x – 0.644)\), check using compound angle formula)
\(x \approx 1.29, 4.99\)
Max = 13, Min = −13

\(\sqrt{29} \cos (x – 1.19)\)
\(10 \sin (x – 0.644)\)（因为 \(6\cos x – 8\sin x = 10 \sin (x – 0.644)\)，请用复合角公式验证）
\(x \approx 1.29, 4.99\)
最大值 = 13，最小值 = −13

12. Summary | 总结

To simplify \(a \cos x \pm b \sin x\):

化简 \(a \cos x \pm b \sin x\) 的方法：

Identify \(a\) and \(b\) from the given expression.
Calculate \(R = \sqrt{a^2 + b^2}\).
Determine \(\alpha\) from \(\tan \alpha = b/a\), respecting the quadrant.
Choose the appropriate form: \(R \cos (x \mp \alpha)\) or \(R \sin (x \pm \beta)\).

从给定表达式中识别 \(a\) 和 \(b\)。
计算 \(R = \sqrt{a^2 + b^2}\)。
根据 \(\tan \alpha = b/a\) 确定 \(\alpha\)，注意象限。
选择合适的形式：\(R \cos (x \mp \alpha)\) 或 \(R \sin (x \pm \beta)\)。

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。

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