Example 6.12.1 – Solving a sin θ + b cos θ = c | 例题6.12.1 – 解 a sin θ + b cos θ = c 型方程

📚 Example 6.12.1 – Solving a sin θ + b cos θ = c | 例题6.12.1 – 解 a sin θ + b cos θ = c 型方程

In this example, we will solve a trigonometric equation of the form a sin θ + b cos θ = c by rewriting the left-hand side as a single sine function using the R-method. This technique is central to AQA A-Level Mathematics, particularly in Paper 2.

在本例题中，我们将通过 R 方法将形如 a sin θ + b cos θ = c 的三角方程左侧改写为单一正弦函数来求解。该技巧是 AQA A-Level 数学的核心内容，尤其在 Paper 2 中非常重要。

1. The Problem | 题目

Solve, for 0° ≤ θ < 360°, the equation

求解方程，其中 0° ≤ θ < 360°：

3 sin θ + 4 cos θ = 2

Give your answers correct to one decimal place.

请将答案精确到一位小数。

2. The Key Idea – The R-Method | 关键思路 – R 方法

Any expression of the form a sin θ + b cos θ can be written as R sin(θ + α), where R = √(a² + b²) and α is an acute angle satisfying certain conditions.

任何形如 a sin θ + b cos θ 的表达式都可以写成 R sin(θ + α) 的形式，其中 R = √(a² + b²)，α 是一个满足特定条件的锐角。

We expand R sin(θ + α) using the compound angle formula:

我们使用复合角公式展开 R sin(θ + α)：

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

Comparing coefficients with a sin θ + b cos θ gives:

与 a sin θ + b cos θ 比较系数可得：

R cos α = a, R sin α = b

Therefore R = √(a² + b²) and tan α = b / a.

因此 R = √(a² + b²) 且 tan α = b / a。

3. Step 1 – Find R and α | 第一步 – 求 R 和 α

For the equation 3 sin θ + 4 cos θ = 2, we have a = 3 and b = 4.

对于方程 3 sin θ + 4 cos θ = 2，我们有 a = 3，b = 4。

First calculate R:

首先计算 R：

R = √(3² + 4²) = √(9 + 16) = √25 = 5

Next find α using tan α = b / a:

然后使用 tan α = b / a 求 α：

tan α = 4 / 3

α = tan⁻¹(4 / 3) ≈ 53.13°

Since a > 0 and b > 0, α is indeed the acute angle in the first quadrant, so no adjustment is needed.

因为 a > 0 且 b > 0，α 确实是第一象限的锐角，因此无需调整。

4. Step 2 – Rewrite the Equation | 第二步 – 重写方程

Substitute R and α into the original equation:

将 R 和 α 代入原方程：

5 sin(θ + 53.13°) = 2

Divide both sides by 5:

两边同时除以 5：

sin(θ + 53.13°) = 2 / 5 = 0.4

Now the equation has been reduced to a basic sine equation, which we can solve directly.

现在方程已经化简为一个基本正弦方程，可以直接求解。

5. Step 3 – Solve sin X = 0.4 | 第三步 – 解 sin X = 0.4

Let X = θ + 53.13°. First find the principal solution:

令 X = θ + 53.13°。首先求主解：

X = sin⁻¹(0.4) ≈ 23.58°

For 0° ≤ θ < 360°, we have:

因为 0° ≤ θ < 360°，所以：

53.13° ≤ X < 413.13°

Within this interval, sin X = 0.4 has two solutions:

在此区间内，sin X = 0.4 有两个解：

X₁ ≈ 23.58° (the principal solution, which lies inside the interval)

X₁ ≈ 23.58°（主解，位于区间内）
X₂ ≈ 180° − 23.58° = 156.42° (sine is positive in the second quadrant)

X₂ ≈ 180° − 23.58° = 156.42°（正弦在第二象限为正）

6. Step 4 – Convert Back to θ | 第四步 – 转换回 θ

Recall that X = θ + 53.13°, so θ = X − 53.13°.

注意 X = θ + 53.13°，因此 θ = X − 53.13°。

For X₁ ≈ 23.58°:

对于 X₁ ≈ 23.58°：

θ = 23.58° − 53.13° = −29.55°

This value is less than 0°, so we add 360°:

该值小于 0°，因此加上 360°：

θ = −29.55° + 360° = 330.45° ≈ 330.5°

For X₂ ≈ 156.42°:

对于 X₂ ≈ 156.42°：

θ = 156.42° − 53.13° = 103.29° ≈ 103.3°

Both values lie within the required range 0° ≤ θ < 360°.

两个值都在要求的范围 0° ≤ θ < 360° 内。

7. Step 5 – Verify the Solutions | 第五步 – 验证解

Always check your solutions by substituting back into the original equation.

务必通过代入原方程来检查你的解。

For θ ≈ 330.5°:

对于 θ ≈ 330.5°：

3 sin 330.5° + 4 cos 330.5° ≈ 3(−0.4924) + 4(0.8704) ≈ −1.477 + 3.482 ≈ 2.005

For θ ≈ 103.3°:

对于 θ ≈ 103.3°：

3 sin 103.3° + 4 cos 103.3° ≈ 3(0.9730) + 4(−0.2306) ≈ 2.919 − 0.922 ≈ 1.997

Both results are close to 2, confirming that the solutions are correct within rounding error.

两个结果都接近 2，确认了解在四舍五入误差范围内是正确的。

8. Using a Different Form – R sin(θ − α) | 使用不同形式 – R sin(θ − α)

Sometimes it is more convenient to write 3 sin θ + 4 cos θ as R cos(θ − α) instead. Both approaches are valid, but the choice affects the solving steps.

有时将 3 sin θ + 4 cos θ 写成 R cos(θ − α) 更方便。两种方法都有效，但选择会影响后续求解步骤。

If we write:

如果我们写成：

3 sin θ + 4 cos θ = 5 cos(θ − α)

Then cos α = 4 / 5 and sin α = 3 / 5, so α = tan⁻¹(3 / 4) ≈ 36.87°. The equation becomes:

则 cos α = 4 / 5 且 sin α = 3 / 5，所以 α = tan⁻¹(3 / 4) ≈ 36.87°。方程变为：

5 cos(θ − 36.87°) = 2

cos(θ − 36.87°) = 0.4

This gives θ − 36.87° = ±66.42° + 360°n, and solving yields the same final answers.

这给出 θ − 36.87° = ±66.42° + 360°n，求解后得到相同的最终答案。

9. Common Mistakes | 常见错误

Students often make errors in this type of question. Here are the most frequent pitfalls.

学生在此类题目中常犯错误。以下是最常见的陷阱。

Forgetting that the interval for X changes when substituting X = θ + α. Always adjust the domain before finding solutions.

忘记代入 X = θ + α 后 X 的区间会变化。务必在求解除之前先调整定义域。
Using tan⁻¹(b / a) without checking which quadrant α lies in. If a or b is negative, α may not be acute.

不检查 α 所在象限就使用 tan⁻¹(b / a)。如果 a 或 b 为负，α 可能不是锐角。
Adding 360° incorrectly when θ is negative. Instead, add 360° once and check whether the result lies in the required range.

当 θ 为负时错误地加 360°。应该加一次 360°，然后检查结果是否在所需范围内。
Forgetting to find the second solution in the correct quadrant. For sin X = k, the second solution is 180° − X.

忘记在正确象限内找到第二个解。对于 sin X = k，第二个解是 180° − X。

10. Practice Questions | 练习题

Try these questions on your own to consolidate the method.

尝试独立完成以下练习题以巩固方法。

Solve 5 sin θ + 12 cos θ = 3 for 0° ≤ θ < 360°.

求解 5 sin θ + 12 cos θ = 3，其中 0° ≤ θ < 360°。
Solve 2 sin θ − 5 cos θ = 1 for 0° ≤ θ < 360°.

求解 2 sin θ − 5 cos θ = 1，其中 0° ≤ θ < 360°。
Find the maximum and minimum values of 3 sin θ + 4 cos θ, and the smallest positive θ at which each occurs.

求 3 sin θ + 4 cos θ 的最大值和最小值，以及各自取得时最小的正 θ。

11. Summary | 总结

The R-method is a powerful tool for solving equations of the form a sin θ + b cos θ = c.

R 方法是求解 a sin θ + b cos θ = c 型方程的强有力工具。

Compute R = √(a² + b²) and find α using tan α = b / a.

计算 R = √(a² + b²)，并使用 tan α = b / a 求 α。
Rewrite the equation as R sin(θ + α) = c, then divide by R.

将方程改写为 R sin(θ + α) = c，然后除以 R。
Adjust the interval for X = θ + α before solving.

在求解之前调整 X = θ + α 的区间。
Find all solutions in the adjusted interval, then convert back to θ.

在调整后的区间内找到所有解，然后转换回 θ。
Always verify your final answers by substitution.

始终通过代入来验证你的最终答案。

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Example 6.12.1 – Solving a sin θ + b cos θ = c | 例题6.12.1 – 解 a sin θ + b cos θ = c 型方程