📚 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
ide 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。
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。
e 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。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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