Examples 6K: Trigonometric Equations & Identities | 例题6K：三角方程与恒等式

📚 Examples 6K: Trigonometric Equations & Identities | 例题6K：三角方程与恒等式

Section 6K in the AQA A-Level Mathematics textbook focuses on solving trigonometric equations using identities, including quadratic forms and double-angle substitutions. This is one of the most heavily examined skills in Pure Mathematics Paper 1 and Paper 2, and mastering it is essential for securing top marks.

在AQA A-Level数学教科书的6K部分中，我们重点学习如何运用恒等式解三角方程，包括二次形式和二倍角代入。这是纯数学试卷1和试卷2中最常考查的技能之一，掌握它对获得高分至关重要。

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1. Essential Identities You Must Know | 必须掌握的基本恒等式

n²θ = 2cos²θ − 1 = 1 − 2sin²θ. The third is the tangent definition: tan θ = sin θ ÷ cos θ.

在尝试本练习集中的任何方程之前，你需要熟练掌握三组恒等式。第一组是毕达哥拉斯恒等式：sin²θ + cos²θ = 1。第二组是二倍角恒等式：sin 2θ = 2 sin θ cos θ，以及 cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ。第三组是正切定义：tan θ = sin θ ÷ cos θ。

sin²θ + cos²θ = 1 tan θ = sin θ / cos θ sin 2θ = 2 sin θ cos θ

These identities allow you to rewrite an equation in terms of a single trigonometric function, which is the key first step in most 6K problems.

这些恒等式允许你将方程改写为只含一个三角函数的表达式，这是解决大多数6K问题的关键第一步。

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2. Solving Basic Equations of the Form sin θ = k | 解形如 sin θ = k 的基本方程

0° − α (or π − α) is also a solution. Continue adding or subtracting the period, 360° (or 2π), to generate further solutions.

当在给定区间内解 sin θ = k 时，你必须找出该区间内的所有解。使用反正弦函数求出主值，然后利用正弦曲线的对称性：如果 θ = α 是一个解，那么 θ = 180° − α（或 π − α）也是解。继续加上或减去周期 360°（或 2π）来生成更多解。

Worked Example 1 | 例题1： Solve 2 sin θ = 1 for 0° ≤ θ ≤ 360°.

sin θ = 1/2 → θ = 30°（主值）

al is only 360° wide, no further values from adding the period fit. Hence θ = 30° or 150°.

区间内的第二个解是 θ = 180° − 30° = 150°。由于区间宽度只有360°，加上周期后的值不再适用。因此 θ = 30° 或 150°。

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3. Solving Equations Involving tan θ | 解含 tan θ 的方程

t, and every other solution is obtained by adding multiples of the period: θ = α + n×180°.

正切函数的周期为 180°（或 π），不同于正弦和余弦的周期 360°（或 2π）。对于 tan θ = k，利用反正切求出主值 α，其他所有解均通过加上周期的整数倍获得：θ = α + n×180°。

Worked Example 2 | 例题2： Solve 5 sin θ − 2 cos θ = 0 for 0° ≤ θ ≤ 360°.

First, rearrange to isolate the ratio of sine to cosine:

首先，重新排列以分离正弦与余弦之比：

5 sin θ = 2 cos θ → sin θ / cos θ = 2/5 → tan θ = 0.4

θ = tan⁻¹(0.4) ≈ 21.8°

period 180°, the second solution is 21.8° + 180° = 201.8°. Therefore θ ≈ 21.8° or 201.8°.

由于正切的周期为180°，第二个解是 21.8° + 180° = 201.8°。因此 θ ≈ 21.8° 或 201.8°。

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4. Quadratic Trigonometric Equations | 二次三角方程

tion exactly as you would a variable in an ordinary quadratic equation: factorise or use the quadratic formula, then solve each resulting linear trigonometric equation separately.

许多6K问题给出的是单一三角函数上的二次方程，例如 2sin²θ + sin θ − 1 = 0。将三角函数视作普通二次方程中的变量来处理：因式分解或使用二次公式，然后分别解每个得到的一次三角方程。

Worked Example 3 | 例题3： Solve 2 sin²θ + sin θ − 1 = 0 for 0° ≤ θ ≤ 360°.

Factor the quadratic:

对二次式进行因式分解：

(2 sin θ − 1)(sin θ + 1) = 0

sin θ = 1/2 或 sin θ = −1

nly solution in the range is θ = 270°. Hence the full solution set is θ = 30°, 150°, 270°.

由 sin θ = 1/2，得 θ = 30° 或 150°。由 sin θ = −1，在区间内的唯一解是 θ = 270°。因此完整解集为 θ = 30°, 150°, 270°。

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5. Using sin²θ + cos²θ = 1 to Convert Equations | 利用 sin²θ + cos²θ = 1 转换方程

1 − sin²θ. This reduces the equation to a single quadratic in one trigonometric function.

当方程同时包含 sin²θ 与 cos θ，或 cos²θ 与 sin θ 时，毕达哥拉斯恒等式可以让你完全替换其中一个平方项。例如，sin²θ = 1 − cos²θ 或 cos²θ = 1 − sin²θ。这会将方程简化为关于一个三角函数的二次方程。

Worked Example 4 | 例题4： Solve 3 sin²θ − 5 cos θ − 1 = 0 for 0° ≤ θ ≤ 360°.

Replace sin²θ with 1 − cos²θ:

用 1 − cos²θ 替换 sin²θ：

3(1 − cos²θ) − 5 cos θ − 1 = 0

3 − 3cos²θ − 5 cos θ − 1 = 0

3cos²θ + 5 cos θ − 2 = 0

1. Therefore θ = cos⁻¹(1/3) ≈ 70.5°, and the symmetric solution θ = 360° − 70.5° = 289.5°.

因式分解：(3 cos θ − 1)(cos θ + 2) = 0。因此 cos θ = 1/3 或 cos θ = −2。第二个选项无效，因为 cos θ 的取值范围始终在 −1 和 1 之间。因此 θ = cos⁻¹(1/3) ≈ 70.5°，以及对称解 θ = 360° − 70.5° = 289.5°。

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6. Equations Involving Double Angles | 涉及二倍角的方程

u first before finding θ. The second method is often cleaner when the argument is exactly a multiple of the variable.

当方程包含 sin 2θ 或 cos 2θ 时，你有两种策略：要么利用第1节的恒等式展开二倍角，要么直接令 u = 2θ，先解出 u 再求 θ。当参数恰为变量的倍数时，第二种方法通常更简洁。

Worked Example 5 | 例题5： Solve cos 2θ + 3 cos θ + 2 = 0 for 0° ≤ θ ≤ 360°.

Use the identity cos 2θ = 2cos²θ − 1:

使用恒等式 cos 2θ = 2cos²θ − 1：

2cos²θ − 1 + 3 cos θ + 2 = 0

2cos²θ + 3 cos θ + 1 = 0

(2 cos θ + 1)(cos θ + 1) = 0

° and θ = 240°. From cos θ = −1, the solution is θ = 180°. Therefore θ = 120°, 180°, 240°.

因此 cos θ = −1/2 或 cos θ = −1。由 cos θ = −1/2，解为 θ = 120° 和 θ = 240°。由 cos θ = −1，解为 θ = 180°。因此 θ = 120°, 180°, 240°。

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7. Equations with Compound Angles | 复合角方程

s to let X equal the entire argument, solve for X across the appropriately stretched interval, then convert back to θ at the end. This avoids sign errors and ensures no solutions are lost.

AQA考官经常考查形如 sin(2θ − 30°) = k 或 cos(3θ + π/4) = k 的方程。关键技巧是令 X 等于整个角参数，在适当扩展的区间内解出 X，最后再转换回 θ。这样可以避免符号错误并确保不遗漏解。

Worked Example 6 | 例题6： Solve sin(2θ − 30°) = √3/2 for 0° ≤ θ ≤ 360°.

Let X = 2θ − 30°. Since 0° ≤ θ ≤ 360°, multiply by 2 and subtract 30°:

令 X = 2θ − 30°。因为 0° ≤ θ ≤ 360°，乘以2再减去30°：

−30° ≤ X ≤ 690°

Now solve sin X = √3 Adding the period gives X = 60°, 120°, 420°, 480°. Now convert back using θ = (X + 30°)/2:

现在在 −30° ≤ X ≤ 690° 上解 sin X = √3/2。主解为 X = 60°。加上周期得 X = 60°, 120°, 420°, 480°。然后用 θ = (X + 30°)/2 转换回去：

X	θ = (X + 30°) ÷ 2
60°	45°
120°	75°
420°	225°
480°	255°

5°, 75°, 225°, 255°. Always check that each converted value lies in the original interval.

因此解为 θ = 45°, 75°, 225°, 255°。务必检查每个转换后的值是否落在原区间内。

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8. Equations Requiring Double-Angle Expansion | 需要展开二倍角的方程

or out the common term. This converts the problem into a product of two simpler equations.

有些方程将 sin 2θ 与 sin θ 或 cos θ 直接结合。在这些情况下，展开 sin 2θ = 2 sin θ cos θ，然后提出公因子。这会将问题转化为两个更简单的方程相乘。

Worked Example 7 | 例题7： Solve sin 2θ = sin θ for 0° ≤ θ ≤ 360°.

2 sin θ cos θ = sin θ

2 sin θ cos θ − sin θ = 0

sin θ (2 cos θ − 1) = 0

θ = 1/2, we get θ = 60°, 300°. The complete solution set is θ = 0°, 60°, 180°, 300°, 360°.

因此 sin θ = 0 或 cos θ = 1/2。由 sin θ = 0，得 θ = 0°, 180°, 360°。由 cos θ = 1/2，得 θ = 60°, 300°。完整解集为 θ = 0°, 60°, 180°, 300°, 360°。

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9. Exam-Style Problem: Using Multiple Identities | 考试风格题：综合运用多个恒等式

The most demanding 6K questions require you to select the correct identity from memory and apply it in sequence. Consider the following typical exam question worth 6 marks.

最难的6K问题需要你从记忆中选出正确的恒等式并按顺序应用。请看下面这道典型的6分考试题。

Worked Example 8 | 例题8： Solve 4 sin θ cos θ = √3 for 0° ≤ θ ≤ 180°.

Recognise that 4 sin θ cos θ = 2(2 sin θ cos θ) = 2 sin 2θ. This transforms the equation:

注意 4 sin θ cos θ = 2(2 sin θ cos θ) = 2 sin 2θ。这可以将方程变形为：

2 sin 2θ = √3 → sin 2θ = √3/2

Let X = 2θ. Since 0° ≤ θ ≤ 180°, we have 0° ≤ X ≤ 360°. Solving sin X = √3/2 gives X = 60° or X = 120°. Converting back, θ = X/2, so θ = 30° or 60°.

令 X = 2θ。因为 0° ≤ θ ≤ 180°，所以 0° ≤ X ≤ 360°。解 sin X = √3/2 得 X = 60° 或 X = 120°。转换回去，θ = X/2，所以 θ = 30° 或 60°。

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10. Common Mistakes and How to Avoid Them | 常见错误及如何避免

alid solution; and failing to adjust the interval when substituting X = 2θ or X = θ − 30°.

学生在这一主题上失分的方式是可预测的。最常见的错误包括：忘记正弦或余弦方程中的第二个解（对称解）；两边同时除以可能为零的三角函数，例如在 sin θ = 0 也是有效解时除以 sin θ；以及在代入 X = 2θ 或 X = θ − 30° 后未能调整区间范围。

• Always | 始终：Write down the stretched interval before solving for X.

  在求解 X 之前，先写出扩展后的区间范围。

• Never | 切勿：Cancel a factor of sin θ or cos θ unless you have checked the zeros separately.

  除非已单独检查过零点，否则不要约去 sin θ 或 cos θ 因子。

• Always | 始终：Verify that every final answer lies within the original interval stated in the question.

  验证每个最终答案都落在题目给定的原始区间内。

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11. Summary of the 6K Method | 6K解题方法总结

The general strategy for every Example 6K problem can be summarised in four steps. Step 1: Use identities to rewrite the equation

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