📚 Solving Trigonometric Equation: sin 2θ = cos θ | 求解三角方程：sin 2θ = cos θ
t of the Edexcel A-Level Pure Mathematics syllabus. This article provides a detailed, step‑by‑step solution to a common exam‑ ind each step, highlight common pitfalls, and reinforce key identities that you must know for the exam.
三角方程是Edexcel A-Level纯数课程的核心内容。本文详细讲解一道常见考题：在 0 ≤ θ ≤ 2π 范围内求解 sin 2θ = cos θ。我们将逐步展示解题思路，强调易错点，并巩固考试必备的重要恒等式。
1. Understanding the Equation | 理解方程
ouble angle on the left and a single angle on the right. To solve it, we need to express both sides in terms of the same trigonometric ratio, allowing us to factorise or use standard solutions.
方程 sin 2θ = cos θ 左边含有二倍角，右边是单角。要解这个方程，我们需要把两边写成同一种三角比的形式，从而进行因式分解或使用标准解。
t would lose solutions where cos θ = 0. We must avoid division by a term that can be zero.
我们可能想两边同除以 cos θ，但这样会丢失 cos θ = 0 的解。必须避免除以可能为零的项。
2. Double‑Angle Identity for sin 2θ | 正弦的二倍角公式
The key identity from the Edexcel formula booklet is:
Edexcel 公式表给出的关键恒等式是：
sin 2θ ≡ 2 sin θ cos θ
This allows us to rewrite the left‑hand side in terms of sin θ and cos θ, matching the cosine term on the right.
这样一来，左边就可以用 sin θ 和 cos θ 表示，与右边的 cos θ 相匹配。
Substituting gives:
代入后得到：
2 sin θ cos θ = cos θ
3. Rearranging to Factorise | 移项与因式分解
Bring all terms to one side so we can factorise:
将所有项移到一边，以便分解因式：
2 sin θ cos θ − cos θ = 0
Factor out the common factor cos θ:
提取公因式 cos θ：
cos θ (2 sin θ − 1) = 0
Now we have a product equal to zero, which gives two separate equations to solve.
现在我们得到乘积等于零的形式，于是可以转化为两个独立的方程求解。
4. Solving the First Factor: cos θ = 0 | 求解第一个因式：cos θ = 0
We need to find all values of θ in the interval 0 ≤ θ ≤ 2π for which cos θ = 0.
我们需要找出在 0 ≤ θ ≤ 2π 区间内使得 cos θ = 0 的所有 θ 值。
From the unit circle or cosine graph, cosine is zero at odd multiples of π/2. Within the given domain, the solutions are:
根据单位圆或余弦图像，余弦在 π/2 的奇数倍处为零。在给定范围内，解为：
- θ = π/2
- θ = 3π/2
5. Solving the Second Factor: 2 sin θ − 1 = 0 | 求解第二个因式：2 sin θ − 1 = 0
Rearrange to isolate sin θ:
整理方程，求出 sin θ：
2 sin θ = 1 ⇒ sin θ = ½
in θ = ½ corresponds to the principal values θ = π/6 and θ = 5π/6 in the range 0 ≤ θ ≤ 2π.
我们知道 sin θ = ½ 在 0 到 2π 范围内的主值为 θ = π/6 和 θ = 5π/6。
6. Verifying the Domain and Quadrants | 验证定义域与象限
The interval 0 ≤ θ ≤ 2π includes all four quadrants. Let us double‑check that all our solutions lie within this closed interval.
区间 0 ≤ θ ≤ 2π 包含全部四个象限。我们来确认所得的解都落在这个闭区间内。
- π/2 ≈ 1.571
- 3π/2 ≈ 4.712
- π/6 ≈ 0.524
- 5π/6 ≈ 2.618
All values are between 0 and 2π inclusive, so they are valid.
这些值都在 0 到 2π（含）之间，因此是有效的。
We also note that sin θ and cos θ are defined for all real θ, so there are no restrictions other than the domain given.
我们还注意到 sin θ 和 cos θ 对所有实数有定义，因此除了题目给定的范围外没有其他限制。
7. Listing All Solutions | 列出所有解
Collecting the results from both factors, the complete solution set for 0 ≤ θ ≤ 2π is:
综合两个因式的结果，在 0 ≤ θ ≤ 2π 内的全部解为：
θ = π/6, π/2, 5π/6, 3π/2
It is good practice to write solutions in ascending order and to include units (radians are assumed for A‑Level examinations unless degrees are specified).
建议按从小到大的顺序书写答案，并标明单位（除非特别说明，A‑Level 考试中默认使用弧度）。
8. Graphical Interpretation (Optional) | 图形解释（选读）
same axes for 0 ≤ θ ≤ 2π, the intersection points occur exactly at the solutions we found.
如果在同一坐标系中画出 y = sin 2θ 和 y = cos θ 在 0 ≤ θ ≤ 2π 的图像，它们的交点正好就是我们找到的解。
This visual check confirms there are no missed solutions. The periodicity of the functions would normally produce infinitely many solutions, but we restrict to the given interval.
这种可视化检验可以确保没有遗漏解。虽然函数的周期性通常会产生无穷多解，但题目限定了区间。
9. Common Mistakes and How to Avoid Them | 常见错误与避免方法
Dividing by cos θ: olutions where cos θ = 0 (π/2 and 3π/2). Always bring all terms to one side and factorise.
除以 cos θ：若将原方程两边除以 cos θ，会丢失 cos θ = 0 的解（π/2 和 3π/2）。切记将所有项移到一边再因式分解。
Forgetting the second sine solution: = ½ also gives 5π/6. Use the CAST diagram or graph to find all values in the given range.
忘记正弦的第二个解：很多学生能写出 π/6，却忘了 sin θ = ½ 还会给出 5π/6。利用 CAST 图或函数图像获取区间内的所有值。
Incorrect double‑angle identity: Some confuse sin 2θ with sin²θ or second derivatives. Memorise sin 2θ ≡ 2 sin θ cos θ from the formula booklet.
二倍角公式用错：有人会把 sin 2θ 错记为 sin²θ 或二阶导数。务必记住公式表中的 sin 2θ ≡ 2 sin θ cos θ。
10. Check Using Exact Values | 用精确值验算
Substitute each solution back into the original equation to verify:
将每个解代回原方程进行验证：
| θ | sin 2θ | cos θ | Equal? |
|---|---|---|---|
| π/6 | sin(π/3) = √3/2 | √3/2 | ✓ |
| π/2 | sin π = 0 | 0 | ✓ |
| 5π/6 | sin(5π/3) = −√3/2 | −√3/2 | ✓ |
| 3π/2 | sin 3π = 0 | 0 | ✓ |
The table confirms that each solution satisfies the equation, so our answer is correct.
表格验证了每个解都满足方程，证明答案正确。
11. Extension: Using cos θ as sin(π/2 − θ) | 拓展：利用 cos θ = sin(π/2 − θ)
An alternative approach uses the complementary angle identity: cos θ ≡ sin(π/2 − θ). Then the equation becomes:
另一种解法利用余角恒等式：cos θ ≡ sin(π/2 − θ)。方程变为：
sin 2θ = sin(π/2 − θ)
torisation method shown earlier is simpler and less prone to errors with periodic solutions.
这样需求解 sin A = sin B，要考虑 A = B + 2kπ 和 A = π − B + 2kπ。此法可行，但前述因式分解法更简单，且不易在处理周期解时出错。
12. Summary and Exam Tips | 总结与应试技巧
To solve trigonometric equations like sin 2θ = cos θ:
解 sin 2θ = cos θ 这类三角方程的要点是：
- Use the double‑angle identity to unify the ratios.
- 使用二倍角恒等式统一三角比。
- Bring all terms to one side and factorise – never divide by a trig function that could be zero.
- 将所有项移到一边并因式分解——绝不要除以可能为零的三角函数。
- Solve each resulting equation within the specified interval.
- 在指定区间内分别求解各个方程。
- Check for all possible angles using symmetry or CAST.
- 利用对称性或 CAST 图求出所有可能的角度。
- Present final answers clearly and in ascending order.
- 清晰、按升序列写出最终答案。
Mastering this technique will prepare you for similar questions on the Edexcel Pure Mathematics paper, whether they involve sin, cos, tan or combinations with other trigonometric identities.
掌握这一方法将使你有能力应对 Edexcel 纯数试卷中类似的题目，无论题目涉及 sin、cos、tan 还是与其他三角恒等式的组合。
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