📚 Solving Trigonometric Identities: Example 6.9.1 Explained | 三角恒等式求解与证明：Example 6.9.1 详解
In AQA A-Level Mathematics, proving trigonometric identities is a core skill that frequently appears in Pure Mathematics exams. It tests your algebraic manipulation, fluency with foundational formulas, and logical structuring. Example 6.9.1 provides a classic illustration of how a complex-looking fractional identity can be proven elegantly using double-angle formulas.
在 AQA A-Level 数学考试中，证明三角恒等式是一项核心技能，经常出现在纯数学试卷中。它考察你的代数化简能力、对基础公式的熟练程度以及逻辑构建能力。Example 6.9.1 是一个经典例题，完美展示了如何使用二倍角公式，优雅地证明一个看似复杂的分数型恒等式。
1. Understanding the Identity | 理解恒等式
An identity is an equation that holds true for all values of the variable within its domain. It is denoted by the triple bar symbol (≡) rather than the ordinary equals sign (=). For example, the statement (1 − cos 2x) / (1 + cos 2x) ≡ tan² x asserts that no matter what value of x you substitute (excluding points where the denominator is zero), the left-hand side (LHS) will always exactly equal the right-hand side (RHS).
恒等式是指在其定义域内对所有变量值都成立的等式。它使用三重横线符号（≡）而不是普通的等号（=）来表示。例如，恒等式 (1 − cos 2x) / (1 + cos 2x) ≡ tan² x 表明，无论你代入任何 x 值（除非分母为零），等号左边 (LHS) 的值始终完全等于等号右边 (RHS) 的值。
The key difference between an equation and an identity is that an equation is only true for specific values (e.g., x = 30°), whereas an identity is universally true. When asked to “prove” an identity, you must demonstrate that one side can be transformed into the other using established mathematical rules, rather than solving for a specific value.
等式与恒等式的关键区别在于，等式仅对特定值成立（例如 x = 30°），而恒等式则普遍成立。当要求“证明”一个恒等式时，你必须展示一边如何利用既定的数学规则转化为另一边，而不是求解一个特定的值。
2. Recalling Core Trigonometric Identities | 回顾核心三角恒等式
To tackle this proof, you must have a ready command of the double-angle formulas. Specifically, the cosine double-angle formula has three equivalent forms, and choosing the correct one is the secret to solving this problem efficiently:
要完成这个证明，你必须熟练掌握二倍角公式。具体来说，余弦的二倍角公式有三种等价形式，而选择正确的形式是高效解决此题的关键：
- cos 2x ≡ cos² x − sin² x (The standard form)
- cos 2x ≡ 1 − 2 sin² x (Used when simplifying expressions with 1 − cos 2x)
- cos 2x ≡ 2 cos² x − 1 (Used when simplifying expressions with 1 + cos 2x)
Additionally, you must be comfortable with the fundamental identity for tangent: tan x ≡ sin x / cos x. The ability to switch between these forms fluently is an explicit requirement of the AQA specification.
此外，你必须熟悉正切函数的基本恒等式：tan x ≡ sin x / cos x。能够流畅地在这些形式之间切换，是 AQA 考纲的明确要求。
3. The Given Example 6.9.1 | 题目 Example 6.9.1
Let us examine the exact problem presented in this section. The question asks us to prove the following trigonometric identity:
让我们来看这一节给出的具体问题。题目要求我们证明以下三角恒等式：
(1 − cos 2x) / (1 + cos 2x) ≡ tan² x
This type of question is worth 4-5 marks in a typical AQA Pure Mathematics paper. The examiner is not just looking for the correct algebra; they are looking for a clear, logical progression from one side of the identity to the other, with all necessary steps explicitly shown.
此类问题在典型的 AQA 纯数学试卷中通常占 4-5 分。阅卷者不仅看代数运算是否正确，更看重是否有一条清晰、逻辑严密地从等式一边推导到另一边的过程，并且所有必要步骤都已明确写出。
4. Strategic Approach | 解题策略
When proving an identity, you generally choose the more complicated side to manipulate, aiming to simplify it down to the simpler side. Here, the LHS (1 − cos 2x) / (1 + cos 2x) is clearly more complex than the RHS tan² x. Therefore, we will exclusively manipulate the LHS.
在证明恒等式时，我们通常选择更为复杂的一边进行化简，目标是将其简化成较简单的一边。在这里，左边 (1 − cos 2x) / (1 + cos 2x) 显然比右边 tan² x 更复杂。因此，我们将只对左边进行变形式处理。
The critical strategic insight is to notice the “1 ∓ cos 2x” structure. The presence of the constant 1 and the double angle 2x strongly suggests using the modified double-angle formulas. The numerator contains “1 − cos 2x”, which pairs perfectly with cos 2x ≡ 1 − 2 sin² x. Conversely, the denominator contains “1 + cos 2x”, which pairs perfectly with cos 2x ≡ 2 cos² x − 1.
关键的策略观察点是“1 ∓ cos 2x”这个结构。常数 1 和倍角 2x 的同时出现，强烈暗示我们要使用变形的二倍角公式。分子包含“1 − cos 2x”，它完美对应公式 cos 2x ≡ 1 − 2 sin² x。相反，分母包含“1 + cos 2x”，它完美对应公式 cos 2x ≡ 2 cos² x − 1。
5. Step-by-Step Solution (Part 1) | 分步解析（第一部分）
Let us begin the proof by focusing entirely on the Left-Hand Side. We write down the LHS and apply the relevant substitutions.
我们开始进行证明，完全专注于等式的左边。先写下左边，然后代入相关的公式。
Step 1: Apply the double-angle formula to the numerator. Using cos 2x ≡ 1 − 2 sin² x, we substitute into the numerator:
第一步：对分子应用二倍角公式。利用 cos 2x ≡ 1 − 2 sin² x，我们将其代入分子：
1 − cos 2x = 1 − (1 − 2 sin² x) = 2 sin² x
Step 2: Apply the double-angle formula to the denominator. Using cos 2x ≡ 2 cos² x − 1, we substitute into the denominator:
第二步：对分母应用二倍角公式。利用 cos 2x ≡ 2 cos² x − 1，我们将其代入分母：
1 + cos 2x = 1 + (2 cos² x − 1) = 2 cos² x
Notice how the “+1” and “−1” cancel perfectly in both the numerator and the denominator, leaving us with a simple 2 sin² x and 2 cos² x respectively. This clean cancellation is exactly why we chose these specific forms of the double-angle formula.
注意，分子和分母中的“+1”和“−1”分别完美抵消，只剩下简单的 2 sin² x 和 2 cos² x。这种清晰的抵消正是我们选择特定形式二倍角公式的原因。
6. Step-by-Step Solution (Part 2) | 分步解析（第二部分）
Now that we have simplified the individual components, we can assemble the entire LHS.
现在我们已经化简了各个组成部分，我们可以重新组装整个等式的左边。
Step 3: Rewrite the LHS. Substitute the simplified numerator and denominator back into the fraction:
第三步：重写左边。将化简后的分子和分母代回原分数中：
LHS = (2 sin² x) / (2 cos² x)
Step 4: Cancel the common factor of 2. The factor of 2 is common to both the numerator and the denominator, so it cancels out completely:
第四步：约去公因数 2。数字 2 是分子和分母的公因数，因此可以直接约掉：
LHS = sin² x / cos² x = (sin x / cos x)²
Step 5: Apply the tangent ratio identity. Since tan x ≡ sin x / cos x, we can substitute the ratio directly:
第五步：应用正切比值恒等式。由于 tan x ≡ sin x / cos x，我们可以直接代入比值：
LHS = (tan x)² = tan² x = RHS
Therefore, we have successfully proven that the LHS simplifies exactly to the RHS, so the identity is true for all valid values of x.
因此，我们成功地证明了等式的左边可以精确地化简为等式的右边，所以该恒等式对所有合法的 x 值均成立。
7. Verifying the Solution | 验证解答
While a formal proof is complete, it is always good practice to verify the result with a specific numerical substitution. This checks for algebraic errors in our manipulation and builds confidence.
虽然正式的证明已经完成，但代入一个具体的数值进行验证始终是一个好习惯。这可以检查我们变换过程中是否存在代数错误，并增强解题信心。
Let us substitute x = 30° into both sides. For the LHS: cos(2 × 30°) = cos 60° = 0.5. Therefore, LHS = (1 − 0.5) / (1 + 0.5) = 0.5 / 1.5 = 1/3. For the RHS: tan² 30° = (1/√3)² = 1/3. Since both sides equal 1/3, the identity holds true for this specific case, confirming the correctness of our proof.
我们代入 x = 30° 来验证两边。对于左边：cos(2 × 30°) = cos 60° = 0.5。因此，左边 = (1 − 0.5) / (1 + 0.5) = 0.5 / 1.5 = 1/3。对于右边：tan² 30° = (1/√3)² = 1/3。由于两边都等于 1/3，该恒等式在此特定情况下成立，验证了我们证明的正确性。
Let us also test an angle in the second quadrant, say x = 120°. LHS: (1 − cos 240°) / (1 + cos 240°). Since cos 240° = −0.5, LHS = (1 + 0.5) / (1 − 0.5) = 1.5 / 0.5 = 3. RHS: tan² 120° = (−√3)² = 3. This works perfectly. Note that we cannot test x = 90° because tan 90° is undefined, and the denominator 1 + cos 180° = 0, which matches the domain restriction.
我们再来测试第二象限的一个角度，比如 x = 120°。左边：(1 − cos 240°) / (1 + cos 240°)。由于 cos 240° = −0.5，所以左边 = (1 + 0.5) / (1 − 0.5) = 1.5 / 0.5 = 3。右边：tan² 120° = (−√3)² = 3。完美成立。注意，我们不能测试 x = 90°，因为 tan 90° 无定义，且分母 1 + cos 180° = 0，这正好符合定义域的限制。
8. Common Pitfalls & Exam Tips | 常见误区与考试技巧
Students frequently lose marks on identity questions due to a few common errors. Being aware of these pitfalls is essential for securing full marks in the exam.
由于一些常见错误，学生经常在恒等式题目上丢分。了解这些陷阱对于在考试中获得满分至关重要。
- Manipulating both sides simultaneously: The hallmark of a correct proof is that you work on one side only. You may not bring terms from the right-hand side over to the left-hand side. Write “LHS =” and work logically downwards.
- 同时处理等号两边：一个正确证明的标志是你只处理一边。你不能将右边的项移到左边。写下“LHS =”然后逻辑清晰地向下推导。
- Dividing by zero: Since the LHS has a denominator, you must implicitly state that the identity is valid as long as cos 2x ≠ −1 (equivalently, cos x ≠ 0). Do not cancel terms like sin x without noting that sin x = 0 is a separate case (though this is more relevant to solving equations than proving identities).
- 除以零：由于左边是分数形式，必须隐含说明恒等式在 cos 2x ≠ −1（即 cos x ≠ 0）时成立。不要随意约去像 sin x 这样的项而不指出 sin x = 0 是特殊情况（尽管这更多关系到解方程而非证明恒等式）。
- Forgetting the concluding statement: You must explicitly write “= RHS” at the end of your proof. Examiners look for this clear comparison, and a final sentence like “Therefore, the identity is proven” secures the final communication mark.
- 忘记结论性陈述：在证明的末尾必须明确写出“= RHS”。阅卷者会寻找这个清晰的对比，并且一句“因此，该恒等式得证”能确保获得最后的沟通分。
9. Extension: Solving the Associated Equation | 拓展：解相关方程
AQA exams frequently link identities to solving equations. Since we have just proven that (1 − cos 2x) / (1 + cos 2x) ≡ tan² x, we can use this result to solve more complex-looking equations very easily.
AQA 考试经常将恒等式与解方程联系起来。既然我们已经证明了 (1 − cos 2x) / (1 + cos 2x) ≡ tan² x，我们就可以利用这个结果来非常轻松地解决看起来更复杂的方程。
Example Question: Solve the equation (1 − cos 2x) / (1 + cos 2x) = 3 for 0° ≤ x ≤ 360°.
例题：解方程 (1 − cos 2x) / (1 + cos 2x) = 3，其中 0° ≤ x ≤ 360°。
Solution: Using our proven identity, we can substitute the LHS directly. This transforms the equation into tan² x = 3. Taking the square root of both sides gives tan x = ±√3. The principal solution for tan x = √3 is x = 60°. Since tan x has a period of 180°, the general solutions for tan x = √3 are 60° and 240°. For the negative case tan x = −√3, the solutions are 120° and 300°. Therefore, the full set of solutions in the given interval is x = 60°, 120°, 240°, 300°.
解答：利用我们已经证明的恒等式，我们可以直接替换左边。这使方程变为 tan² x = 3。两边开平方根得到 tan x = ±√3。tan x = √3 的主解是 x = 60°。由于 tan x 的周期是 180°，tan x = √3 的通解为 60° 和 240°。对于负的情形 tan x = −√3，解为 120° 和 300°。因此，在给定区间内的完整解集为 x = 60°, 120°, 240°, 300°。
10. Practice Questions | 实战练习
To fully master this topic, it is essential to attempt similar problems independently. Here are two practice questions that mirror the exact style of Example 6.9.1.
为了完全掌握这个知识点，独立练习相似题型是必不可少的。以下是两道与 Example 6.9.1 风格完全一致的练习题。
Question 1: Prove the identity (1 + cos 2x) / (1 − cos 2x) ≡ cot² x.
题目 1：证明恒等式 (1 + cos 2x) / (1 − cos 2x) ≡ cot² x。
Hint: Follow the exact same strategy. Use cos 2x ≡ 2 cos² x − 1 for the numerator, and cos 2x ≡ 1 − 2 sin² x for the denominator. You will end up with cos² x /
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