📚 A-Level Mathematics: Strategies for Proving Trigonometric Identities | A-Level数学：三角恒等式的证明思路
Proving trigonometric identities is a classic topic in A-Level Mathematics. Unlike solving equations, where you seek unknown values, an identity is a statement that is true for all permitted values of the variable. The key skill is not memorising every formula, but recognising which transformation moves you closer to a simpler, equivalent form. This article presents a structured set of strategies, illustrated with worked examples, to help you approach identity proofs with confidence.
证明三角恒等式是 A-Level 数学中的经典题型。与解方程不同，解方程需要求未知值，而恒等式是对变量的所有允许取值都成立的命题。关键能力不在于记住每一条公式，而在于识别哪种变形能让你更接近一个更简单、等价的形式。本文将提供一套结构化的证明思路，并通过例题演示，帮助你自信地应对恒等式证明。
1. Core Identities You Must Know | 必须掌握的基本恒等式
Before planning a proof, you need a fluent command of the fundamental identities. The Pythagorean identities are the backbone: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ. You also need the compound angle formulas: sin(A ± B), cos(A ± B), tan(A ± B), and their double-angle special cases. In addition, the factor formulas (sum-to-product) and product-to-sum formulas appear frequently in A-Level papers.
在规划证明之前，你需要熟练掌握基本的恒等式。毕达哥拉斯恒等式是骨干：sin²θ + cos²θ = 1，1 + tan²θ = sec²θ，以及 1 + cot²θ = cosec²θ。你还需要和角公式：sin(A ± B)、cos(A ± B)、tan(A ± B)，以及它们的二倍角特例。此外，和差化积公式与积化和差公式在 A-Level 试卷中也经常出现。
2. General Principle: Work from the Complicated Side | 基本原则：从复杂的一侧入手
If you are asked to prove that expression L equals expression R, a reliable first move is to start with the side that looks more complicated and simplify it until it becomes the other side. This is usually easier than randomly manipulating both sides, because it gives you a clear target. If both sides look equally complex, pick the one with more fractions, products, or trigonometric functions.
如果题目要求证明表达式 L 等于表达式 R，一个可靠的第一步是从看起来更复杂的一侧开始，将其化简为另一侧。这通常比盲目地两边同时变形更容易，因为你有一个明确的目标。如果两边复杂度相当，就选择包含更多分式、乘积或三角函数的那一侧。
Example: Prove that (1 + tan²θ) cos²θ = 1.
Start with the left side. Since 1 + tan²θ = sec²θ, the left side becomes sec²θ · cos²θ = (1/cos²θ) · cos²θ = 1. The proof is one line. The key was to recognise the Pythagorean form.
从左侧开始。由于 1 + tan²θ = sec²θ，左边变为 sec²θ · cos²θ = (1/cos²θ) · cos²θ = 1。证明只有一行，关键是识别出毕达哥拉斯形式。
3. Convert Everything to Sine and Cosine | 将一切化为正弦和余弦
When you see tan, sec, cosec, or cot, consider rewriting them in terms of sin and cos. This often reveals hidden common factors or makes algebraic cancellation possible. For example, tanθ = sinθ/cosθ, secθ = 1/cosθ, cosecθ = 1/sinθ, and cotθ = cosθ/sinθ. This strategy is especially useful when the identity involves mixed ratios.
当你看到 tan、sec、cosec 或 cot 时，考虑用 sin 和 cos 重写它们。这往往会暴露出隐藏的公因式，或使代数约分成为可能。例如，tanθ = sinθ/cosθ，secθ = 1/cosθ，cosecθ = 1/sinθ，cotθ = cosθ/sinθ。当恒等式中混合了多种比值时，此策略特别有效。
Example: Prove that cotθ + tanθ = cosecθ secθ.
Write the left side as cosθ/sinθ + sinθ/cosθ. Combine the fractions: (cos²θ + sin²θ)/(sinθ cosθ) = 1/(sinθ cosθ). But 1/(sinθ cosθ) = (1/sinθ)(1/cosθ) = cosecθ secθ, which is the right side.
将左边写为 cosθ/sinθ + sinθ/cosθ。合并分式：(cos²θ + sin²θ)/(sinθ cosθ) = 1/(sinθ cosθ)。而 1/(sinθ cosθ) = (1/sinθ)(1/cosθ) = cosecθ secθ，即右边。
4. Use Pythagorean Substitutions with ‘1’ | 利用’1’的毕达哥拉斯替换
The number 1 is surprisingly powerful. Because sin²θ + cos²θ = 1, you can replace 1 with sin²θ + cos²θ, or replace sin²θ with 1 – cos²θ, and vice versa. Similarly, sec²θ – tan²θ = 1 and cosec²θ – cot²θ = 1. These substitutions often enable factorisation, especially when you have a difference of squares.
数字 1 具有惊人的作用。因为 sin²θ + cos²θ = 1，你可以将 1 替换为 sin²θ + cos²θ，或将 sin²θ 替换为 1 – cos²θ，反之亦然。同样，sec²θ – tan²θ = 1，cosec²θ – cot²θ = 1。这些替换常能促成因式分解，尤其是当你遇到平方差时。
Example: Prove that cos⁴θ – sin⁴θ = cos²θ – sin²θ.
Factor the left side as (cos²θ – sin²θ)(cos²θ + sin²θ). Since cos²θ + sin²θ = 1, the left side reduces to cos²θ – sin²θ. This is a direct application of factoring with the Pythagorean identity.
将左边因式分解为 (cos²θ – sin²θ)(cos²θ + sin²θ)。由于 cos²θ + sin²θ = 1，左边化简为 cos²θ – sin²θ。这是利用毕达哥拉斯恒等式进行因式分解的直接应用。
5. Combine Fractions by Finding a Common Denominator | 通过通分合并分式
Many identities involve sums or differences of rational expressions. A standard algebraic step is to write them over a common denominator. After expanding the numerator, look for opportunities to use sin²θ + cos²θ = 1 or other identities to simplify the numerator into a single term that matches the denominator.
许多恒等式涉及有理表达式的和或差。一个标准的代数步骤是将它们通分。展开分子后，寻找机会使用 sin²θ + cos²θ = 1 或其他恒等式将分子化简为与分母匹配的单一表达式。
Example: Prove that 1/(1 + cosθ) + 1/(1 – cosθ) = 2 cosec²θ.
Combine the left side: [(1 – cosθ) + (1 + cosθ)] / [(1 + cosθ)(1 – cosθ)] = 2/(1 – cos²θ). Using 1 – cos²θ = sin²θ, this becomes 2/sin²θ = 2 cosec²θ. Notice how the numerator collapsed to a constant.
合并左边：[(1 – cosθ) + (1 + cosθ)] / [(1 + cosθ)(1 – cosθ)] = 2/(1 – cos²θ)。利用 1 – cos²θ = sin²θ，变为 2/sin²θ = 2 cosec²θ。注意分子如何化简为常数。
6. Factorise to Reveal Shared Structure | 通过因式分解揭示共享结构
If an expression contains four terms or a difference of powers, try grouping and factorising. For instance, a² – b² = (a – b)(a + b), and a³ ± b³ = (a ± b)(a² ∓ ab + b²). When trigonometric expressions mimic these algebraic patterns, factorisation can cancel entire brackets or simplify them to 1. Always look for common factors such as sinθ, cosθ, or (1 + sinθ).
如果表达式包含四项或幂差，尝试分组与因式分解。例如，a² – b² = (a – b)(a + b)，a³ ± b³ = (a ± b)(a² ∓ ab + b²)。当三角表达式模仿这些代数模式时，因式分解可以约掉整个括号或将其化简为 1。始终留意 sinθ、cosθ 或 (1 + sinθ) 等公因式。
Example: Prove that (sinθ + cosθ)² + (sinθ – cosθ)² = 2.
Expand both squares: sin²θ + 2sinθ cosθ + cos²θ + sin²θ – 2sinθ cosθ + cos²θ. The cross terms cancel, leaving 2sin²θ + 2cos²θ = 2(sin²θ + cos²θ) = 2. Factoring out the 2 is the key final step.
展开两个平方：sin²θ + 2sinθ cosθ + cos²θ + sin²θ – 2sinθ cosθ + cos²θ。交叉项相消，剩下 2sin²θ + 2cos²θ = 2(sin²θ + cos²θ) = 2。提出公因数 2 是关键的最后一步。
7. Use Double-Angle Formulas to Reduce Powers | 使用二倍角公式降幂
Double-angle formulas are essential when you see expressions like sin2θ, cos2θ, or powers such as sin²θ. The identities sin2θ = 2sinθ cosθ and cos2θ = cos²θ – sin²θ = 2cos²θ – 1 = 1 – 2sin²θ allow you to rewrite quadratic powers as linear expressions. This is particularly useful when an identity mixes single angles and double angles.
当你看到 sin2θ、cos2θ 或 sin²θ 这样的幂次时，二倍角公式必不可少。sin2θ = 2sinθ cosθ 和 cos2θ = cos²θ – sin²θ = 2cos²θ – 1 = 1 – 2sin²θ 允许你将二次幂改写为一次表达式。当恒等式混合了单角与二倍角时，这一工具尤其有用。
Example: Prove that cos2θ = 1 – 2sin²θ.
This is actually one of the standard forms of the double-angle formula. To prove it from the compound angle formula, write cos2θ = cos(θ + θ) = cosθ cosθ – sinθ sinθ = cos²θ – sin²θ. Then replace cos²θ with 1 – sin²θ to obtain 1 – 2sin²θ.
这实际上是二倍角公式的标准形式之一。为了从和角公式推导它，写 cos2θ = cos(θ + θ) = cosθ cosθ – sinθ sinθ = cos²θ – sin²θ。然后将 cos²θ 替换为 1 – sin²θ，得到 1 – 2sin²θ。
8. Apply Sum-to-Product and Product-to-Sum Formulas | 运用和差化积与积化和差公式
When an identity contains sums of sines or cosines, such as sinA + sinB or cosA – cosB, the sum-to-product formulas are often the fastest route. These formulas convert sums into products, which then cancel or simplify. Conversely, product-to-sum formulas convert products like 2sinA cosB into sums, which may be easier to integrate or compare.
当恒等式中含有正弦或余弦的和，如 sinA + sinB 或 cosA – cosB 时，和差化积公式通常是最快的路径。这些公式将和转化为积，然后可以约分或化简。相反，积化和差公式将 2sinA cosB 这样的积转化为和，这可能更易于比较或化简。
Example: Prove that (sinA + sinB)² + (cosA + cosB)² = 4 cos²((A – B)/2).
Using sum-to-product, sinA + sinB = 2 sin((A+B)/2) cos((A-B)/2) and cosA + cosB = 2 cos((A+B)/2) cos((A-B)/2). Squaring and adding gives 4 cos²((A-B)/2)[sin²((A+B)/2) + cos²((A+B)/2)] = 4 cos²((A-B)/2). The Pythagorean identity collapses the bracket to 1.
利用和差化积，sinA + sinB = 2 sin((A+B)/2) cos((A-B)/2)，cosA + cosB = 2 cos((A+B)/2) cos((A-B)/2)。平方后相加得到 4 cos²((A-B)/2)[sin²((A+B)/2) + cos²((A+B)/2)] = 4 cos²((A-B)/2)。毕达哥拉斯恒等式将括号化简为 1。
9. Multiply by a Conjugate to Simplify Denominators | 乘以共轭式化简分母
If a denominator contains 1 + sinθ or 1 – cosθ, multiplying both numerator and denominator by the conjugate (1 – sinθ or 1 + cosθ) can transform the denominator into a single term via the difference of squares. This technique is especially useful when the other side is a simple reciprocal or a single trigonometric function.
如果分母含有 1 + sinθ 或 1 – cosθ，将分子分母同时乘以共轭式 (1 – sinθ 或 1 + cosθ)，可通过平方差将分母化为单一表达式。当另一边是简单的倒数或单个三角函数时，此技巧特别有用。
Example: Prove that sinθ/(1 + cosθ) = (1 – cosθ)/sinθ.
Start with the left side. Multiply numerator and denominator by 1 – cosθ: sinθ(1 – cosθ)/[(1 + cosθ)(1 – cosθ)] = sinθ(1 – cosθ)/(1 – cos²θ) = sinθ(1 – cosθ)/sin²θ = (1 – cosθ)/sinθ. This proves the identity and also shows the useful half-angle relation tan(θ/2).
从左边开始，将分子分母同乘 1 – cosθ：sinθ(1 – cosθ)/[(1 + cosθ)(1 – cosθ)] = sinθ(1 – cosθ)/(1 – cos²θ) = sinθ(1 – cosθ)/sin²θ = (1 – cosθ)/sinθ。这就证明了恒等式，同时也显示了有用的半角关系 tan(θ/2)。
10. Work Backwards from the Target | 从目标反向推导
If you cannot see an immediate path from the left side to the right side, try starting from the right side and simplifying it to the left. Because equality is symmetric, this is a fully valid proof method. Sometimes the ‘simpler’ side is not actually simpler; the other side may factor more naturally once you know what to aim for.
如果你无法立即看出从左边到右边的路径，可以尝试从右边开始化简到左边。因为等式具有对称性，这是一种完全有效的证明方法。有时看起来’更简单’的一侧实际上并不简单；一旦你知道目标，另一侧可能更容易自然分解。
Example: Prove that secθ – tanθ = cosθ/(1 + sinθ).
Working from the right side, multiply numerator and denominator by 1 – sinθ: cosθ(1 – sinθ)/(1 – sin²θ) = cosθ(1 – sinθ)/cos²θ = (1 – sinθ)/cosθ = 1/cosθ – sinθ/cosθ = secθ – tanθ. This is a clean reverse proof.
从右边开始，将分子分母同乘 1 – sinθ：cosθ(1 – sinθ)/(1 – sin²θ) = cosθ(1 – sinθ)/cos²θ = (1 – sinθ)/cosθ = 1/cosθ – sinθ/cosθ = secθ – tanθ。这是一条清晰的反向证明。
11. Common Mistakes and Final Checks | 常见错误与最终检查
A frequent mistake is treating identities as equations: you cannot move terms across the equals sign and then ‘solve’ as if both sides are known. Another error is forgetting domain restrictions, such as when dividing by cosθ, which is invalid if cosθ = 0. Always state any restriction if the manipulation requires division by a variable expression. Finally, check your proof by substituting a specific angle, say θ = 30° or a simple value, to ensure both sides agree numerically.
一个常见错误是将恒等式当作方程来解：你不能把项移到等号另一边，然后像两边都已知那样’求解’。另一个错误是忘记定义域限制，例如除以 cosθ 时，如果 cosθ = 0 则无效。如果变形需要除以含变量的表达式，请务必说明限制条件。最后，代入一个具体角度（如 θ = 30° 或某个简单值）来检查证明，确保两边数值一致。
- Do not divide by sinθ or cosθ unless you have justified that it is non-zero.
- Keep expressions factorised as long as possible to reveal cancellation.
- Write down every step; skipped steps often hide sign errors.
- For A-Level, use only approved formulas; many are given in the formula book.
除非你已证明 sinθ 或 cosθ 不为零，否则不要除以它们。尽量保持表达式因式分解状态，以便发现约分。写出每一步；跳步往往隐藏符号错误。对于 A-Level，只使用允许的公式；许多公式在公式书中已给出。
12. Conclusion | 总结
Proving trigonometric identities is not about random guesswork. By following a systematic approach — starting with the complicated side, converting to sine and cosine, using Pythagorean substitutions, combining fractions, factoring, and applying angle formulas — you can turn almost any identity into a short sequence of algebraic steps. With regular practice, you will begin to see patterns immediately, saving time and avoiding errors in the exam.
证明三角恒等式不是盲目猜测。通过系统化的方法——从复杂的一侧开始，化为正弦与余弦，使用毕达哥拉斯替换，合并分式，因式分解，以及运用角度公式——你可以将几乎任何恒等式转化为一系列简短的代数步骤。通过经常练习，你会开始即时识别模式，从而在考试中节省时间并避免错误。
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