📚 Trigonometric Identities and Equations | 三角恒等式与方程
In Edexcel A-Level Mathematics, trigonometric identities and equations form one of the most heavily examined Pure Mathematics topics. You need to use identities flexibly to simplify expressions, prove statements, and solve equations in a given interval. This revision guide covers the key identities, the compound and double angle formulas, the R-formula, and systematic equation-solving strategies.
在 Edexcel A-Level 数学中，三角恒等式与方程是纯数学部分考查最频繁的主题之一。你需要灵活运用恒等式来化简表达式、证明命题，并在给定区间内求解方程。本复习指南涵盖基本恒等式、复合角公式、倍角公式、R 公式以及系统的方程求解策略。
1. Fundamental Identities | 基本恒等式
The most basic identities come directly from the definitions of sine, cosine and tangent on the unit circle. Since tan θ = sin θ / cos θ, reciprocal functions are defined by cosec θ = 1/sin θ, sec θ = 1/cos θ and cot θ = 1/tan θ = cos θ / sin θ. These allow you to convert between different trigonometric functions.
最基本的恒等式直接来自单位圆上正弦、余弦和正切的定义。因为 tan θ = sin θ / cos θ，倒数函数定义为 cosec θ = 1/sin θ、sec θ = 1/cos θ 以及 cot θ = 1/tan θ = cos θ / sin θ。这些恒等式让你能够在不同三角函数之间转换。
For Edexcel papers, you are expected to move confidently between sin, cos, tan and their reciprocals. A common starting step is to write everything in terms of sin θ and cos θ, especially when proving identities.
在 Edexcel 考试中，你需要熟练地在 sin、cos、tan 及其倒数之间转换。一个常见的起始步骤是将所有函数都写成 sin θ 和 cos θ 的形式，尤其是在证明恒等式时。
2. Pythagorean Identities and Their Uses | 勾股恒等式及其应用
The Pythagorean identity is sin²θ + cos²θ = 1. Dividing this equation by cos²θ gives tan²θ + 1 = sec²θ, and dividing by sin²θ gives 1 + cot²θ = cosec²θ. These three forms are essential for simplifying expressions and solving equations that mix different functions.
勾股恒等式为 sin²θ + cos²θ = 1。将此式两边同时除以 cos²θ，得到 tan²θ + 1 = sec²θ；两边同时除以 sin²θ，得到 1 + cot²θ = cosec²θ。这三种形式对于化简表达式和求解混合不同三角函数的方程至关重要。
sin²θ + cos²θ = 1 | tan²θ + 1 = sec²θ | 1 + cot²θ = cosec²θ
A typical application is finding the exact value of another trigonometric function when one is known. For example, if sin θ = 3/5 and θ is acute, then cos θ = √(1 – 9/25) = 4/5, so tan θ = 3/4. Remember to consider the sign of the root according to the quadrant.
一个典型应用是已知一个三角函数值，求另一个三角函数值的精确值。例如，若 sin θ = 3/5 且 θ 为锐角，则 cos θ = √(1 – 9/25) = 4/5，因此 tan θ = 3/4。记住要根据象限考虑根号的正负号。
3. Compound Angle Formulas | 复合角公式
The compound angle formulas express trigonometric functions of sums or differences. For sine and cosine, you must be careful with the signs: sin(A ± B) = sin A cos B ± cos A sin B, and cos(A ± B) = cos A cos B ∓ sin A sin B. The tangent version is tan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B).
复合角公式表示两个角的和或差的三角函数。对于正弦和余弦，必须特别注意符号：sin(A ± B) = sin A cos B ± cos A sin B，而 cos(A ± B) = cos A cos B ∓ sin A sin B。正切的复合角公式为 tan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B)。
sin(A ± B) = sin A cos B ± cos A sin B
cos(A ± B) = cos A cos B ∓ sin A sin B
These formulas allow exact values such as sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4. They also underpin the double angle formulas.
这些公式可以求非特殊角的精确值，例如 sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4。它们也是倍角公式的基础。
4. Double Angle Formulas | 倍角公式
Setting B = A in the compound angle formulas gives the double angle formulas. The most important are sin 2θ = 2 sin θ cos θ, cos 2θ = cos²θ – sin²θ = 2cos²θ – 1 = 1 – 2sin²θ, and tan 2θ = 2tan θ / (1 – tan²θ).
在复合角公式中令 B = A，即可得到倍角公式。最重要的倍角公式为 sin 2θ = 2 sin θ cos θ，cos 2θ = cos²θ – sin²θ = 2cos²θ – 1 = 1 – 2sin²θ，以及 tan 2θ = 2tan θ / (1 – tan²θ)。
The three equivalent forms of cos 2θ are particularly useful. For example, cos 2θ = 2cos²θ – 1 rearranges to cos²θ = (1 + cos 2θ)/2, which is widely used in integration. In equation solving, choose the form that matches the rest of the expression.
cos 2θ 的三种等价形式特别有用。例如，cos 2θ = 2cos²θ – 1 可变形为 cos²θ = (1 + cos 2θ)/2，这在积分中广泛使用。在解方程时，应选择与表达式其余部分相匹配的形式。
5. Factor Formulae | 和差化积公式
Factor formulae convert sums or differences of trigonometric functions into products. They include sin P + sin Q = 2 sin((P + Q)/2) cos((P – Q)/2), sin P – sin Q = 2 cos((P + Q)/2) sin((P – Q)/2), cos P + cos Q = 2 cos((P + Q)/2) cos((P – Q)/2), and cos P – cos Q = -2 sin((P + Q)/2) sin((P – Q)/2).
和差化积公式将三角函数的和或差转化为乘积。这些公式包括 sin P + sin Q = 2 sin((P + Q)/2) cos((P – Q)/2)，sin P – sin Q = 2 cos((P + Q)/2) sin((P – Q)/2)，cos P + cos Q = 2 cos((P + Q)/2) cos((P – Q)/2)，以及 cos P – cos Q = -2 sin((P + Q)/2) sin((P – Q)/2)。
These are useful when solving equations such as sin 3θ + sin θ = 0. Applying the first formula gives 2 sin 2θ cos θ = 0, leading to sin 2θ = 0 or cos θ = 0, which is much easier to solve. Watch the negative sign in the cos P – cos Q formula.
这些公式在求解如 sin 3θ + sin θ = 0 这样的方程时非常有用。应用第一个公式得到 2 sin 2θ cos θ = 0，从而得到 sin 2θ = 0 或 cos θ = 0，这更容易求解。注意 cos P – cos Q 公式中的负号。
6. The R-Formula and Harmonic Form | R 公式与辅助角形式
The R-formula expresses a sin θ + b cos θ as a single sine or cosine function. For example, a sin θ + b cos θ = R sin(θ + α), where R = √(a² + b²) and α is chosen so that cos α = a/R and sin α = b/R. Equivalently, it can be written as R cos(θ – α).
R 公式将 a sin θ + b cos θ 表达为单个正弦或余弦函数。例如，a sin θ + b cos θ = R sin(θ + α)，其中 R = √(a² + b²)，且 α 满足 cos α = a/R 和 sin α = b/R。它也可以等价地写成 R cos(θ – α)。
a sin θ + b cos θ = R sin(θ + α), R = √(a² + b²), tan α = b/a
This method is used to solve equations of the form a sin θ + b cos θ = c and to find maximum and minimum values. Since R sin(θ + α) has a maximum of R and a minimum of -R, the original expression has the same range. Edexcel questions often ask for the smallest positive value of α or exact values of α.
这种方法用于求解形如 a sin θ + b cos θ = c 的方程，以及求最大值和最小值。由于 R sin(θ + α) 的最大值为 R，最小值为 -R，原表达式具有相同的值域。Edexcel 题目常要求求出 α 的最小正值或 α 的精确值。
7. Strategy for Solving Trigonometric Equations | 解三角方程的策略
A reliable strategy is essential for solving trigonometric equations under timed conditions. First identify the required interval, such as 0 ≤ θ < 2π or 0° ≤ θ < 360°. Then use identities to reduce the equation to a single trigonometric function. Factorise if possible, and solve each factor separately.
在限时条件下求解三角方程，可靠的解题策略至关重要。首先确定要求的区间，例如 0 ≤ θ < 2π 或 0° ≤ θ < 360°。然后使用恒等式将方程化为单一三角函数。如果可能，进行因式分解，并分别求解每个因子。
For example, sin θ = 0.5 in the interval 0 ≤ θ < 2π has principal solution θ = π/6. Using the symmetry of the sine curve or a CAST diagram, the second solution is θ = 5π/6. Always check whether the question expects answers in radians or degrees.
例如，在区间 0 ≤ θ < 2π 内，sin θ = 0.5 的主解为 θ = π/6。利用正弦曲线的对称性或 CAST 图，第二个解为 θ = 5π/6。务必检查题目要求答案使用弧度还是角度。
8. Solving Equations with Multiple Angles | 多倍角方程求解
When the equation involves sin 2θ, cos 3θ or tan(θ/2), use a substitution. For example, to solve sin 2θ = 0.5 for 0 ≤ θ < 2π, let u = 2θ. Then the interval for u becomes 0 ≤ u < 4π, which is twice as large. Solve sin u = 0.5 in this expanded interval, then convert back with θ = u/2.
当方程涉及 sin 2θ、cos 3θ 或 tan(θ/2) 时，应使用换元。例如，要在 0 ≤ θ < 2π 内求解 sin 2θ = 0.5，设 u = 2θ。此时 u 的区间变为 0 ≤ u < 4π，扩大了一倍。在这个扩大的区间内解 sin u = 0.5，然后用 θ = u/2 换回。
The most common error is forgetting to expand the interval. If sin u = 0.5 has solutions u = π/6, 5π/6, 13π/6, 17π/6 in 0 ≤ u < 4π, then θ = π/12, 5π/12, 13π/12, 17π/12. Always list all possible values before dividing by the coefficient.
最常见的错误是忘记扩大区间。若 sin u = 0.5 在 0 ≤ u < 4π 内有解 u = π/6, 5π/6, 13π/6, 17π/6，则 θ = π/12, 5π/12, 13π/12, 17π/12。在除以系数之前，始终先列出所有可能的值。
9. Quadratic Forms and Hidden Quadratics | 二次型与隐二次方程
Many trigonometric equations are quadratic in disguise. For instance, 2 sin²θ + 3 sin θ – 2 = 0 can be solved by letting x = sin θ, giving 2x² + 3x – 2 = 0. Factorising gives (2x – 1)(x + 2) = 0, so sin θ = 1/2 or sin θ = -2. The second solution is impossible because -1 ≤ sin θ ≤ 1, so discard it.
许多三角方程实际上是隐藏的二次方程。例如，2 sin²θ + 3 sin θ – 2 = 0 可以令 x = sin θ 来求解，得到 2x² + 3x – 2 = 0。因式分解得 (2x – 1)(x + 2) = 0，所以 sin θ = 1/2 或 sin θ = -2。第二个解不可能，因为 -1 ≤ sin θ ≤ 1，因此舍去。
Hidden quadratics often arise when an equation contains both cos θ and sin²θ, or both sin θ and cos²θ. Use sin²θ + cos²θ = 1 to rewrite one function in terms of the other. For example, 2 cos²θ + sin θ = 1 becomes 2(1 – sin²θ) + sin θ = 1, leading to 2 sin²θ – sin θ – 1 = 0.
当方程同时包含 cos θ 和 sin²θ，或同时包含 sin θ 和 cos²θ 时，常出现隐二次方程。利用 sin²θ + cos²θ = 1 将其中一个函数用另一个表示。例如，2 cos²θ + sin θ = 1 化为 2(1 – sin²θ) + sin θ = 1，得到 2 sin²θ – sin θ – 1 = 0。
10. Equations Involving Reciprocal Functions | 涉及倒数函数的方程
Equations with cosec θ, sec θ or cot θ can usually be converted using reciprocal identities. For instance, cot θ = √3 is equivalent to tan θ = 1/√3, provided sin θ and cos θ are both non-zero. The equation cosec θ = 2 becomes sin θ = 1/2, but you must ensure sin θ ≠ 0.
含有 cosec θ、sec θ 或 cot θ 的方程通常可以用倒数恒等式进行转换。例如，cot θ = √3 等价于 tan θ = 1/√3，前提是 sin θ 和 cos θ 均不为零。方程 cosec θ = 2 化为 sin θ = 1/2，但必须确保 sin θ ≠ 0。
Always check for extraneous solutions after multiplying by a reciprocal function. For example, if an equation is multiplied by cos θ, any value of θ that makes cos θ = 0 must be excluded from the final answer unless it also satisfied the original equation.
在乘以倒数函数后，务必检查是否引入额外解。例如，如果方程两边同时乘以 cos θ，任何使 cos θ = 0 的 θ 值都必须从最终答案中排除，除非它也满足原方程。
11. Proof and Identities Verification | 证明与恒等式验证
When proving a trigonometric identity, start from the more complicated side and transform it until it matches the other side. Use foundational identities first: express tan, cosec, sec and cot in terms of sin and cos, then apply sin²θ + cos²θ = 1, double angle formulas, or factorisation.
在证明三角恒等式时，从较复杂的一边入手，将其变换直至与另一边相同。首先使用基本恒等式：将 tan、cosec、sec 和 cot 用 sin 和 cos 表示，然后应用 sin²θ + cos²θ = 1、倍角公式或因式分解。
For example, to prove (sin θ + cos θ)² = 1 + sin 2θ, expand the left side: sin²θ + 2 sin θ cos θ + cos²θ. By the Pythagorean identity this becomes 1 + 2 sin θ cos θ, and by the double angle formula it equals 1 + sin 2θ. Never assume the result in your working.
例如，要证明 (sin θ + cos θ)² = 1 + sin 2θ，展开左边：sin²θ + 2 sin θ cos θ + cos²θ。根据勾股恒等式，它变为 1 + 2 sin θ cos θ，再根据倍角公式等于 1 + sin 2θ。在推导过程中切勿假设结论成立。
12. Exam Tips and Common Pitfalls | 考试技巧与常见错误
Edexcel examiners frequently note mistakes such as losing solutions when dividing by a factor like sin θ, forgetting to expand the interval for multiple angles, mixing up radians and degrees, and incorrectly assuming sin(A + B) = sin A + sin B. Avoid dividing by a trigonometric expression unless you are certain it cannot be zero.
Edexcel 考官经常指出的错误包括：除以 sin θ 等因子时丢失解；多倍角问题中忘记扩大区间；混淆弧度和角度；以及错误地认为 sin(A + B) = sin A + sin B。除非确定三角表达式不可能为零，否则不要除以该表达式。
Use a CAST diagram or graph of the function to find all solutions in the required interval. After solving, substitute one or two values back into the original equation as a quick check. In proof questions, write every step clearly and justify each transformation.
使用 CAST 图或函数图像来找到所需区间内的所有解。求解后，将一两个值代回原方程进行快速检验。在证明题中，清晰写出每一步并说明每一步变换的理由。
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