📚 Investigation 2: Compound Angle Formulae | 探究二：复角公式
In the study of trigonometry, you have already met the sine, cosine and tangent of a single angle. However, many real‑world problems — from wave interference to rotating machinery — require expressions like sin(A + B) or cos(2θ). The compound angle formulae (also called addition formulae) are the set of identities that allow us to expand trigonometric functions of sums or differences of angles. In this investigation, you will discover where these identities come from, prove them using geometric and algebraic methods, and apply them to solve equations, simplify expressions and find exact values. By the end, you will understand why these formulae are fundamental tools in the IB Mathematics: Analysis and Approaches and Applications and Interpretation courses.
在三角学中，你已经熟悉了单角的正弦、余弦和正切。然而，许多实际问题——从波的干涉到旋转机械——都需要处理 sin(A + B) 或 cos(2θ) 这样的表达式。复角公式（也称和角公式）是一组能将角度和差的正弦、余弦、正切展开成单角三角函数组合的恒等式。在这项探究中，你将发现这些恒等式的来源，用几何和代数方法证明它们，并应用它们来解方程、化简表达式和求精确值。通过本探究，你将理解为什么这些公式是 IB 数学分析与方法和应用与解释课程中的基础工具。
1. What Are Compound Angles? | 什么是复角？
A compound angle is formed by adding or subtracting two or more angles. For example, 75° can be expressed as 45° + 30°, and 15° can be written as 45° − 30°. If we know the sine and cosine of the building‑block angles such as 30°, 45° and 60°, we intuitively want to find sin(45°+30°) directly from sin45° and sin30°. However, trigonometric functions are not linear: sin(A+B) is not equal to sinA + sinB. This makes compound angle problems both interesting and challenging.
复角是由两个或多个角度相加或相减形成的角度。例如，75° 可以表示为 45° + 30°，15° 可以写成 45° − 30°。如果我们知道了诸如 30°、45° 和 60° 这些基本角的正弦和余弦，就自然希望直接从 sin45° 和 sin30° 求出 sin(45°+30°)。然而，三角函数不是线性的：sin(A+B) 并不等于 sinA + sinB。这使得复角问题既有趣又具有挑战性。
2. The Need for New Identities | 为何需要新的恒等式
e claim sin(30°+60°) = sin30° + sin60°. The left side is sin90° = 1. The right side is ½ + √3 B) accurately, we need a set of identities that are logically derived, not guessed. The compound angle formulae fill exactly this gap.
考虑命题 sin(30°+60°) = sin30° + sin60°。左边是 sin90° = 1。右边是 ½ + √3/2 ≈ 1.366。两者明显不同。这个简单的反例说明，一个和角的正弦不能被天真地拆分成正弦的和。为了正确处理 sin(A ± B) 和 cos(A ± B) 这类表达式，我们需要一组经过逻辑推导而非猜测的恒等式。复角公式正好填补了这一空白。
3. Deriving cos(A − B) from the Unit Circle | 从单位圆推导 cos(A − B)
One of the most elegant derivations uses the distance formula on the unit circle. Draw a circle of radius 1 centred at the origin. Let points P and Q be at angles A and B respectively. Their coordinates are P(cosA, sinA) and Q(cosB, sinB). The squared distance PQ² can be computed in two different ways: directly from the coordinates, and by rotating the whole configuration so that Q moves to (1,0) and P moves to an angle of (A−B).
最优雅的推导方法之一是利用单位圆上的距离公式。画一个以原点为圆心、半径为 1 的圆。设点 P 和 Q 分别位于角度 A 和 B 处，坐标分别为 P(cosA, sinA) 和 Q(cosB, sinB)。平方距离 PQ² 可以用两种不同方式计算：直接利用坐标，以及通过旋转整个图形，使 Q 移动到 (1,0)，同时 P 移动到角度 (A−B) 处。
After rotation, the new coordinates of P become (cos(A−B), sin(A−B)). Equating the two expressions for PQ² gives:
旋转后，P 的新坐标变为 (cos(A−B), sin(A−B))。令两个 PQ² 表达式相等，得到：
(cosA − cosB)² + (sinA − sinB)² = (cos(A−B) − 1)² + (sin(A−B) − 0)²
es and using the Pythagorean identity sin²θ + cos²θ = 1 leads to the foundational formula:
展开两边并利用恒等式 sin²θ + cos²θ = 1，便可得到基础公式：
cos(A − B) = cosA cosB + sinA sinB
This identity is the building block for all other compound angle expansions. Note the plus sign on the right‑hand side, which is a very common point of confusion.
这个恒等式是所有其他复角展开式的基石。请注意右侧的加号，这是一个非常容易混淆的地方。
4. Deriving cos(A + B) | 推导 cos(A + B)
with −B and use the negative angle identities cos(−B)=cosB and sin(−B)=−sinB. This yields:
一旦确立了 cos(A − B)，我们可以用 −B 替换 B，并利用负角恒等式 cos(−B)=cosB 以及 sin(−B)=−sinB。由此得到：
cos(A + B) = cosA cosB − sinA sinB
Observe the sign change: the minus sign for cosine of a sum is crucial. Together, the two cosine compound angle formulae are often remembered as: “cos of sum has minus, cos of difference has plus”.
注意符号变化：两角和的余弦中的减号非常关键。这两个余弦复角公式往往被记作：“和的余弦带减号，差的余弦带加号”。
5. Deriving sin(A ± B) | 推导 sin(A ± B)
The sine addition formulae can be obtained from the cosine identities using the co‑function relationship sinθ = cos(π/2 − θ). For sin(A+B), write:
正弦的和角公式可以利用互余关系 sinθ = cos(π/2 − θ) 从余弦恒等式得到。对于 sin(A+B)，可以写为：
sin(A + B) = cos(π/2 − (A+B)) = cos((π/2 − A) − B)
Apply the cos(A−B) formula with angles (π/2−A) and B, then use co‑function identities again to turn the expression back into sines and cosines of A and B. The result is:
对角度 (π/2−A) 和 B 应用 cos(A−B) 公式，然后再次利用互余关系将表达式变回关于 A 和 B 的正弦和余弦。最终结果是：
sin(A + B) = sinA cosB + cosA sinB
Replacing B with −B gives the difference formula:
将 B 替换成 −B 便得到差角公式：
sin(A − B) = sinA cosB − cosA sinB
These identities show that sine of a sum or difference mixes sines and cosines symmetrically, with the sign matching the operation between angles.
这些恒等式表明，正弦的和差对称地混合了正弦与余弦，且符号与角度之间的运算符号一致。
6. Deriving tan(A ± B) | 推导 tan(A ± B)
are derived by dividing the sine and cosine results. Since tanθ = sinθ/cosθ, we can write:
正切的复角公式通过将正弦和余弦的结果相除而推导出来。因为 tanθ = sinθ/cosθ，我们可以写出：
tan(A + B) = (sinA cosB + cosA sinB) / (cosA cosB − sinA sinB)
Divide numerator and denominator by cosA cosB (provided neither cosine is zero) to obtain the compact form:
将分子和分母同时除以 cosA cosB（假设两个余弦均不为零），得到简洁形式：
tan(A + B) = (tanA + tanB) / (1 − tanA tanB)
Similarly, substituting −B into the formula yields:
类似地，代入 −B 得到：
tan(A − B) = (tanA − tanB) / (1 + tanA tanB)
These rational expressions are extremely useful when dealing with slopes, angular bisectors and calculus problems involving derivatives of inverse trigonometric functions.
这些有理表达式在处理斜率、角平分线以及涉及反三角函数导数的微积分问题时极为有用。
7. Double Angle Formulae as Special Cases | 倍角公式作为特例
und angle formulae simplify to the double angle formulae. Setting A = B in sin(A+B) gives:
当 A = B 时，复角公式简化为倍角公式。在 sin(A+B) 中令 A = B 可得：
sin(2A) = 2 sinA cosA
From cos(A+B) with A = B:
在 cos(A+B) 中令 A = B：
cos(2A) = cos²A − sin²A
his can be further rewritten as cos(2A) = 2cos²A − 1 or cos(2A) = 1 − 2sin²A. For tangent:
利用 sin²A + cos²A = 1，这可以进一步改写为 cos(2A) = 2cos²A − 1 或 cos(2A) = 1 − 2sin²A。对于正切：
tan(2A) = 2 tanA / (1 − tan²A)
These double angle identities appear frequently in integration, solving trigonometric equations and simplifying expressions. They are also a direct consequence of the compound angle investigation.
这些倍角恒等式在积分、解三角方程和化简表达式中频繁出现。它们也是复角探究的直接推论。
8. Geometric Proofs and Visual Meaning | 几何证明与直观含义
Beyond the unit circle, you can also prove the compound angle formulae by constructing a right‑angled triangle and decomposing an angle. For instance, draw a triangle with an angle A+B, then drop perpendiculars to create smaller triangles with angles A and B. By expressing side lengths in terms of sine and cosine, you can derive the same identities. This geometric approach reinforces the idea that these are not just algebraic tricks — they are statements about lengths in the plane.
除了单位圆方法，你还可以通过构造直角三角形并分解一个角度来证明复角公式。例如，画一个包含角度 A+B 的三角形，然后作垂线创造出具有角度 A 和 B 的较小三角形。通过用正弦和余弦表示边长，可以推导出同样的恒等式。这种几何方法强化了一个观念：这些不仅仅是代数技巧，它们是关于平面中线段的陈述。
Visual learners often benefit from labelling the sides of the composite triangle clearly. The identities for sin(A+B) and cos(A+B) emerge naturally from the ratios of those sides.
视觉型学习者往往受益于清晰地标记组合三角形的各边。sin(A+B) 和 cos(A+B) 的恒等式会从这些边的比值中自然而然地显现出来。
9. Exact Values of Non‑Standard Angles | 非标准角的精确值
tric values for angles such as 15°, 75°, 105° and so on. For example, 75° = 45° + 30°, so:
复角公式的一个直接应用是计算诸如 15°、75°、105° 等角度的精确三角函数值。例如，75° = 45° + 30°，所以：
sin75° = sin(45°+30°) = sin45°cos30° + cos45°sin30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4
Similarly, using the difference formula we can find cos15° = cos(45°−30°) = (√6 + √2)/4, a memorable symmetry. These exact values are extremely helpful for simplifying radicals in the solution of equations and integrals.
类似地，利用差角公式我们可以求出 cos15° = cos(45°−30°) = (√6 + √2)/4，这是一个容易记忆的对称形式。这些精确值对于在解方程和积分中化简根式极有帮助。
10. Simplifying Expressions | 化简表达式
The compound angle identities are often used to condense trigonometric expressions into a single term. For example, 2 sinx cosx is instantly recognised as sin(2x). An expression like ½ cosθ − √3 (π/6 − θ) = −sin(θ − π/6). Recognising these patterns is a key skill assessed in IB exams.
复角恒等式常被用来将三角表达式压缩成一个单项。例如，2 sinx cosx 能立即被识别为 sin(2x)。类似 ½ cosθ − √3/2 sinθ 这样的表达式可能看起来杂乱，但如果我们把它改写为 sin(π/6) cosθ − cos(π/6) sinθ，就会发现它与 sin(π/6 − θ) = −sin(θ − π/6) 的形式匹配。识别这些模式是 IB 考试中考查的关键技能。
When simplifying, always check whether coefficients correspond to a known sine or cosine of a special angle. This “reverse” use of the compound angle formulae is just as important as the forward expansion.
化简时，始终要检查系数是否对应于某个特殊角的正弦或余弦值。这种“逆向”使用复角公式的方法与正向展开同等重要。
11. Solving Equations Using Compound Angle Formulae | 利用复角公式解方程
mpound angle formulae can turn the equation into a simple linear combination of sinx and cosx. For the example above:
三角方程常常包含诸如 sin(x+π/3) + sin(x−π/3) = 1 的和。使用复角公式展开每一项，可以将方程转化为 sinx 和 cosx 的简单线性组合。对于上述例子：
[sinx·½ + cosx·√3/2] + [sinx·½ − cosx·√3/2] = sinx = 1
Without the formulae, such problems would be extremely difficult to solve systematically.
因此 x = π/2 + 2kπ。如果没有这些公式，这类问题将极难系统地求解。
of a variable, such as solving tan(2x) = cot(x) by expressing both sides in terms of tanx.
当三角函数内部的自变量是变量的倍数时，复角恒等式也能提供帮助，例如通过用 tanx 表示两边来解 tan(2x) = cot(x) 这类方程。
12. Common Pitfalls and Exam Tips | 常见陷阱与应试技巧
compute both sides and check. In IB exams, always show the substitution step, even if you are using a formula book. This demonstrates understanding and safeguards against mark loss from simple sign slips.
符号错误是首要问题。学生经常将 cos(A+B) 错记为 cosA cosB + sinA sinB，或者将 sin(A−B) 错记为 sinA cosB + cosA sinB。一个有效的记忆方法是使用简单数值进行检验：设 A=30°，B=60°，计算两边并核对。在 IB 考试中，即使你使用公式手册，也要展示代入步骤。这能体现理解过程，并防止因简单的符号差错而失分。
tfall is forgetting the denominator condition in tan(A±B). The formula is undefined when 1 ∓ tions, remember that compound angle expansions can introduce extraneous solutions if squaring occurs, so verify your answers.
另一个常见陷阱是忘记 tan(A±B) 中分母的条件。当 1 ∓ tanA tanB = 0 时公式无定义，这对应于 A±B 是 π/2 的奇数倍。在适用的情况下，务必说明定义域限制。最后，在解方程时，记住如果涉及平方，复角展开可能会引入增根，因此要验证答案。
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