📚 Deriving the Angle Addition Formulas and Their Classic Applications | 和角公式的推导与典型应用
The angle addition formulas — also known as the sum and difference identities — form the backbone of trigonometry. They express sin(A ± B), cos(A ± B), and tan(A ± B) in terms of the sines, cosines, and tangents of the individual angles A and B. These identities are not just abstract relationships; they are the key to simplifying complex trigonometric expressions, solving equations, and evaluating exact values of angles that are not on the unit circle’s standard list.
和角公式——也称为两角和与差的恒等式——是整个三角学的基石。它们将 sin(A ± B)、cos(A ± B) 和 tan(A ± B) 表示为 A 和 B 各自的正弦、余弦与正切的组合。这些恒等式不仅仅是抽象的数学关系；它们是化简复杂三角表达式、求解三角方程以及计算非标准特殊角精确值的关键工具。
1. Why Do We Need the Angle Addition Formulas? | 为什么需要和角公式？
Before diving into derivations, it is worth asking why these formulas matter. Suppose you need to find the exact value of sin 75°. Since 75° = 45° + 30°, and the exact values of sin and cos for 45° and 30° are known, the angle addition formula allows you to combine them into a single exact expression. Without these formulas, you would be limited to decimal approximations.
在深入推导之前，值得先问一句：这些公式为什么重要？假设你需要求 sin 75° 的精确值。由于 75° = 45° + 30°，而我们熟悉 45° 和 30° 的正弦、余弦精确值，和角公式就能把它们组合成一个精确的表达式。没有这些公式，我们只能得到小数近似值。
Beyond exact evaluation, the formulas are essential in calculus — the derivative of sin x is derived using the identity for sin(x + h). They also appear in physics when adding waves of different phases, in engineering when analysing alternating currents, and in computer graphics when rotating coordinates in the plane.
除了精确求值之外，这些公式在微积分中同样不可或缺——sin x 的导数正是利用 sin(x + h) 的展开式推导的。在物理学中，叠加不同相位的波动需要用到它；在工程中分析交流电离不开它；在计算机图形学中旋转平面坐标也要依靠它。
2. Geometric Derivation for cos(A – B) | cos(A – B) 的几何推导
The most elegant and intuitive derivation of the addition formulas begins with cos(A – B). Consider the unit circle with centre O. Place two points P and Q on the circle such that the angle between the positive x-axis and OP is A, and the angle between the positive x-axis and OQ is B. The coordinates of P are (cos A, sin A), and the coordinates of Q are (cos B, sin B).
和角公式中最优雅、最直观的推导从 cos(A – B) 入手。考虑以 O 为圆心的单位圆。在圆上取两点 P 和 Q，设 OP 与 x 轴正方向的夹角为 A，OQ 与 x 轴正方向的夹角为 B。那么 P 的坐标为 (cos A, sin A)，Q 的坐标为 (cos B, sin B)。
Now compute the distance between P and Q using the distance formula. The square of the distance PQ is:
现在用距离公式计算 P 和 Q 之间的距离。距离 PQ 的平方为：
PQ² = (cos A − cos B)² + (sin A − sin B)²
Expanding this gives cos²A − 2cosAcosB + cos²B + sin²A − 2sinAsinB + sin²B. Using the Pythagorean identity, the squared sine and cosine terms combine to 2, so:
展开后得到 cos²A − 2cosAcosB + cos²B + sin²A − 2sinAsinB + sin²B。利用毕达哥拉斯恒等式，正弦与余弦的平方项合并为 2，因此：
PQ² = 2 − 2(cosAcosB + sinAsinB)
Now rotate the coordinate system so that Q lies on the x-axis. In this rotated frame, P has angle A – B, so its coordinates are (cos(A – B), sin(A – B)) and Q has coordinates (1, 0). The distance squared is then:
现在旋转坐标系，使 Q 落在 x 轴上。在这个旋转后的坐标系中，P 的角度为 A − B，所以它的坐标为 (cos(A − B), sin(A − B))，Q 的坐标为 (1, 0)。此时距离的平方为：
PQ² = (cos(A − B) − 1)² + sin²(A − B) = 2 − 2cos(A − B)
Equating the two expressions for PQ² gives 2 − 2cos(A − B) = 2 − 2(cosAcosB + sinAsinB), which simplifies to:
令两个 PQ² 的表达式相等，得到 2 − 2cos(A − B) = 2 − 2(cosAcosB + sinAsinB)，化简后即为：
cos(A − B) = cosAcosB + sinAsinB
3. Deriving cos(A + B) and sin(A – B) | 推导 cos(A + B) 与 sin(A – B)
Once cos(A – B) is established, the other three identities follow quickly. To derive cos(A + B), replace B with -B in the formula for cos(A – B). Since cos(-B) = cos B and sin(-B) = -sin B, we obtain:
一旦 cos(A − B) 成立，其余三个恒等式可以迅速推出。要推导 cos(A + B)，只需将 cos(A − B) 公式中的 B 替换为 −B。因为 cos(−B) = cos B，sin(−B) = −sin B，于是得到：
cos(A + B) = cosAcosB − sinAsinB
For sin(A – B), we use the cofunction identity sin θ = cos(90° − θ). Let θ = A – B, so sin(A – B) = cos(90° − A + B) = cos((90° − A) + B). Applying the cos(A + B) formula with A replaced by (90° − A) and B unchanged:
对于 sin(A − B)，我们利用余函数恒等式 sin θ = cos(90° − θ)。令 θ = A − B，则 sin(A − B) = cos(90° − A + B) = cos((90° − A) + B)。在 cos(A + B) 公式中将 A 替换为 (90° − A)，B 不变：
sin(A − B) = sinAcosB − cosAsinB
Finally, for sin(A + B), substitute B → -B into the sin(A – B) formula. The term containing sin(-B) flips sign, giving:
最后，对于 sin(A + B)，把 B 替换为 −B 代入 sin(A − B) 公式。含 sin(−B) 的项改变符号，得到：
sin(A + B) = sinAcosB + cosAsinB
4. Deriving tan(A ± B) | 推导 tan(A ± B)
The tangent addition formulas follow directly from the sine and cosine formulas. Since tan(A + B) = sin(A + B) / cos(A + B), we write:
正切的和角公式可以直接从正弦、余弦公式推出。由于 tan(A + B) = sin(A + B) / cos(A + B)，我们可以写成：
tan(A + B) = (sinAcosB + cosAsinB) / (cosAcosB − sinAsinB)
Divide the numerator and denominator by cosAcosB, assuming cosAcosB ≠ 0. The terms become:
在 cosAcosB ≠ 0 的前提下，分子分母同时除以 cosAcosB，各项变为：
tan(A + B) = (tanA + tanB) / (1 − tanAtanB)
Similarly, using sin(A – B) and cos(A – B):
类似地，利用 sin(A − B) 和 cos(A − B)：
tan(A − B) = (tanA − tanB) / (1 + tanAtanB)
Note the sign changes: the numerator of tan(A – B) has a minus sign, while the denominator has a plus sign. A common student error is mixing up these signs, so it is worth memorising them as a pair.
注意符号变化：tan(A − B) 的分子是减号，而分母是加号。学生常见的错误就是把这两个符号搞混，所以建议把这两个公式成对记忆。
5. Exact Value Calculations | 精确值计算
One of the most direct applications of the angle addition formulas is computing exact values for angles like 15°, 75°, and 105°. Let us work through the classic example of sin 75°.
和角公式最直接的应用之一就是计算像 15°、75°、105° 这类角度的精确值。让我们以 sin 75° 这个经典例子为例。
Notice that 75° = 45° + 30°. Using sin(A + B) = sinAcosB + cosAsinB with A = 45° and B = 30°:
注意 75° = 45° + 30°。利用 sin(A + B) = sinAcosB + cosAsinB，令 A = 45°，B = 30°：
sin 75° = sin45°cos30° + cos45°sin30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4
Similarly, cos 15° = cos(45° – 30°) = cos45°cos30° + sin45°sin30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4. It is a beautiful coincidence that sin 75° and cos 15° are equal — a direct reflection of the cofunction identity sin75° = cos15°.
类似地，cos 15° = cos(45° − 30°) = cos45°cos30° + sin45°sin30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4。有趣的是 sin 75° 恰好等于 cos 15°——这正是余函数恒等式 sin75° = cos15° 的直接体现。
For tangent, consider tan 105° = tan(60° + 45°):
再看正切，求 tan 105° = tan(60° + 45°)：
tan 105° = (tan60° + tan45°) / (1 − tan60°tan45°) = (√3 + 1) / (1 − √3)
This expression can be rationalised to −(√3 + 1)²/2 = −(4 + 2√3)/2 = −2 − √3. Always rationalise the denominator when presenting final exact answers.
这个表达式可以通过有理化分母化为 −(√3 + 1)²/2 = −(4 + 2√3)/2 = −2 − √3。在给出最终精确答案时，务必对分母进行有理化。
6. Simplifying Trigonometric Expressions | 化简三角函数表达式
The angle addition formulas can also be applied in reverse to simplify expressions. For instance, consider the expression sin x cos 30° + cos x sin 30°. Recognising that this has the form sin(A + B) with A = x and B = 30°, we can collapse it instantly to sin(x + 30°).
和角公式反过来运用也可以化简表达式。例如，考虑表达式 sin x cos 30° + cos x sin 30°。注意到它的形式是 sin(A + B)，其中 A = x，B = 30°，于是可以立刻合并为 sin(x + 30°)。
This technique is especially powerful when dealing with expressions involving different amplitudes and phases. Consider the expression √3 sin x + cos x. Can we write it as a single sine function? We want the form R sin(x + α), where R > 0. Expanding R sin(x + α) gives R sin x cos α + R cos x sin α. Matching coefficients:
这种技术在处理包含不同振幅和相位的表达式时尤为强大。考虑表达式 √3 sin x + cos x。我们能把它写成单个正弦函数吗？我们想写成 R sin(x + α) 的形式，其中 R > 0。展开 R sin(x + α) 得到 R sin x cos α + R cos x sin α。对比系数：
R cos α = √3, R sin α = 1
Squaring and adding gives R² = 3 + 1 = 4, so R = 2. Dividing the two equations gives tan α = 1/√3, so α = 30°. Therefore:
将两式平方后相加得到 R² = 3 + 1 = 4，所以 R = 2。两式相除得 tan α = 1/√3，所以 α = 30°。因此：
√3 sin x + cos x = 2 sin(x + 30°)
This single-wave form makes it much easier to sketch the graph or solve equations involving the original expression. The general method — writing a sin x + b cos x as R sin(x + α) — is an exam favourite across all major boards.
这种单波形式极大地方便了作图或求解包含原表达式的方程。将 a sin x + b cos x 写成 R sin(x + α) 的通用方法是各大考试局的高频考点。
7. Solving Trigonometric Equations | 解三角方程
The addition formulas often unlock equations that initially look intractable. Consider the equation:
和角公式常常能解开看似棘手的方程。考虑方程：
cos x cos 40° + sin x sin 40° = 0.5, 0° ≤ x ≤ 360°
The left-hand side is exactly cos(x – 40°) by the reverse of the cos(A – B) formula. So the equation becomes:
左边恰好是 cos(x − 40°)，这是 cos(A − B) 公式的逆用。因此方程变为：
cos(x − 40°) = 0.5
The general solution for cos θ = 0.5 is θ = 60° + 360°n or θ = 300° + 360°n. Thus x − 40° = 60° or 300° (within the range), giving x = 100° or x = 340°. The addition formula transformed a clumsy equation into a trivial one.
cos θ = 0.5 的通解是 θ = 60° + 360°n 或 θ = 300° + 360°n。因此在给定范围内 x − 40° = 60° 或 300°，解得 x = 100° 或 x = 340°。和角公式把一个笨拙的方程变成了一个简单问题。
8. Double-Angle and Half-Angle Connections | 二倍角与半角公式的联系
A major consequence of the addition formulas is the double-angle identities. Setting B = A in sin(A + B), we get:
和角公式最重要的推论之一就是二倍角公式。在 sin(A + B) 中令 B = A，得到：
sin 2A = 2 sinA cosA
Similarly, setting B = A in cos(A + B):
类似地，在 cos(A + B) 中令 B = A：
cos 2A = cos²A − sin²A
Using the Pythagorean identity, this can be rewritten in two additional forms:
利用毕达哥拉斯恒等式，这个公式还可以写成另外两种形式：
cos 2A = 2cos²A − 1 = 1 − 2sin²A
These alternative forms are the gateway to half-angle identities. Rearranging 2cos²A = 1 + cos 2A gives cos²A = (1 + cos 2A)/2. Replacing A with θ/2 yields:
这些变形是通往半角公式的桥梁。由 2cos²A = 1 + cos 2A 变形得 cos²A = (1 + cos 2A)/2。将 A 替换为 θ/2 得到：
cos²(θ/2) = (1 + cos θ)/2, sin²(θ/2) = (1 − cos θ)/2
Thus every half-angle formula is ultimately a shadow of the original angle addition formulas. In calculus, this relationship is essential for integrating even powers of sine and cosine.
因此，每一个半角公式归根结底都是和角公式的影子。在微积分中，这一关系对于积分正弦、余弦的偶次幂至关重要。
9. Applications in Geometry and Calculus | 在几何与微积分中的应用
In geometry, the addition formulas allow us to find the angle between two lines from their gradients. If two lines have gradients m₁ and m₂, the tangent of the angle θ between them is:
在几何中，和角公式让我们能够通过两条直线的斜率求出它们的夹角。若两条直线的斜率分别为 m₁ 和 m₂，则它们的夹角 θ 满足：
tan θ = |(m₂ − m₁) / (1 + m₁m₂)|
This formula is simply tan(B – A) adapted to slopes, since the angle of a line relative to the x-axis is related to its slope by m = tan φ.
这个公式本质上是 tan(B − A) 的斜率版本，因为直线与 x 轴的夹角 φ 满足 m = tan φ。
In calculus, the derivative of sin x is derived by evaluating the limit of [sin(x + h) − sin x] / h as h approaches 0. Expanding sin(x + h) using the addition formula:
在微积分中，sin x 的导数通过求 [sin(x + h) − sin x] / h 在 h 趋于 0 时的极限得到。用和角公式展开 sin(x + h)：
sin(x + h) = sin x cos h + cos x sin h
So the difference quotient becomes sin x(cos h − 1)/h + cos x(sin h / h). Using the standard limits, this approaches sin x · 0 + cos x · 1 = cos x. Without the addition formula, this derivation would be impossible.
差商于是变为 sin x(cos h − 1)/h + cos x(sin h / h)。利用标准极限，它趋近于 sin x · 0 + cos x · 1 = cos x。没有和角公式，这一推导就无从谈起。
10. Common Mistakes and Exam Tips | 常见错误与应试建议
Several errors recur among students when applying the addition formulas. First, mixing up the signs: cos(A + B) has a minus sign between the terms, while sin(A + B) has a plus sign. Second, forgetting that the formulas work for any real angles, not just acute ones; the geometric derivation assumes positive acute angles, but the identities are valid universally. Third, applying the tangent formula when tan A or tan B is undefined — for example when A or B is 90° — because the denominator cosAcosB becomes zero.
学生在应用和角公式时，有几个反复出现的错误。第一，混淆符号：cos(A + B) 中间是减号，而 sin(A + B) 中间是加号。第二，忘记公式对任意实数角都成立，而不仅仅是锐角；几何推导假设的是正的锐角，但恒等式普遍成立。第三，当 tan A 或 tan B 不存在时——例如 A 或 B 为 90° 时——仍然使用正切公式，因为此时 cosAcosB 为零。
For exams, always check whether the question asks for an exact value or a decimal approximation. When simplifying, write your final answer with rationalised denominators and a positive sign in front of the surd if possible. Practice recognising the reverse form: any expression of the form sin x cos y + cos x sin y can be collapsed to sin(x + y). This recognition saves valuable time in an exam setting.
考试时，务必看清题目要求的是精确值还是小数近似值。化简时，最终答案要有理化分母，并在可能的情况下将根号前的符号整理为正号。同时要练习识别逆向形式：任何形如 sin x cos y + cos x sin y 的表达式都可以合并为 sin(x + y)。这种辨识能力在考场上能为你节省宝贵的时间。
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