Sum and Difference Formulas in Trigonometry | IB数学：和差角公式与推导

📚 Sum and Difference Formulas in Trigonometry | IB数学：和差角公式与推导

The sum and difference formulas are essential tools in trigonometry, allowing us to express trigonometric functions of combined angles in terms of functions of the original angles. They appear frequently in IB Mathematics Analysis and Approaches (AA) and are the foundation for many identities, calculus derivatives, and solving equations.

和差角公式是三角学中的重要工具，它让我们能够用原来角度的三角函数来表示复合角度的三角函数。在IB数学分析与方法（AA）中，这些公式频繁出现，也是许多恒等式、微积分求导以及解方程的基础。

1. The Sum and Difference Formulas | 和差角公式一览

There are six core formulas, grouped by sine, cosine, and tangent:

和差角公式共有六个核心式子，按正弦、余弦和正切分组：

cos(A − B) = cos A cos B + sin A sin B
cos(A + B) = cos A cos B − sin A sin B
sin(A + B) = sin A cos B + cos A sin B
sin(A − B) = sin A cos B − cos A sin B
tan(A + B) = (tan A + tan B) / (1 − tan A tan B)
tan(A − B) = (tan A − tan B) / (1 + tan A tan B)

Notice the pattern: the cosine formula has a minus sign inside and a “co-function” structure, while the sine formula keeps the same sign. Many students find the cosine formula the easiest to memorise if they remember: “cos cos, sin sin, with the opposite sign.”

注意其中的规律：余弦公式内部是减号时，展开用加号；内部是加号时，展开用减号。而正弦公式内部符号与展开符号一致。许多同学觉得余弦公式最好记：”cos cos, sin sin，符号相反”。

2. Derivation of cos(A − B) via the Unit Circle | 利用单位圆推导 cos(A − B)

The most elegant derivation of cos(A − B) uses two points on the unit circle. Consider angles A and B measured from the positive x-axis, and place points P and Q on the unit circle:

推导 cos(A − B) 最优雅的方法是利用单位圆上的两个点。考虑从 x 轴正方向量起的角 A 和 B，在单位圆上取点 P 和 Q：

P = (cos A, sin A), Q = (cos B, sin B)

We can compute the distance squared between P and Q using the ordinary coordinate distance formula:

我们可以用普通坐标距离公式计算 P 和 Q 之间距离的平方：

PQ² = (cos A − cos B)² + (sin A − sin B)²

Expanding gives cos²A − 2cosA cosB + cos²B + sin²A − 2sinA sinB + sin²B. Since sin²θ + cos²θ = 1, this simplifies to 2 − 2(cosA cosB + sinA sinB).

展开后得到 cos²A − 2cosA cosB + cos²B + sin²A − 2sinA sinB + sin²B。因为 sin²θ + cos²θ = 1，这个式子化简为 2 − 2(cosA cosB + sinA sinB)。

Now rotate the coordinate axes so that one point lies on the x-axis. The distance PQ is unchanged by rotation. If we rotate by −B, point Q moves to angle 0, and point P moves to angle A − B. Their coordinates become Q’ = (1, 0) and P’ = (cos(A − B), sin(A − B)).

现在将坐标轴旋转，使其中一个点落在 x 轴上。旋转不改变 PQ 的长度。如果我们旋转 −B，点 Q 移到角 0 处，点 P 移到角 A − B 处。它们的坐标变为 Q’ = (1, 0) 和 P’ = (cos(A − B), sin(A − B))。

The distance squared between these rotated points is:

旋转后两点之间距离的平方为：

PQ² = (cos(A−B) − 1)² + (sin(A−B) − 0)²

This simplifies to cos²(A−B) − 2cos(A−B) + 1 + sin²(A−B) = 2 − 2cos(A−B). Equating the two expressions for PQ² gives:

这化简为 cos²(A−B) − 2cos(A−B) + 1 + sin²(A−B) = 2 − 2cos(A−B)。让两个 PQ² 表达式相等，得到：

2 − 2cos(A−B) = 2 − 2(cosA cosB + sinA sinB)

Therefore cos(A − B) = cos A cos B + sin A sin B. This derivation is elegant because it relies only on the distance formula and the unit circle definition of sine and cosine.

因此 cos(A − B) = cos A cos B + sin A sin B。这个推导非常优美，因为它只依赖距离公式以及正弦、余弦在单位圆上的定义。

3. Deriving cos(A + B) | 推导 cos(A + B)

Once cos(A − B) is known, cos(A + B) follows immediately by replacing B with −B:

一旦知道 cos(A − B)，只需将 B 替换为 −B 即可得到 cos(A + B)：

cos(A + B) = cos(A − (−B)) = cos A cos(−B) + sin A sin(−B)

Using cos(−B) = cos B and sin(−B) = −sin B, we obtain:

利用 cos(−B) = cos B 和 sin(−B) = −sin B，我们得到：

cos(A + B) = cos A cos B − sin A sin B

This sign change is critical: addition in the angle becomes subtraction in the product terms involving sine.

这里的符号变化非常关键：角度相加时，涉及正弦的乘积项要变为相减。

4. Deriving sin(A ± B) via Cofunction Identities | 用余角恒等式推导 sin(A ± B)

We can derive the sine formulas from the cosine formulas using the cofunction identity sin θ = cos(π/2 − θ). Let us find sin(A + B):

我们可以利用余角恒等式 sin θ = cos(π/2 − θ) 从余弦公式推导正弦公式。先求 sin(A + B)：

sin(A + B) = cos(π/2 − (A + B)) = cos((π/2 − A) − B)

Apply the cos(A − B) formula with “A” replaced by (π/2 − A) and “B” replaced by B:

将 cos(A − B) 公式中的 A 替换为 (π/2 − A)，B 替换为 B：

= cos(π/2 − A) cos B + sin(π/2 − A) sin B

Now cos(π/2 − A) = sin A and sin(π/2 − A) = cos A, so:

现在 cos(π/2 − A) = sin A，sin(π/2 − A) = cos A，因此：

sin(A + B) = sin A cos B + cos A sin B

Replacing B with −B gives sin(A − B) = sin A cos B − cos A sin B, because sin(−B) = −sin B and cos(−B) = cos B.

将 B 替换为 −B 得到 sin(A − B) = sin A cos B − cos A sin B，因为 sin(−B) = −sin B，cos(−B) = cos B。

5. Deriving tan(A ± B) | 推导 tan(A ± B)

The tangent formulas are derived by writing tan(A + B) = sin(A + B) / cos(A + B) and dividing numerator and denominator by cos A cos B. This is a classic IB derivation that you should be able to reproduce:

正切公式的推导方法是写出 tan(A + B) = sin(A + B) / cos(A + B)，然后分子分母同除以 cos A cos B。这是一个经典的IB推导，你应该能够自己复现：

tan(A + B) = (sin A cos B + cos A sin B) / (cos A cos B − sin A sin B)

Dividing each term by cos A cos B gives:

每一项都除以 cos A cos B，得到：

tan(A + B) = (tan A + tan B) / (1 − tan A tan B)

Similarly, tan(A − B) = (tan A − tan B) / (1 + tan A tan B). The denominator sign is opposite to the angle sign: for A − B we have 1 + tanA tanB.

类似地，tan(A − B) = (tan A − tan B) / (1 + tan A tan B)。分母的符号与角度符号相反：例如 A − B 时，分母为 1 + tanA tanB。

6. Key Consequences: Double-Angle and Product-to-Sum | 重要推论：二倍角与积化和差

By setting A = B in the sum formulas, we immediately obtain the double-angle identities:

在和角公式中令 A = B，我们立刻得到二倍角公式：

sin(2A) = 2 sin A cos A

cos(2A) = cos² A − sin² A = 2 cos² A − 1 = 1 − 2 sin² A

tan(2A) = 2 tan A / (1 − tan² A)

The product-to-sum formulas can also be derived by adding or subtracting sum formulas. These are often used in IB to rewrite products like sin A sin B as sums, which is especially helpful in integration.

积化和差公式也可以通过将和差公式相加或相减推导出来。在IB中，这类公式常用于将 sin A sin B 这样的乘积改写为和，在积分时尤其有用。

7. Worked Example 1: Exact Value of cos 15° | 例题1：求 cos 15° 的精确值

We know that 15° = 45° − 30°. Using the cosine difference formula:

我们知道 15° = 45° − 30°。使用余弦差角公式：

cos 15° = cos(45° − 30°) = cos 45° cos 30° + sin 45° sin 30°

Substituting the exact values:

代入精确值：

= (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2) / 4

This is the exact value of cos 15°. Finding such exact values is a common exam question type in both paper 1 and paper 2.

这就是 cos 15° 的精确值。这类求精确值的问题是试卷1和试卷2中的常见题型。

8. Worked Example 2: Solving an Equation with sin(A + B) | 例题2：利用 sin(A + B) 解方程

Solve sin(x + π/4) = √2 / 2 for 0 ≤ x ≤ 2π.

解方程 sin(x + π/4) = √2 / 2，其中 0 ≤ x ≤ 2π。

Expand using the sine sum formula:

用正弦和角公式展开：

sin x cos(π/4) + cos x sin(π/4) = √2 / 2

Since cos(π/4) = sin(π/4) = √2/2, the equation becomes:

由于 cos(π/4) = sin(π/4) = √2/2，方程变为：

(√2/2)(sin x + cos x) = √2 / 2 → sin x + cos x = 1

One way to solve this is to rewrite sin x + cos x = √2 sin(x + π/4). So √2 sin(x + π/4) = 1, giving sin(x + π/4) = √2/2 again. This confirms the original equation. The solutions are found from sin θ = √2/2:

我们可以把 sin x + cos x = √2 sin(x + π/4) 改写。于是 √2 sin(x + π/4) = 1，即 sin(x + π/4) = √2/2，又回到了原方程。解 sin θ = √2/2 可得：

x + π/4 = π/4, 3π/4, 9π/4, … → x = 0, π/2, 2π

Within 0 ≤ x ≤ 2π, the solutions are x = 0, π/2, 2π. This example shows that simplifying using sum formulas can make a seemingly complicated equation more transparent.

在 0 ≤ x ≤ 2π 内，解为 x = 0, π/2, 2π。这个例子说明使用和差角公式化简能让看似复杂的方程变得更加清晰。

9. Common Pitfalls and How to Avoid Them | 常见错误与避免方法

Pitfall 1: Writing sin(A + B) = sin A + sin B. This is generally false. Always remember the full expansion.

易错点1： 把 sin(A + B) 写成 sin A + sin B。这通常是错误的。一定要记住完整的展开式。

Pitfall 2: Sign errors in cos(A ± B). The formula for cos(A + B) has a minus sign in the middle. A common memory aid: for “plus” angles, cosine uses “minus”; for “minus” angles, cosine uses “plus”.

易错点2： cos(A ± B) 的符号出错。cos(A + B) 展开式中间是减号。常用记忆口诀：角度是加，余弦用减；角度是减，余弦用加。

Pitfall 3: Forgetting restrictions on tangent. The tan(A + B) formula is undefined if cos A cos B = 0, or if tan A tan B = 1. In IB you must state where the formula is valid.

易错点3： 忘记正切公式的限制条件。当 cos A cos B = 0 或 tan A tan B = 1 时，tan(A + B) 公式无定义。在IB中你需要说明公式的有效范围。

Pitfall 4: Mixing up radian and degree mode in exact value problems. Always check the question; IB papers often combine π and degree measures in the same section.

易错点4： 在精确值问题中混淆弧度与角度制。请务必检查题目；IB试卷经常在同一部分混合使用 π 和角度制。

10. Exam Tips and Practice Strategy | 考试技巧与练习策略

In IB Mathematics, sum and difference formulas are not just for memorisation – they must be applied in proofs, solving equations, and even in calculus. Here are some practical tips:

在IB数学中，和差角公式不只是用来记忆的——它们在证明、解方程甚至微积分中都会用到。以下是一些实用建议：

Memorise the six core formulas exactly. Write them out from memory every day for a week before the exam.
记住六个核心公式。 考试前一周每天默写一遍。
Practise deriving them. Many IB questions ask you to “show that” or “prove” an identity. If you can derive the formulas, you can always recover them if you forget.
练习推导过程。 许多IB题目要求你”证明”某个恒等式。如果你会推导公式，即使忘了也能现场推出来。
Connect to other topics. Use these formulas to derive double-angle, half-angle, and sum-to-product identities. This strengthens your overall trigonometry network.
结合其他知识点。 用和差角公式推导二倍角、半角以及和差化积公式，这样能加强你整个三角知识网络。
Use them in calculus. The derivative of sin x is cos x only when x is in radians. Sum formulas help with limits like lim (sin(x+h) − sin x)/h.
在微积分中使用它们。 只有在弧度制下，sin x 的导数才是 cos x。和差公式有助于处理类似 lim (sin(x+h) − sin x)/h 的极限。

11. Summary | 总结

The sum and difference formulas are logical consequences of the unit circle definition of trigonometric functions. Mastering their derivations gives you both intuition and memory fallback. These formulas appear in nearly every IB trigonometry unit, from solving equations to proving identities, and they are essential for success in both Analysis and Approaches and Applications and Interpretation.

和差角公式是三角函数单位圆定义的逻辑推论。掌握它们的推导过程既能加深理解，也能在遗忘时为你提供备用方案。这些公式几乎出现在IB三角学的每一个单元中，从解方程到证明恒等式，对分析与方法以及应用与解释两个方向的考试成功都至关重要。

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Sum and Difference Formulas in Trigonometry | IB数学：和差角公式与推导