📚 Double Angle Formulas and Their Applications | IB数学：二倍角公式及其应用
The double angle formulas are among the most frequently used identities in IB Mathematics, appearing in trigonometry, calculus, and geometry. They express trigonometric functions of 2θ in terms of functions of θ, allowing us to simplify expressions, solve equations, and evaluate integrals.
二倍角公式是IB数学中最常用的恒等式之一，出现在三角学、微积分和几何中。它们将 2θ 的三角函数用 θ 的三角函数表示，帮助我们化简表达式、解方程和计算积分。
1. The Basic Double Angle Formulas | 二倍角公式的基本形式
The three fundamental double angle formulas are:
三个基本的二倍角公式是：
sin 2θ = 2 sin θ cos θ
cos 2θ = cos² θ − sin² θ
tan 2θ = 2 tan θ / (1 − tan² θ)
of θ for which both sides are defined. For tan 2θ, we require θ ≠ π/2 + kπ and tan θ ≠ ±1.
这些恒等式对所有使两边都有定义的 θ 成立。对于 tan 2θ，需要 θ ≠ π/2 + kπ 且 tan θ ≠ ±1。
2. The Sine Double Angle Formula | 正弦二倍角公式
The formula sin 2θ = 2sinθcosθ is derived directly from the sine addition formula sin(A+B) = sin Acos B + cos Asin B by setting A = B = θ.
公式 sin 2θ = 2sinθcosθ 直接由正弦和角公式 sin(A+B) = sin Acos B + cos Asin B 令 A = B = θ 得到。
This identity is especially useful in integration when we need to convert a product of sine and cosine into a single sine function.
这个恒等式在积分中特别有用，因为我们可以将正弦和余弦的乘积转化为单一的正弦函数。
For example, ∫ sin 2x dx = -12cos 2x + C.
例如，∫ sin 2x dx = -12cos 2x + C。
3. The Three Forms of Cos 2θ | cos 2θ 的三种形式
The cosine double angle formula has three equivalent forms, each useful in different contexts:
余弦二倍角公式有三种等价形式，每种形式在不同场合下各有用途：
Form 1: cos 2θ = cos2θ - sin2θ — the original form, useful for simplifying expressions with both squares.
形式1： cos 2θ = cos2θ - sin2θ —— 原始形式，适用于含两个平方的表达式。
Form 2: cos 2θ = 2cos2θ - 1 — this gives the power-reduction identity cos2θ = 1+cos 2θ2.
形式2： cos 2θ = 2cos2θ - 1 —— 由此得到降幂公式 cos2θ = 1+cos 2θ2。
Form 3: cos 2θ = 1 - 2sin2θ — this gives sin2θ = 1-cos 2θ2.
形式3： cos 2θ = 1 - 2sin2θ —— 由此得到 sin2θ = 1-cos 2θ2。
These forms are essential for integration of even powers of sine and cosine.
这些形式对计算正弦和余弦偶数次幂的积分至关重要。
4. The Tangent Double Angle Formula | 正切二倍角公式
The tangent double angle formula is derived from the tangent addition formula:
正切二倍角公式由正切和角公式推导得到：
tan 2θ = 2 tan θ / (1 − tan² θ)
This formula is valid when θ ≠ π4 + kπ2 and θ ≠ π2 + kπ. It is commonly used in solving trigonometric equations and in coordinate geometry to find slopes of lines.
该公式在 θ ≠ π4 + kπ2 且 θ ≠ π2 + kπ 时成立。它常用于解三角方程以及解析几何中求直线的斜率。
For example, if tan θ = 12, then tan 2θ = 2 × 12⁄1 - (12)2 = 13/4 = 43.
例如，若 tan θ = 12，则 tan 2θ = 2 × 12⁄1 - (12)2 = 13/4 = 43。
5. Derivation from Angle Addition Formulas | 由和角公式推导
Understanding the derivation helps memorisation and reveals the structure of these identities.
理解推导过程有助于记忆，并能揭示这些恒等式的结构。
Start with the sine addition formula:
从正弦和角公式出发：
sin(A+B) = sin A cos B + cos A sin B
Set A = B = θ to obtain sin 2θ = 2sinθcosθ.
令 A = B = θ，得到 sin 2θ = 2sinθcosθ。
Similarly, cos(A+B) = cos Acos B - sin Asin B gives cos 2θ = cos2θ - sin2θ.
同理，cos(A+B) = cos Acos B - sin Asin B 给出 cos 2θ = cos2θ - sin2θ。
Finally, using tan(A+B) = tan A + tan B1-tan Atan B with A=B=θ yields the tangent double angle formula.
最后，令 tan(A+B) = tan A + tan B1-tan Atan B 中的 A=B=θ，即得正切二倍角公式。
6. Simplifying Trigonometric Expressions | 化简三角表达式
Double angle formulas are powerful tools for rewriting complicated trigonometric expressions into simpler forms.
二倍角公式是将复杂三角表达式改写为简单形式的有力工具。
Example: Simplify sin 2x1+cos 2x.
例：化简 sin 2x1+cos 2x。
Using sin 2x = 2sin xcos x and 1+cos 2x = 1 + (2cos2 x - 1) = 2cos2 x, we get:
使用 sin 2x = 2sin xcos x 和 1+cos 2x = 1 + (2cos2 x - 1) = 2cos2 x，得到：
2sin xcos x⁄2cos2 x = sin xcos x = tan x
Thus the whole expression reduces to tan x, which is much easier to analyse.
因此整个表达式化简为 tan x，更便于分析。
7. Solving Trigonometric Equations | 解三角方程
Double angle formulas often transform equations involving 2x into equations involving x, making them solvable by factoring or using basic trigonometric values.
二倍角公式常把含 2x 的方程转化为含 x 的方程，从而可以通过因式分解或基本三角函数值求解。
Example: Solve cos 2x = cos x for 0 ≤ x < 2π.
例：解方程 cos 2x = cos x，其中 0 ≤ x < 2π。
Using cos 2x = 2cos2 x - 1, the equation becomes:
利用 cos 2x = 2cos2 x - 1，方程变为：
2 cos² x − 1 = cos x → 2 cos² x − cos x − 1 = 0
Factor: (2cos x + 1)(cos x - 1) = 0. Thus cos x = 1 or cos x = -12.
因式分解：(2cos x + 1)(cos x - 1) = 0。因此 cos x = 1 或 cos x = -12。
The solutions are x = 0, 2π3, 4π3 in the given interval.
在给定区间内，解为 x = 0, 2π3, 4π3。
8. Applications in Integration | 在积分中的应用
One of the most common uses of double angle formulas is in evaluating integrals of squared trigonometric functions.
二倍角公式最常见的用途之一是计算三角函数的平方积分。
For example, ∫ sin2 x dx can be evaluated using sin2 x = 1-cos 2x2:
例如，∫ sin2 x dx 可用 sin2 x = 1-cos 2x2 计算：
∫ sin² x dx = ∫ (1 − cos 2x)/2 dx = x/2 − sin 2x/4 + C
Similarly, ∫ cos2 x dx = ∫ (1+cos 2x)/2 dx = x/2 + sin 2x/4 + C.
同理，∫ cos2 x dx = ∫ (1+cos 2x)/2 dx = x/2 + sin 2x/4 + C。
These integrals appear frequently in IB HL calculus questions and in finding areas under curves.
这些积分在IB高级水平微积分题目以及求曲线面积时频繁出现。
9. Geometric and Physical Applications | 几何与物理应用
Double angle formulas are not just abstract identities; they have concrete applications in geometry and physics.
二倍角公式不是抽象的恒等式，它们在几何和物理中有具体应用。
In projectile motion, the range R of a projectile launched with speed u at angle θ is given by R = u2 sin 2θ⁄g. Using sin 2θ = 2sinθcosθ or recognizing that maximum range occurs when sin 2θ = 1 (i.e., θ = 45circ) is a direct application.
在抛体运动中，以速度 u、角度 θ 发射的物体射程为 R = u2 sin 2θ⁄g。利用 sin 2θ = 2sinθcosθ 或识别最大射程出现在 sin 2θ = 1 即 θ = 45circ 时，就是直接应用。
In geometry, the area of an isosceles triangle with equal sides r and included angle 2θ is 12r2sin 2θ = r2sinθcosθ.
在几何中，腰长为 r、顶角为 2θ 的等腰三角形面积为 12r2sin 2θ = r2sinθcosθ。
These examples show how the double angle formula simplifies real-world calculations.
这些例子展示了二倍角公式如何简化现实中的计算。
10. Common Mistakes and Tips | 常见错误与技巧
Students often make mistakes when applying double angle formulas. Here are some common pitfalls and how to avoid them:
学生在应用二倍角公式时常犯错。以下是一些常见陷阱及避免方法：
Mistake: Writing sin 2θ = 2sinθ without the cosθ factor. Remember the factor of cosθ is essential.
错误：将 sin 2θ 写成 2sinθ 而漏掉 cosθ。记住 cosθ 因子必不可少。
Mistake: Confusing cos 2θ = cos2θ - sin2θ with cos2θ + sin2θ = 1. The plus sign is only for Pythagorean identity.
错误：将 cos 2θ = cos2θ - sin2θ 与 cos2θ + sin2θ = 1 混淆。加号只属于勾股恒等式。
Tip: When solving equations, always check the domain of θ and whether tan 2θ is undefined.
技巧：解方程时，始终检查 θ 的定义域以及 tan 2θ 是否无定义。
Tip: For integration of sin2 x or cos2 x, always use power-reduction forms derived from cos 2θ.
技巧：计算 sin2 x 或 cos2 x 的积分时，务必使用由 cos 2θ 导出的降幂公式。
11. Practice Problems | 练习与答案
Test your understanding with these selected problems. Solutions are provided below.
请用以下精选题目测试你的理解。答案附在下方。
| Problem 1 | If sin θ = 35 and θ is in quadrant II, find sin 2θ. |
| 问题1 | 已知 sin θ = 35 且 θ 在第二象限，求 sin 2θ。 |
| Problem 2 | Solve tan 2x = 1 for 0 ≤ x < π. |
| 问题2 | 解方程 tan 2x = 1，其中 0 ≤ x < π。 |
| Problem 3 | Evaluate ∫0π/2 cos2 x dx. |
| 问题3 | 计算 ∫0π/2 cos2 x dx。 |
Solutions:
答案：
1. Since θ is in quadrant II, cos θ = -√1-sin2θ = -45. Thus sin 2θ = 2 · 35 · (-45) = -2425.
1. 因为 θ 在第二象限，cos θ = -√1-sin2θ = -45。所以 sin 2θ = 2 · 35 · (-45) = -2425。
2. tan 2x = 1 implies 2x = π4 + kπ, so x = π8 + kπ2. Within 0 ≤ x < π, the solutions are x = π8, 5π8.
2. tan 2x = 1 得 2x = π4 + kπ，所以 x = π8 + kπ2。在 0 ≤ x < π 内，解为 x = π8, 5π8。
3. ∫0π/2 cos2 x dx = ∫0π/2 1+cos 2x2 dx = [x2 + sin 2x4]0π/2 = π4.
3. ∫0π/2 cos2 x dx = ∫0π/2 1+cos 2x2 dx = [x2 + sin 2x4]0π/2 = π4。
12. Summary | 总结
The double angle formulas are indispensable in IB Mathematics. They connect trigonometric functions of 2θ to those of θ, enabling simplification, equation solving, and integration. Mastery of these identities and their derivations will save time in exams and build a strong foundation for higher-level mathematics.
二倍角公式在IB数学中不可或缺。它们将 2θ 的三角函数与 θ 的三角函数联系起来，帮助化简、解方程和积分。熟练掌握这些恒等式及其推导过程，将在考试中节省时间，并为更高阶的数学打下坚实基础。
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图