📚 Investigation 1 - Double angle formulae | 探究1 - 二倍角公式
Welcome to this guided investigation into the double angle formulae, a fundamental set of identities in IB Mathematics. Whether you are preparing for an exam, crafting an internal assessment, or simply deepening your understanding of trigonometry, mastering these relationships will enable you to simplify expressions, solve equations, and connect geometric and algebraic reasoning with confidence.
欢迎来到关于二倍角公式的引导式探究，这是 IB 数学中一组基本恒等式。无论你是在备考、撰写内部评估，还是仅仅想加深对三角学的理解，掌握这些关系都将使你能够自信地化简表达式、求解方程，并将几何推理与代数推理联系起来。
1. Introduction to Double Angle Formulae | 二倍角公式简介
Double angle formulae express trigonometric functions of twice an angle, such as sin(2θ), cos(2θ), and tan(2θ), in terms of functions of the original angle θ. They are derived directly from the compound angle identities and appear throughout the IB syllabus, from solving trigonometric equations to analysing periodic motion.
二倍角公式将两倍角的三角函数，如 sin(2θ)、cos(2θ) 和 tan(2θ)，用原角 θ 的三角函数表示。它们直接由和角恒等式推导而来，并贯穿整个 IB 课程，从解三角方程到分析周期性运动都有其身影。
2. Prerequisite: Compound Angle Identities | 预备知识：和角恒等式
an A + tan B) / (1 - tan A tan B). The double angle formulae emerge when we set A = B = θ.
在探究二倍角结果之前，我们必须熟练掌握和角（和与差）恒等式。对于两个角 A 和 B，它们分别是：sin(A + B) = sin A cos B + cos A sin B，cos(A + B) = cos A cos B - sin A sin B，以及 tan(A + B) = (tan A + tan B) / (1 - tan A tan B)。当我们令 A = B = θ 时，二倍角公式便随之出现。
3. Deriving sin(2θ) | 推导 sin 2θ
s θ + cos θ sin θ. Combining the two identical terms gives the clean double angle formula:
sin 2θ = 2 sin θ cos θ
从 sin(A + B) 出发，令 A = θ, B = θ，我们得到 sin(θ + θ) = sin θ cos θ + cos θ sin θ。合并两个完全相同的项，得到简洁的二倍角公式：
sin 2θ = 2 sin θ cos θ
This identity is valid for all real θ and shows that the sine of a double angle is twice the product of sine and cosine of the original angle. It is particularly useful when an argument contains both a sine and a cosine factor.
这个恒等式对所有实数 θ 都成立，它表明两倍角的正弦等于原角正弦与余弦乘积的两倍。当表达式中同时含有正弦和余弦因子时，它特别有用。
4. Deriving cos(2θ) | 推导 cos 2θ
A = B = θ gives cos(θ + θ) = cos θ cos θ - sin θ sin θ. This simplifies to the first form:
cos 2θ = cos²θ - sin²θ
利用 cos(A + B) = cos A cos B - sin A sin B 并代入 A = B = θ，得到 cos(θ + θ) = cos θ cos θ - sin θ sin θ。化简后得到第一种形式：
cos 2θ = cos²θ - sin²θ
We will explore alternative versions of this identity shortly, but this form directly highlights the relationship to the Pythagorean identity. Notice that you can also obtain it from the compound cosine identity with a minus sign.
我们很快会探究这个恒等式的其他形式，但这一形式直接凸显了与勾股恒等式的关系。注意，你也可以从带有减号的余弦和角恒等式得到它。
5. Deriving tan(2θ) | 推导 tan 2θ
ting A = B = θ gives tan(θ + θ) = (tan θ + tan θ) / (1 - tan θ tan θ). This simplifies to:
tan 2θ = 2 tan θ / (1 - tan²θ)
正切的二倍角公式可由和角恒等式 tan(A + B) = (tan A + tan B) / (1 - tan A tan B) 得到。令 A = B = θ，得到 tan(θ + θ) = (tan θ + tan θ) / (1 - tan θ tan θ)。化简后得到：
tan 2θ = 2 tan θ / (1 - tan²θ)
This formula is valid only when tan θ and tan 2θ are defined, i.e., θ ≠ π/2 + kπ and 2θ ≠ π/2 + kπ, where k is an integer. The denominator reveals an important restriction: tan²θ ≠ 1, so θ ≠ π/4 + kπ/2.
该公式仅在 tan θ 和 tan 2θ 有定义时成立，即 θ ≠ π/2 + kπ 且 2θ ≠ π/2 + kπ，其中 k 为整数。分母揭示了一个重要限制：tan²θ ≠ 1，因此 θ ≠ π/4 + kπ/2。
6. Geometric Interpretation | 几何解释
A geometric insight can make the double angle identities more tangible. Consider an isosceles triangle or a unit circle: doubling the angle corresponds to rotating a point through twice the arc. For sin 2θ, the formula 2 sin θ cos θ can be understood as twice the area of a right triangle with legs sin θ and cos θ on the unit circle. Similarly, cos 2θ represents the horizontal shift after rotation through 2θ, decomposable via the cosine rule for the triangle formed by two chords.
几何直观可以让二倍角恒等式变得更具体。考虑一个等腰三角形或单位圆：角度加倍对应于将点旋转过两倍的弧长。对于 sin 2θ，公式 2 sin θ cos θ 可以理解为单位圆上以 sin θ 和 cos θ 为直角边的直角三角形面积的两倍。类似地，cos 2θ 表示旋转 2θ 后的水平位移，可通过两条弦构成的三角形的余弦定理进行分解。
Drawing a unit circle and labelling the coordinates (cos θ, sin θ) and (cos 2θ, sin 2θ) helps visualise why the product 2 sin θ cos θ matches the y-coordinate after doubling the angle. This geometric connection reinforces the algebraic derivation.
绘制单位圆并标出坐标 (cos θ, sin θ) 和 (cos 2θ, sin 2θ)，有助于直观理解为何 2 sin θ cos θ 恰好等于角度加倍后的 y 坐标。这种几何联系强化了代数推导的意义。
7. Alternative Forms of cos(2θ) | cos 2θ 的几种等价形式
tten using the Pythagorean identity sin²θ + cos²θ = 1. Expressing sin²θ = 1 - cos²θ gives:
cos 2θ = 2 cos²θ - 1
Conversely, substituting cos²θ = 1 - sin²θ yields:cos 2θ = 1 - 2 sin²θ
利用勾股恒等式 sin²θ + cos²θ = 1，可以将 cos 2θ = cos²θ - sin²θ 改写成其他形式。将 sin²θ = 1 - cos²θ 代入得到：
cos 2θ = 2 cos²θ - 1
反之，将 cos²θ = 1 - sin²θ 代入得到：cos 2θ = 1 - 2 sin²θ
These three forms are interchangeable and are chosen based on the context: the first emphasises the difference of squares, the second is useful for expressing cos 2θ in terms of cosine alone, and the third is handy when sine appears squared. They are essential when integrating or solving equations.
这三种形式可以互换，并根据上下文选用：第一种形式强调平方差，第二种便于只用余弦表示 cos 2θ，而第三种在出现平方正弦时很方便。在积分或求解方程时，它们至关重要。
8. Solving Trigonometric Equations | 解三角方程
either sin θ = 0 or cos θ = 1/2, leading to general solutions within the required domain.
二倍角公式经常出现在诸如 sin 2θ = sin θ 或 cos 2θ + cos θ = 0 的方程中。代入恒等式可将方程转化为单一角度形式。对于 sin 2θ = sin θ，用 2 sin θ cos θ 替换 sin 2θ，得到 2 sin θ cos θ - sin θ = 0，即 sin θ (2 cos θ - 1) = 0。于是 sin θ = 0 或 cos θ = 1/2，从而在给定定义域内求出通解。
- 1 = 0 gives cos θ = 1/2 or cos θ = -1, providing a systematic route to the solution set.
处理 cos 2θ + cos θ = 0 时，利用 cos 2θ = 2 cos²θ - 1，得到 2 cos²θ - 1 + cos θ = 0，这是关于 cos θ 的二次方程。解 2c² + c - 1 = 0 得 cos θ = 1/2 或 cos θ = -1，为求解解集提供了系统路径。
9. Proving Trigonometric Identities | 证明三角恒等式
. The fraction simplifies to (2 sin²θ) / (2 sin θ cos θ) = tan θ, confirming the identity.
二倍角恒等式是证明复杂等式的有力工具。例如，要验证 (1 - cos 2θ) / sin 2θ = tan θ，从左边开始。将 cos 2θ 替换为 1 - 2 sin²θ，sin 2θ 替换为 2 sin θ cos θ：分子变为 1 - (1 - 2 sin²θ) = 2 sin²θ；分母为 2 sin θ cos θ。分式化简为 (2 sin²θ) / (2 sin θ cos θ) = tan θ，从而证得恒等式。
= cos 2θ. This elegantly uses both the difference of squares and the Pythagorean identity.
另一个常见恒等式是 cos⁴θ - sin⁴θ = cos 2θ。将左边因式分解为 (cos²θ - sin²θ)(cos²θ + sin²θ) = (cos²θ - sin²θ)·1 = cos 2θ。这巧妙地将平方差公式和勾股恒等式结合使用。
10. Half-Angle Connections | 半角关联
om cos 2θ = 1 - 2 sin²θ we get sin²(θ/2) = (1 - cos θ)/2, so sin(θ/2) = ±√[(1 - cos θ)/2].
对余弦二倍角形式进行变形可导出半角恒等式。由 cos 2θ = 2 cos²θ - 1，解出 cos²θ：cos²θ = (1 + cos 2θ)/2。将 θ 替换为 θ/2，得到 cos²(θ/2) = (1 + cos θ)/2，因此 cos(θ/2) = ±√[(1 + cos θ)/2]。类似地，由 cos 2θ = 1 - 2 sin²θ 可得 sin²(θ/2) = (1 - cos θ)/2，所以 sin(θ/2) = ±√[(1 - cos θ)/2]。
These half-angle forms are particularly useful in integration and in IB investigations involving nested trigonometric expressions. The sign is chosen based on the quadrant of θ/2.
这些半角形式在积分以及涉及嵌套三角表达式的 IB 探究中特别有用。符号根据 θ/2 所在的象限选取。
11. Common Mistakes and Pitfalls | 常见错误与陷阱
ine. Another pitfall is losing solutions when dividing both sides of an equation by a trigonometric expression; always factorise instead.
一个常见错误是误用正弦二倍角公式：学生可能写成 sin 2θ = 2 sin θ，而漏掉了余弦因子。务必记住 sin 2θ = 2 sin θ cos θ，而不仅仅是正弦的两倍。另一个陷阱是在方程两边同时除以一个三角表达式时丢失解；应始终采用因式分解。
y determining the quadrant can cause sign errors. Always verify your answers by substitution or sketching the unit circle.
对于正切公式，忽视定义域限制 tan²θ ≠ 1 可能引入增根。此外，在使用半角公式时，忘记 ± 号或错误判断象限会导致符号错误。务必通过代入或描绘单位圆来检验答案。
12. Investigation Tasks and Guided Practice | 探究任务与引导练习
To deepen your understanding, conduct a mini-investigation. Task 1: θ and applying double and compound angle identities. Show that sin 3θ = 3 sin θ - 4 sin³θ. Task 2: 2θ = cot θ for 0 ≤ θ < 2π. Transform cot θ into 1/tan θ and solve the resulting equation.
为加深理解，请进行一次小型探究。任务 1： 通过将 3θ 写作 2θ + θ，并运用二倍角和和角恒等式，推导出 sin(3θ) 仅用 sin θ 表示的形式。证明 sin 3θ = 3 sin θ - 4 sin³θ。任务 2： 利用二倍角公式求出 tan 2θ = cot θ 在 0 ≤ θ < 2π 内的所有解。将 cot θ 转换为 1/tan θ 并求解所得方程。
Task 3 (Geometric investigation): On a unit circle, plot points corresponding to θ and 2θ for several values of θ. Verify visually that sin 2θ equals twice the product of sin θ and cos θ by measuring coordinates. Task 4: in 2θ and discuss how this identity can be used to solve equations like sin θ + cos θ = 1/√2. Record your reasoning in a step-by-step format as you would for an IB IA.
任务 3（几何探究）： 在单位圆上，针对若干 θ 值，标出对应于 θ 和 2θ 的点。通过测量坐标，直观验证 sin 2θ 等于 sin θ 与 cos θ 乘积的两倍。任务 4： 证明 (sin θ + cos θ)² = 1 + sin 2θ，并讨论如何利用该恒等式求解诸如 sin θ + cos θ = 1/√2 的方程。按照 IB 内部评估的格式，逐步记录你的推理过程。
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