Double-Angle Formulae | 二倍角公式

📚 Double-Angle Formulae | 二倍角公式

The double-angle formulae are fundamental identities in A-Level Mathematics that express trigonometric functions of 2θ in terms of functions of θ. Derived directly from the compound angle (addition) formulae, they are essential for solving trigonometric equations, proving complex identities, and performing integrations that involve powers of sine and cosine. A thorough understanding of these identities, particularly the three equivalent forms of cos 2θ, is critical for success in the Edexcel A-Level Mathematics specification, where they feature prominently in Pure Mathematics 2 and further pure units.

二倍角公式是 A-Level 数学中将 2θ 的三角函数用 θ 的三角函数表示的基本恒等式。它们直接由和角公式推导而来，是求解三角方程、证明复杂恒等式以及进行含正弦和余弦幂次积分的重要工具。透彻理解这些恒等式，尤其是 cos 2θ 的三种等价形式，对于在 Edexcel A-Level 数学考试中取得好成绩至关重要，相关内容在纯数学 2 及后续纯数单元中占有突出地位。

1. Derivation from Addition Formulae | 从和角公式推导

n B becomes sin(θ + θ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ. Hence, the first key identity is sin 2θ ≡ os B − sin A sin B gives cos(θ + θ) = cos θ cos θ − sin θ sin θ = cos²θ − sin²θ, so cos 2θ ≡ 1 − tan A tan B) yields tan(θ + θ) = (tan θ + tan θ) / (1 − tan θ tan θ), therefore tan 2θ ≡ (2 tan θ) / (1 − tan²θ).

要推导二倍角公式，我们从和角恒等式出发并令 A = B = θ。对于正弦函数：sin(A + B) = sin A cos B + cos A sin B 变为 sin(θ + θ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ。因此第一个关键恒等式是 sin 2θ ≡ 2 sin θ cos θ。对于余弦：cos(A + B) = cos A cos B − sin A sin B 给出 cos(θ + θ) = cos θ cos θ − sin θ sin θ = cos²θ − sin²θ，故 cos 2θ ≡ cos²θ − sin²θ。对于正切：tan(A + B) = (tan A + tan B) / (1 − tan A tan B) 得到 tan(θ + θ) = (tan θ + tan θ) / (1 − tan θ tan θ)，因此 tan 2θ ≡ (2 tan θ) / (1 − tan²θ)。

sin 2θ ≡ 2 sin θ cos θ

cos 2θ ≡ cos²θ − sin²θ

tan 2θ ≡ 2 tan θ / (1 − tan²θ)

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2. sin 2θ Identities and Properties | sin 2θ 恒等式及其性质

s particularly useful because it can be reversed to express a product as a single sine: sin θ cos θ = ½ sin 2θ. This half-angle product form is invaluable when integrating products of sine and cosine. For example, sin 3θ cos 3θ = ½ sin 6θ. It also demonstrates symmetry: sin 2θ attains its maximum when sin θ and cos θ share the same sign, i.e., in the first and third quadrants.

恒等式 sin 2θ = 2 sin θ cos θ 表明两倍角的正弦等于原角正弦与余弦乘积的两倍。这种形式特别有用，因为它可以反转为用单个正弦表示乘积：sin θ cos θ = ½ sin 2θ。这种半角乘积形式在积分正弦和余弦乘积时极具价值。例如，sin 3θ cos 3θ = ½ sin 6θ。它还展示了对称性：当 sin θ 和 cos θ 同号时，即第一和第三象限， sin 2θ 取得最大值。

Verification with exact values confirms the identity: for θ = 30°, sin 60° = √3/2, and 2 sin 30° cos 30° = 2 × (1/2) × (√3/2) = √3/2. Similarly, for θ = 45°, sin 90° = 1 and 2 sin 45° cos 45° = 2 × (√2/2) × (√2/2) = 1.

用精确值验证可以确认该恒等式：对于 θ = 30°，sin 60° = √3/2，而 2 sin 30° cos 30° = 2 × (1/2) × (√3/2) = √3/2。类似地，对于 θ = 45°，sin 90° = 1，且 2 sin 45° cos 45° = 2 × (√2/2) × (√2/2) = 1。

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3. The Three Forms of cos 2θ | cos 2θ 的三种形式

ms. Substituting sin²θ = 1 − cos²θ gives cos 2θ = cos²θ − (1 − cos²θ) = 2 cos²θ − 1. Alternatively, replacing cos²θ with 1 − sin²θ yields cos 2θ = (1 − sin²θ) − sin²θ = 1 − 2 sin²θ. These three equivalent expressions are:

利用恒等式 sin²θ + cos²θ = 1，基本的余弦二倍角公式 cos 2θ = cos²θ − sin²θ 可改写成另外两种形式。将 sin²θ = 1 − cos²θ 代入得 cos 2θ = cos²θ − (1 − cos²θ) = 2 cos²θ − 1。或者，将 cos²θ 替换为 1 − sin²θ 得 cos 2θ = (1 − sin²θ) − sin²θ = 1 − 2 sin²θ。这三种等价表达式为：

cos 2θ = cos²θ − sin²θ

cos 2θ = 2 cos²θ − 1

cos 2θ = 1 − 2 sin²θ

nate cosine. The mixed form cos²θ − sin²θ often helps in factorising or in conjunction with the difference of squares. In integration, we invert these to express sin²θ or cos²θ in terms of cos 2θ, leading to the half-angle formulae: sin²θ = ½(1 − cos 2θ) and cos²θ = ½(1 + cos 2θ).

选择合适的形式是一项关键技能。如果希望仅用正弦表示 cos 2θ，形式 1 − 2 sin²θ 最适用；而 2 cos²θ − 1 用于消除余弦。混合形式 cos²θ − sin²θ 通常在因式分解或结合平方差时有用。在积分中，我们将这些公式反置，用 cos 2θ 表示 sin²θ 或 cos²θ，从而得到半角公式：sin²θ = ½(1 − cos 2θ) 和 cos²θ = ½(1 + cos 2θ)。

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4. tan 2θ Identity and Domain Restrictions | tan 2θ 恒等式及定义域限制

. The denominator 1 − tan²θ = 0 implies tan θ = ±1, i.e., θ = 45° + k·90° (π/4 + k·π/2). At these points tan 2θ is undefined, which corresponds to the asymptotes of the tangent function when 2θ = 90° + n·180°. Additionally, if tan θ itself is undefined (θ = 90° + m·180°), then 2θ = 180° + 2m·180°, where tan 2θ = 0, and the formula can still be interpreted as a limit.

正切的二倍角公式为 tan 2θ = (2 tan θ) / (1 − tan²θ)。该恒等式仅在 tan θ 有定义且分母不为零时成立。分母 1 − tan²θ = 0 意味着 tan θ = ±1，即 θ = 45° + k·90°（π/4 + k·π/2）。在这些点 tan 2θ 无定义，这与正切函数在 2θ = 90° + n·180° 处的渐近线相对应。此外，若 tan θ 本身无定义（θ = 90° + m·180°），则 2θ = 180° + 2m·180°，此时 tan 2θ = 0，且该公式仍可作为极限情况理解。

he extra solutions introduced by the double angle, as well as exclude the forbidden values where the expression becomes undefined.

该恒等式常被用来求解两边都含正切的方程。例如，在求解 tan 2θ = tan θ 时，必须考虑两倍角带来的额外解，并排除表达式无定义的禁用值。

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5. Solving Equations using Double-Angle Formulae | 使用二倍角公式解方程

Double-angle formulae are essential for solving trigonometric equations where terms involve different multiples of the angle. The strategy is to rewrite all trigonometric functions in terms of the same angle, usually θ, and then solve the resulting equation.

二倍角公式对于求解包含不同角度倍数的三角方程至关重要。解题策略是将所有三角函数改写为同一角度（通常为 θ）的函数，然后求解所得方程。

Example 1: Solve sin 2θ = cos θ for 0° ≤ θ ≤ 360°.
the given interval are θ = 90°, 270° (from cos θ = 0) and θ = 30°, 150° (from sin θ = ½).

例 1： 在 0° ≤ θ ≤ 360° 内解 sin 2θ = cos θ。
利用 sin 2θ = 2 sin θ cos θ，得 2 sin θ cos θ = cos θ。移项：cos θ (2 sin θ − 1) = 0。因此 cos θ = 0 或 sin θ = ½。给定区间内的解为 θ = 90°, 270°（来自 cos θ = 0）以及 θ = 30°, 150°（来自 sin θ = ½）。

Example 2: Solve cos 2θ = sin θ for 0 ≤ θ ≤ 2π.
n θ = ½ or sin θ = −1. The solutions are θ = π/6, 5π/6 (sin θ = ½) and θ = 3π/2 (sin θ = −1). All values lie within 0 ≤ θ ≤ 2π.

例 2： 在 0 ≤ θ ≤ 2π 内解 cos 2θ = sin θ。
选用形式 cos 2θ = 1 − 2 sin²θ，得到 1 − 2 sin²θ = sin θ。整理得 2 sin²θ + sin θ − 1 = 0，因式分解为 (2 sin θ − 1)(sin θ + 1) = 0。于是 sin θ = ½ 或 sin θ = −1。解为 θ = π/6, 5π/6（来自 sin θ = ½）以及 θ = 3π/2（来自 sin θ = −1）。所有值均在 0 ≤ θ ≤ 2π 范围内。

When solving such equations, always check that the solutions do not make any original expressions undefined, especially when tangent is involved.

解此类方程时，始终要检查解是否会使原表达式无定义，尤其在涉及正切时更应留意。

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6. Proving Trigonometric Identities | 证明三角恒等式

Double-angle formulae are powerful tools for proving more complex identities. The approach is typically to start with the more complicated side and reduce it to a simpler form using one or more double-angle substitutions.

二倍角公式是证明更复杂恒等式的有力工具。通常从较复杂的一边入手，运用一个或多个二倍角代换将其化简为更简单的形式。

Proof 1: Prove that (1 − cos 2θ) / (1 + cos 2θ) ≡ tan²θ.
cos²θ − 1) = 2 cos²θ. Hence the fraction becomes 2 sin²θ / (2 cos²θ) = tan²θ, as required.

证明 1： 证明 (1 − cos 2θ) / (1 + cos 2θ) ≡ tan²θ。
从左式开始，分子用 cos 2θ = 1 − 2 sin²θ：1 − (1 − 2 sin²θ) = 2 sin²θ。分母用 cos 2θ = 2 cos²θ − 1：1 + (2 cos²θ − 1) = 2 cos²θ。因此分式化为 2 sin²θ / (2 cos²θ) = tan²θ，得证。

Proof 2: Show that sin 2θ / (1 + cos 2θ) ≡ tan θ.
Substitute sin 2θ = 2 sin θ cos θ and cos 2

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