Deriving Trigonometric Identities with Complex Numbers | 利用复数推导三角恒等式

📚 Deriving Trigonometric Identities with Complex Numbers | 利用复数推导三角恒等式

Complex numbers provide an elegant bridge between algebra and trigonometry. By representing trigonometric functions through Euler's formula, many identities that are usually memorised can be derived in a few lines.

复数在代数与三角之间架起了一座优雅的桥梁。借助欧拉公式将三角函数表示为复数形式，许多通常需要死记硬背的恒等式只需几行便可推导出来。

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1. Euler's Formula and the Unit Circle | 欧拉公式与单位圆

Euler's formula states that for any real angle θ: e^iθ = cos θ + i sin θ. This single equation encodes the coordinates of a point on the unit circle.

欧拉公式指出，对任意实数角 θ：e^iθ = cos θ + i sin θ。这一个等式就包含了单位圆上一点的坐标信息。

Because |e^iθ| = 1, every complex number of the form e^iθ lies on the unit circle. The real part gives cos θ and the imaginary part gives sin θ.

因为 |e^iθ| = 1，形如 e^iθ 的每个复数都位于单位圆上。其实部给出 cos θ，虚部给出 sin θ。

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2. Sum of Angles via Complex Multiplication | 通过复数乘法推导和角公式

If we multiply two unit complex numbers e^iα and e^iβ, the exponents add: e^iα · e^iβ = e^(i(α+β)). Expanding both sides gives the sum formulae.

若将两个单位复数 e^iα 与 e^iβ 相乘，指数相加：e^iα · e^iβ = e^(i(α+β))。将两边展开即可得到和角公式。

cos β − sin α sin β) + i(sin α cos β + cos α sin β). On the right, cos(α+β) + i sin(α+β).

左边展开为 α + i sin α)(cos β + i sin β) = (cos α cos β − sin α sin β) + i(sin α cos β + cos α sin β)。右边为 cos(α+β) + i sin(α+β)。

Equating real and imaginary parts yields:

比较实部与虚部可得：

cos(α+β) = cos α cos β − sin α sin β
sin(α+β) = sin α cos β + cos α sin β

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3. De Moivre's Theorem | 棣莫弗定理

Applying Euler's formula to (e^iθ)ⁿ gives e^inθ. Therefore (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). This is De Moivre's theorem, and it holds for all integer n.

将欧拉公式用于 (e^iθ)ⁿ，得到 e^inθ。因此 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。这就是棣莫弗定理，对所有整数 n 均成立。

This theorem is the backbone for deriving multiple-angle identities. When n is large, the left side can be expanded binomially, and the real/imaginary parts give compact expressions for cos(nθ) and sin(nθ).

该定理是推导多倍角公式的核心工具。当 n 较大时，左边可用二项式定理展开，实部与虚部即可给出 cos(nθ) 和 sin(nθ) 的紧凑表达式。

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4. Double Angle Identities | 二倍角公式

Take n = 2 in De Moivre's theorem: (cos θ + i sin θ)² = cos 2θ + i sin 2θ. Expanding the square gives:

在棣莫弗定理中取 n = 2：(cos θ + i sin θ)² = cos 2θ + i sin 2θ。展开平方可得：

(cos² θ − sin² θ) + i(2 sin θ cos θ) = cos 2θ + i sin 2θ

Equating real and imaginary parts immediately yields the double-angle identities:

比较实部与虚部，立刻得到二倍角公式：

• cos 2θ = cos² θ − sin² θ
• sin 2θ = 2 sin θ cos θ

= 1, the first identity may be rewritten as cos 2θ = 2cos² θ − 1 or cos 2θ = 1 − 2sin² θ.

结合 cos² θ + sin² θ = 1，第一个恒等式还可以改写为 cos 2θ = 2cos² θ − 1 或 cos 2θ = 1 − 2sin² θ。

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5. Triple Angle Identities | 三倍角公式

For n = 3, De Moivre's theorem gives (cos θ + i sin θ)³ = cos 3θ + i sin 3θ. The binomial expansion is:

当 n = 3 时，棣莫弗定理给出 (cos θ + i sin θ)³ = cos 3θ + i sin 3θ。二项式展开为：

cos³ θ + 3i cos² θ sin θ − 3 cos θ sin² θ − i sin³ θ

ping real and imaginary parts and using sin² θ = 1 − cos² θ and cos² θ = 1 − sin² θ gives:

合并实部与虚部，并利用 sin² θ = 1 − cos² θ、cos² θ = 1 − sin² θ，可得：

• cos 3θ = 4cos³ θ − 3cos θ
• sin 3θ = 3sin θ − 4sin³ θ

These expressions are very helpful for solving cubic equations and for simplifying trigonometric integrals.

这些表达式在求解三次方程以及化简三角积分时非常有用。

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6. Extracting Real and Imaginary Parts | 提取实部与虚部

m a + ib, then equate a = a and b = b. Each equality is a separate trigonometric identity.

一般策略很简单：将复数方程两端都写成 a + ib 的形式，然后令实部等于实部、虚部等于虚部。每一个等式都是一条独立的三角恒等式。

ith n = 5, the binomial expansion of (cos θ + i sin θ)⁵ yields cos 5θ and sin 5θ directly.

例如，取 n = 5，将 (cos θ + i sin θ)⁵ 作二项式展开，可直接得到 cos 5θ 和 sin 5θ。

This method avoids memorising long lists of identities. It also shows why the real part of (cos θ + i sin θ)ⁿ is always a polynomial in cos θ, while the imaginary part is sin θ times a polynomial in cos θ.

这种方法避免了记忆冗长的公式表。同时它也揭示了为什么 (cos θ + i sin θ)ⁿ 的实部总是 cos θ 的多项式，而虚部是 sin θ 乘以 cos θ 的多项式。

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7. Product-to-Sum Identities | 积化和差公式

Start from Euler's formula and add or subtract equations for α and β:

从欧拉公式出发，对 α 与 β 的表达式进行加或减：

e^(i(α+β)) = e^iα · e^iβ

Taking the real part of both sides already gives the sum formula. But if we instead expand the right side using Euler's formula, we obtain:

取两边的实部已经得到和角公式。但如果改用欧拉公式展开右边，会得到：

cos(α+β) + i sin(α+β) = (cos α + i sin α)(cos β + i sin β)

Multiplying out and rearranging both real and imaginary equations leads to:

展开并分别整理实部方程与虚部方程，可得：

• sin α cos β = ½[sin(α+β) + sin(α−β)]
• cos α cos β = ½[cos(α+β) + cos(α−β)]
• sin α sin β = ½[cos(α−β) − cos(α+β)]

These product-to-sum identities are essential in integration, where products of sine and cosine must be converted into sums.

这些积化和差公式在积分中至关重要，因为需要将正弦、余弦的乘积转化为和差形式。

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8. Sum-to-Product Identities | 和差化积公式

To derive sum-to-product identities, factor e^iθ terms. For example, consider e^iA + e^iB.

为了推导和差化积公式，可以对 e^iθ 项提取公因式。例如考虑 e^iA + e^iB。

e^iA + e^iB = e^(i(A+B)/2) · ( e^(i(A−B)/2) + e^(−i(A−B)/2) )

Notice that e^ix + e^−ix = 2cos x. Also e^ix − e^−ix = 2i sin x. Hence:

注意到 e^ix + e^−ix = 2cos x，以及 e^ix − e^−ix = 2i sin x。因此：

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

These identities are useful for solving trigonometric equations because they convert sums into products, making zero sets easier to analyse.

这些公式在解三角方程时非常有用，因为将和差化为乘积后，更容易分析零点集。

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9. Using Conjugates | 利用共轭复数

The conjugate of e^iθ is e^−iθ. Adding and subtracting a complex number and its conjugate extracts real and imaginary parts elegantly:

e^iθ 的共轭为 e^−iθ。将复数与其共轭相加或相减，可以优雅地提取实部和虚部：

cos θ = (e^iθ + e^−iθ) / 2
sin θ = (e^iθ − e^−iθ) / (2i)

n³ θ = (3sin θ − sin 3θ)/4, express sin θ in exponential form, cube it, and combine terms.

这些指数形式可以简化许多证明。例如，要证明恒等式 sin³ θ = (3sin θ − sin 3θ)/4，只需将 sin θ 写成指数形式，求立方，再合并同类项。

This approach is particularly powerful for evaluating trigonometric powers and integrals in IB Paper 3 style questions.

这种方法对于处理 IB 试卷第三部分风格题目中的三角幂次与积分问题尤其强大。

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10. Worked Example: cos³ x and sin³ x | 例题：cos³ x 与 sin³ x

Question: Derive an identity for cos³ x in terms of cos x and cos 3x.

问题：推导 cos³ x 用 cos x 和 cos 3x 表示的恒等式。

Using Euler's formula:

利用欧拉公式：

cos x = (e^ix + e^−ix) / 2

Cubing both sides:

两边立方：

cos³ x = (1/8)( e^3ix + 3e^ix + 3e^−ix + e^−3ix )

Grouping the conjugate pairs:

将共轭对分组：

cos³ x = (1/4)( e^3ix + e^−3ix ) + (3/4)( e^ix + e^−ix )

Therefore:

因此：

cos³ x = (3cos x + cos 3x) / 4

This identity is often used in integration to reduce the power of a trigonometric function.

该恒等式常用于积分中降低三角函数的次数。

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11. Worked Example: Solving Equations with Multiple Angles | 例题：用多倍角解方程

Question: Solve cos 3θ = cos θ, for 0 ≤ θ < 2π.

问题：解方程 cos 3θ = cos θ，其中 0 ≤ θ < 2π。

Using cos 3θ = 4cos³ θ − 3cos θ, the equation becomes:

利用 cos 3θ = 4cos³ θ − 3cos θ，原方程化为：

4cos³ θ − 4cos θ = 0

Factor:

因式分解：

4cos θ (cos² θ − 1) = 0

So cos θ = 0 or cos θ = ±1. Within the interval, θ = 0, π/2, π, 3π/2.

因此 cos θ = 0 或 cos θ = ±1。在给定区间内，θ = 0, π/2, π, 3π/2。

Notice how expressing everything in terms of cos θ transforms a trigonometric equation into an algebraic one.

注意，将所有项统一为 cos θ 后，三角方程就转化成了代数方程。

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12. Summary and IB Exam Tips | 总结与IB考试建议

The complex-number approach to trigonometric identities relies on three tools: Euler's formula, De Moivre's theorem, and the extraction of real/imaginary parts.

用复数推导三角恒等式的核心工具只有三件：欧拉公式、棣莫弗定理，以及提取实部与虚部。

• Write both sides in standard form. Always convert e^iθ into cos θ + i sin θ before comparing.
• Use symmetry. Pair terms with e^iθ and e^−iθ to obtain cosine; subtract them to obtain sine.
• Practice with small n first. Start with n = 2 and n = 3 before attempting n = 5 or general polynomials.
• Check domains. When solving equations, remember to list all solutions in the required interval.

中文要点：

• 两边写为标准形式。比较前，务必先将 e^iθ 化为 cos θ + i sin θ。
• 利用对称性。将 e^iθ 与 e^−iθ 相加得余弦，相减得正弦。
• 先从小 n 练起。先尝试 n = 2、n = 3，再处理 n = 5 或一般多项式。
• 注意定义域。解方程时，记得在要求的区间内列出所有解。

Mastering these techniques not only saves time in exams but also deepens your understanding of the unity in mathematics.

掌握这些技巧不仅能让你在考试中节省时间，更能加深你对数学内在统一性的理解。

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