📚 Application of de Moivre's Theorem in Establishing Trigonometric Identities | 棣莫弗定理在建立三角恒等式中的应用
1. Introduction | 引言
De Moivre's theorem is one of the most elegant results in the AQA A-Level Further Mathematics Core Pure syllabus. It connects complex numbers with circular functions, enabling us to derive a remarkably wide range of trigonometric identities using only binomial expansion and elementary algebra.
棣莫弗定理是AQA A-Level进阶数学Core Pure课程中最优雅的结论之一。它将复数与三角函数联系起来，使我们只需借助二项式展开和初等代数，就能推导出种类繁多的三角恒等式。
Students who master this technique can quickly obtain identities for cos(nθ), sin(nθ), tan(nθ), and also express powers such as cosⁿθ or sinⁿθ as linear combinations of multiple-angle terms. This article develops the method step by step, with fully worked examples and exam-style practice.
掌握这一技巧的学生可以快速获得cos(nθ)、sin(nθ)、tan(nθ)的恒等式，还能将cosⁿθ或sinⁿθ等幂次形式表示为倍角项的线性组合。本文循序渐进地讲解该方法，包含完整例题与考试风格的练习。
2. De Moivre's Theorem and Key Conjugate Forms | 棣莫弗定理与关键的共轭形式
Theorem. For any integer n and any real angle θ,
(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)
定理：对任意整数 n 和任意实数角 θ，有
(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)
Writing z = cos θ + i sin θ = eiθ, we obtain zⁿ = einθ in θ, since the modulus of z is 1. These conjugate relationships are the engine behind most derivations in this topic.
记 z = cos θ + i sin θ = eiθ，则 zⁿ = einθ = cos(nθ) + i sin(nθ)。其余模为1，故取倒数得 1/z = cos θ − i sin θ。这些共轭关系是本节大多数推导的核心动力。
Two combinations appear constantly and are worth memorising:
两个常用组合反复出现，值得牢记：
- z + 1/z = 2 cos θ
- z − 1/z = 2 i sin θ
These allow us to convert between powers of z and multiple-angle trigonometric functions, which is the basis of Sections 8 and 9.
它们使我们能够在z的幂与倍角三角函数之间相互转换，这是第8、9节的基础。
3. The General Method: Expanding (cos θ + i sin θ)ⁿ | 一般方法：展开(cos θ + i sin θ)ⁿ
The core strategy for deriving cos(nθ) or sin(nθ) is to expand the left-hand side of de Moivre's theorem using the binomial theorem, then equate real and imaginary parts with the right-hand side.
推导cos(nθ)或sin(nθ)的核心策略是：用二项式定理展开棣莫弗定理的左端，然后令实部与虚部分别等于右端的对应部分。
For general n, the binomial expansion gives the following alternating patterns:
对一般的 n，二项式展开给出如下交替模式：
cos(nθ) = cosⁿθ − C(n,2) cosⁿ⁻²θ sin²θ + C(n,4) cosⁿ⁻⁴θ sin⁴θ − …
sin(nθ) = C(n,1) cosⁿ⁻¹θ sinθ − C(n,3) cosⁿ⁻³θ sin³θ + C(n,5) cosⁿ⁻⁵θ sin⁵θ − …
Notice that the series for cos(nθ) contains only even powers of sinθ, while the series for sin(nθ) contains only odd powers of sinθ. This is because i raised to an even power is real, while i raised to an odd power is purely imaginary.
注意：cos(nθ)的级数中只含sinθ的偶次幂，而sin(nθ)的级数中只含sinθ的奇次幂。这是因为 i 的偶次幂为实数，奇次幂为纯虚数。
4. Worked Example: n = 3 | 完整例题：n = 3
Expand (cos θ + i sin θ)³ using the binomial theorem:
用二项式定理展开(cos θ + i sin θ)³：
(cos θ + i sin θ)³ = cos³θ + 3i cos²θ sinθ − 3 cosθ sin²θ − i sin³θ
ting real parts with cos(3θ) gives cos 3θ = cos³θ − 3 cosθ sin²θ. Using sin²θ = 1 − cos²θ:
令实部等于cos(3θ)，得 cos 3θ = cos³θ − 3 cosθ sin²θ。利用 sin²θ = 1 − cos²θ：
cos 3θ = cos³θ − 3 cosθ(1 − cos²θ) = 4 cos³θ − 3 cosθ
Equating imaginary parts gives sin 3θ = 3 cos²θ sinθ − sin³θ. Using cos²θ = 1 − sin²θ:
令虚部等于sin(3θ)，得 sin 3θ = 3 cos²θ sinθ − sin³θ。利用 cos²θ = 1 − sin²θ：
sin 3θ = 3(1 − sin²θ) sinθ − sin³θ = 3 sinθ − 4 sin³θ
These two identities are standard results in the AQA formula book but are frequently re-derived in exam questions to test understanding of the binomial method.
这两个恒等式是AQA公式册中的标准结果，但考试题常要求重新推导，以考查对二项式方法的理解。
5. Expressing cos(nθ) in Terms of cos θ Only | 只用cos θ表示cos(nθ)
ower of sinθ can be replaced, so cos(nθ) can always be written as a polynomial in cosθ alone.
因为 sin²θ = 1 − cos²θ，sinθ的每个偶次幂都可以被替换，因此cos(nθ)总可以写成只含cosθ的多项式。
For n = 4, we derive the important quartic formula:
对 n = 4，我们推导重要的四次公式：
cos 4θ = cos⁴θ − 6 cos²θ sin²θ + sin⁴θ
= cos⁴θ − 6 cos²θ(1 − cos²θ) + (1 − cos²θ)² = 8 cos⁴θ − 8 cos²θ + 1
cos 0 = 1. At θ = π/4, it gives 8(1/4) − 8(1/2) + 1 = 2 − 4 + 1 = −1, matching cos π = −1.
验证：当 θ = 0 时，公式给出 8 − 8 + 1 = 1，与 cos 0 = 1 相符；当 θ = π/4 时，给出 8(1/4) − 8(1/2) + 1 = 2 − 4 + 1 = −1，与 cos π = −1 相符。
tor results such as cos 5θ = 16 cos⁵θ − 20 cos³θ + 5 cosθ = cosθ(16 cos⁴θ − 20 cos²θ + 5).
一般规律是：cos(nθ)是cosθ的n次多项式。当n为奇数时只含cosθ的奇次幂；当n为偶数时只含偶次幂。正因如此，才能分解出如 cos 5θ = 16 cos⁵θ − 20 cos³θ + 5 cosθ = cosθ(16 cos⁴θ − 20 cos²θ + 5) 的因式。
6. Expressing sin(nθ) in Terms of sinθ and cosθ | 用sinθ与cosθ表示sin(nθ)
For sin(nθ), the expansion contains exactly one factor of sinθ in every term, so we can factor out sinθ:
对于sin(nθ)，展开式的每一项都恰好含有一个sinθ因子，因此可以提取公因式sinθ：
sin(nθ
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