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CIE 9231 2025 June Paper 21 Q3

A Level / CIE / FP2

CIE 9231 2025 June Paper 21 Paper · Question 3

题目

Problem

By considering the binomial expansion of (z−1z)5\big(z - \frac{1}{z}\big)^5, where z=cos⁡θ+isin⁡θz = \cos\theta + \mathrm{i}\sin\theta, use de Moivre’s theorem to show that

cosec⁡5θ=asin⁡5θ+bsin⁡3θ+csin⁡θ,\cosec^5\theta = \frac{a}{\sin 5\theta + b\sin 3\theta + c\sin\theta},

where aa, bb and cc are integers to be determined.

[6]
题目中文翻译

通过考虑 (z−1z)5\big(z - \frac{1}{z}\big)^5 的二项展开,其中 z=cos⁡θ+isin⁡θz = \cos\theta + \mathrm{i}\sin\theta,使用 de Moivre 定理证明

cosec⁡5θ=asin⁡5θ+bsin⁡3θ+csin⁡θ,\cosec^5\theta = \frac{a}{\sin 5\theta + b\sin 3\theta + c\sin\theta},

其中 aa、bb 和 cc 是需要确定的整数。

(6)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

展开

展开 (z−z−1)5(z-z^{-1})^5,把指数互为相反数的项配对。用 de Moivre 定理把这些差改写成正弦,最后在 sin⁡θ≠0\sin\theta\ne0 时取倒数。

答题过程

展开

By de Moivre’s theorem zm−z−m=2isin⁡(mθ)z^m-z^{-m}=2\mathrm{i}\sin(m\theta) for integer mm. The binomial expansion is

(z−z−1)5= z5−5z3+10z−10z−1 +5z−3−z−5= (z5−z−5)−5(z3−z−3) +10(z−z−1).\begin{align*} (z-z^{-1})^5=&\,z^5-5z^3+10z-10z^{-1}\\ &\,\hspace{2pt}+5z^{-3}-z^{-5}\\ =&\,(z^5-z^{-5})-5(z^3-z^{-3})\\ &\,\hspace{2pt}+10(z-z^{-1}). \end{align*}

Since z−z−1=2isin⁡θz-z^{-1}=2\mathrm{i}\sin\theta, dividing by 2i2\mathrm{i} gives

16sin⁡5θ=sin⁡5θ−5sin⁡3θ+10sin⁡θ.16\sin^5\theta=\sin5\theta-5\sin3\theta+10\sin\theta.

For sin⁡θ≠0\sin\theta\ne0 we therefore have

csc⁡5θ=16sin⁡5θ−5sin⁡3θ+10sin⁡θ,(a,b,c)=(16,−5,10).\csc^5\theta=\frac{16}{\sin5\theta-5\sin3\theta+10\sin\theta}, \qquad (a,b,c)=(16,-5,10).