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CIE 9231 2024 November Paper 21 Q8

A Level / CIE / FP2

CIE 9231 2024 Nov Paper 21 Paper · Question 8

题目

Problem

(a) By considering the binomial expansion of (z+1z)7\left(z + \frac{1}{z}\right)^7, where z=cosθ+isinθz = \cos\theta + \mathrm{i}\sin\theta, use de Moivre’s theorem to show that

cos7θ=acos7θ+bcos5θ+ccos3θ+dcosθ,\cos^7\theta = a\cos 7\theta + b\cos 5\theta + c\cos 3\theta + d\cos\theta,

where aa, bb, cc and dd are constants to be determined.

[5]

Let

In=014πcosnθdθ.I_n = \int_0^{\frac{1}{4}\pi} \cos^n\theta\,\mathrm{d}\theta.

(b) Show that

nIn=212n+(n1)In2.nI_n = 2^{-\frac{1}{2}n} + (n - 1)I_{n - 2}.
[4]

(c) Using the results given in parts (a) and (b), find the exact value of I9I_9.

[5]
题目中文翻译

(a) 考虑二项式展开 (z+1z)7\left(z + \frac{1}{z}\right)^7,其中 z=cosθ+isinθz = \cos\theta + \mathrm{i}\sin\theta,并使用 De Moivre 定理证明

cos7θ=acos7θ+bcos5θ+ccos3θ+dcosθ\cos^7\theta = a\cos 7\theta + b\cos 5\theta + c\cos 3\theta + d\cos\theta

其中 aabbccdd 为待确定的常数。

In=014πcosnθdθI_n = \int_0^{\frac{1}{4}\pi} \cos^n\theta\,\mathrm{d}\theta

(b) 证明

nIn=212n+(n1)In2nI_n = 2^{-\frac{1}{2}n} + (n - 1)I_{n - 2}

(c) 使用 (a) 和 (b) 部分给出的结果,求 I9I_9 的精确值。

解答