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FP2 2025 June Q9

A Level / Edexcel / FP2

IAL 2025 June Paper · Question 9

题目

Problem

9.

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

(a) Use de Moivre’s theorem to show that

sin⁡5θ≡ asin⁡5θ+bsin⁡3θ+csin⁡θ\begin{align*} \sin5\theta \equiv&\, a\sin^5\theta+b\sin^3\theta+c\sin\theta \end{align*}

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

(5)

(b) Hence determine the possible exact values of sin⁡2(kπ5)\sin^2\bigg(\frac{k\pi}{5}\bigg) where k∈Zk\in\mathbb{Z}

(4)
题目中文翻译

本题必须展示所有计算步骤。

完全依赖计算器技术得出的解答不予接受。

(a) 使用棣莫弗定理证明

sin⁡5θ≡ asin⁡5θ+bsin⁡3θ+csin⁡θ,\begin{align*} \sin5\theta \equiv&\,a\sin^5\theta+b\sin^3\theta+c\sin\theta, \end{align*}

并求出整数 aa、bb、cc 的值。

(5)

(b) 由此求 sin⁡2(kπ5)\sin^2\big(\frac{k\pi}5\big) 的所有可能精确值，其中 k∈Zk\in\mathbb Z。

(4)

解答

(a)

解法一

思路

展开

用 De Moivre’s theorem 展开 (cos⁡θ+isin⁡θ)5(\cos\theta+\mathrm{i}\sin\theta)^5，取虚部得到 sin⁡5θ\sin5\theta。然后用 cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta 把式子全部写成 sin⁡θ\sin\theta 的幂。

答题过程

展开

By De Moivre’s theorem,

(cos⁡θ+isin⁡θ)5=cos⁡5θ+isin⁡5θ\begin{align*} (\cos\theta+\mathrm{i}\sin\theta)^5 =\cos5\theta+\mathrm{i}\sin5\theta \end{align*}

Taking imaginary parts after expanding,

sin⁡5θ= 5cos⁡4θsin⁡θ−10cos⁡2θsin⁡3θ+sin⁡5θ\begin{align*} \sin5\theta =&\, 5\cos^4\theta\sin\theta -10\cos^2\theta\sin^3\theta +\sin^5\theta \end{align*}

Use

cos⁡2θ=1−sin⁡2θ\begin{align*} \cos^2\theta=1-\sin^2\theta \end{align*}

Then

sin⁡5θ= 5(1−sin⁡2θ)2sin⁡θ −10(1−sin⁡2θ)sin⁡3θ +sin⁡5θ= 5(1−2sin⁡2θ+sin⁡4θ)sin⁡θ −10sin⁡3θ+10sin⁡5θ+sin⁡5θ= 5sin⁡θ−10sin⁡3θ+5sin⁡5θ −10sin⁡3θ+11sin⁡5θ= 16sin⁡5θ−20sin⁡3θ+5sin⁡θ\begin{align*} \sin5\theta =&\, 5(1-\sin^2\theta)^2\sin\theta\\[2mm] &\,\hspace{2pt}-10(1-\sin^2\theta)\sin^3\theta\\[2mm] &\,\hspace{4pt}+\sin^5\theta\\[3mm] =&\, 5(1-2\sin^2\theta+\sin^4\theta)\sin\theta\\[2mm] &\,\hspace{2pt}-10\sin^3\theta +10\sin^5\theta +\sin^5\theta\\[3mm] =&\, 5\sin\theta-10\sin^3\theta+5\sin^5\theta\\[2mm] &\,\hspace{2pt}-10\sin^3\theta +11\sin^5\theta\\[3mm] =&\, 16\sin^5\theta-20\sin^3\theta+5\sin\theta \end{align*}

Therefore

a=16,b=−20,c=5\begin{align*} \boxed{a=16,\quad b=-20,\quad c=5} \end{align*}

(b)

解法一

思路

展开

令 θ=kπ5\theta=\frac{k\pi}{5}，则 5θ=kπ5\theta=k\pi，所以 sin⁡5θ=0\sin5\theta=0。把 (a) 的式子设为 00，提取 sin⁡θ\sin\theta 后，剩下是关于 sin⁡2θ\sin^2\theta 的二次方程。

答题过程

展开

Let

θ=kπ5\begin{align*} \theta=\frac{k\pi}{5} \end{align*}

where k∈Zk\in\mathbb{Z}. Then

sin⁡5θ=sin⁡(kπ)=0\begin{align*} \sin5\theta=\sin(k\pi)=0 \end{align*}

Using part (a),

16sin⁡5θ−20sin⁡3θ+5sin⁡θ= 0sin⁡θ(16sin⁡4θ−20sin⁡2θ+5)= 0\begin{align*} 16\sin^5\theta-20\sin^3\theta+5\sin\theta =&\, 0\\[2mm] \sin\theta \bigg( 16\sin^4\theta-20\sin^2\theta+5 \bigg) =&\, 0 \end{align*}

One possibility is

sin⁡2θ=0\begin{align*} \sin^2\theta=0 \end{align*}

For the other possibilities, let

u=sin⁡2θ\begin{align*} u=\sin^2\theta \end{align*}

Then

16u2−20u+5=0\begin{align*} 16u^2-20u+5=0 \end{align*}

Solve:

u= 20±(−20)2−4(16)(5)32= 20±8032= 20±4532= 5±58\begin{align*} u =&\, \frac{20\pm\sqrt{(-20)^2-4(16)(5)}}{32}\\[3mm] =&\, \frac{20\pm\sqrt{80}}{32}\\[3mm] =&\, \frac{20\pm4\sqrt5}{32}\\[3mm] =&\, \frac{5\pm\sqrt5}{8} \end{align*}

Therefore the possible exact values are

0,5−58,5+58\begin{align*} \boxed{ 0,\quad \frac{5-\sqrt5}{8},\quad \frac{5+\sqrt5}{8} } \end{align*}