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IAL 2025 June Q9

A Level / Edexcel / FP2

IAL 2025 June Paper · Question 9

题目

Problem

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\left(\frac{k\pi}{5}\right) where 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 \left( 16\sin^4\theta-20\sin^2\theta+5 \right) ={}& 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*}