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IAL 2024 June Q9

A Level / Edexcel / FP2

IAL 2024 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

cos⁡6θ≡32cos⁡6θ−48cos⁡4θ+18cos⁡2θ−1\begin{align*} \cos 6\theta \equiv{}& 32\cos^6 \theta - 48\cos^4 \theta + 18\cos^2 \theta - 1 \end{align*}
(4)

(b) Hence determine the smallest positive root of the equation

48x6−72x4+27x2−1=0\begin{align*} 48x^6 - 72x^4 + 27x^2 - 1 = 0 \end{align*}

giving your answer to 33 decimal places.

(4)

解答

(a)

解法一

思路

展开

用 De Moivre’s theorem 展开 (cos⁡θ+isin⁡θ)6(\cos\theta+\mathrm{i}\sin\theta)^6，取实部。再用 sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta 把所有项化成 cos⁡θ\cos\theta。

答题过程

展开

By De Moivre’s theorem,

(cos⁡θ+isin⁡θ)6=cos⁡6θ+isin⁡6θ\begin{align*} (\cos\theta+\mathrm{i}\sin\theta)^6 =\cos6\theta+\mathrm{i}\sin6\theta \end{align*}

Taking real parts after binomial expansion:

cos⁡6θ= cos⁡6θ−15cos⁡4θsin⁡2θ +15cos⁡2θsin⁡4θ−sin⁡6θ\begin{align*} \cos6\theta =&\, \cos^6\theta -15\cos^4\theta\sin^2\theta\\[2mm] &\,\hspace{2pt}+15\cos^2\theta\sin^4\theta -\sin^6\theta \end{align*}

Use sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta:

cos⁡6θ= cos⁡6θ−15cos⁡4θ(1−cos⁡2θ) +15cos⁡2θ(1−cos⁡2θ)2 −(1−cos⁡2θ)3\begin{align*} \cos6\theta =&\, \cos^6\theta -15\cos^4\theta(1-\cos^2\theta)\\[2mm] &\,\hspace{2pt}+15\cos^2\theta(1-\cos^2\theta)^2\\[2mm] &\,\hspace{4pt}-(1-\cos^2\theta)^3 \end{align*}

Now expand:

cos⁡6θ= cos⁡6θ−15cos⁡4θ+15cos⁡6θ +15cos⁡2θ(1−2cos⁡2θ+cos⁡4θ) −(1−3cos⁡2θ+3cos⁡4θ−cos⁡6θ)= 32cos⁡6θ−48cos⁡4θ+18cos⁡2θ−1\begin{align*} \cos6\theta =&\, \cos^6\theta-15\cos^4\theta+15\cos^6\theta\\[2mm] &\,\hspace{2pt}+15\cos^2\theta(1-2\cos^2\theta+\cos^4\theta)\\[2mm] &\,\hspace{4pt}-(1-3\cos^2\theta+3\cos^4\theta-\cos^6\theta)\\[4mm] =&\, 32\cos^6\theta-48\cos^4\theta+18\cos^2\theta-1 \end{align*}

(b)

解法一

思路

展开

令 x=cos⁡θx=\cos\theta。题目中的多项式与 (a) 的结果相差一个倍数，因此可转化成 cos⁡6θ=−13\cos6\theta=-\frac13。要找最小正根，需要从得到的多个 θ\theta 中选使 cos⁡θ\cos\theta 最小但仍为正的那个。

答题过程

展开

Let

x=cos⁡θ\begin{align*} x=\cos\theta \end{align*}

Using part (a),

48x6−72x4+27x2−1= 32(32x6−48x4+18x2−1)+12= 32cos⁡6θ+12\begin{align*} 48x^6-72x^4+27x^2-1 =&\, \frac32(32x^6-48x^4+18x^2-1)+\frac12\\[4mm] =&\, \frac32\cos6\theta+\frac12 \end{align*}

So the equation becomes

32cos⁡6θ+12= 0cos⁡6θ= −13\begin{align*} \frac32\cos6\theta+\frac12=&\,0\\[2mm] \cos6\theta=&\,-\frac13 \end{align*}

The positive roots for x=cos⁡θx=\cos\theta come from the corresponding values of θ\theta. The values nearest to π2\frac{\pi}{2} while still giving positive cosine are obtained from

6θ=2π+arccos⁡(−13)\begin{align*} 6\theta=2\pi+\arccos\left(-\frac13\right) \end{align*}

Thus

θ= 2π+arccos⁡(−13)6x= cos⁡(2π+arccos⁡(−13)6)= 0.2039…\begin{align*} \theta =&\, \frac{2\pi+\arccos\left(-\frac13\right)}{6}\\[4mm] x =&\, \cos\left( \frac{2\pi+\arccos\left(-\frac13\right)}{6} \right)\\[4mm] =&\, 0.2039\ldots \end{align*}

Therefore the smallest positive root is

0.204\begin{align*} \boxed{0.204} \end{align*}