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FP2 2024 June Q9

A Level / Edexcel / FP2

IAL 2024 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

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)
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

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

(b) 由此求方程

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

的最小正根，答案保留 33 位小数。

(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*}

解法二

思路

展开

令 z=cos⁡θ+isin⁡θz=\cos\theta+\mathrm{i}\sin\theta，则由 De Moivre 定理，zn+z−n=2cos⁡nθz^n+z^{-n}=2\cos n\theta。展开 (z+z−1)6(z+z^{-1})^6 并把对称项配对，可以得到含 cos⁡6θ\cos6\theta、cos⁡4θ\cos4\theta 和 cos⁡2θ\cos2\theta 的关系，再用倍角公式整理。

答题过程

展开

Let z=cos⁡θ+isin⁡θz=\cos\theta+\mathrm{i}\sin\theta. By De Moivre’s theorem,

zn+z−n= 2cos⁡nθ\begin{align*} z^n+z^{-n}=&\,2\cos n\theta \end{align*}

In particular, z+z−1=2cos⁡θz+z^{-1}=2\cos\theta. Expanding and pairing terms gives

 (z+z−1)6= z6+6z4+15z2+20 +15z−2+6z−4+z−6= (z6+z−6)+6(z4+z−4) +15(z2+z−2)+20\begin{align*} &\,(z+z^{-1})^6\\[2mm] =&\,z^6+6z^4+15z^2+20\\[2mm] &\,\hspace{2pt}+15z^{-2}+6z^{-4}+z^{-6}\\[4mm] =&\,(z^6+z^{-6})+6(z^4+z^{-4})\\[2mm] &\,\hspace{2pt}+15(z^2+z^{-2})+20 \end{align*}

Therefore

64cos⁡6θ= 2cos⁡6θ+12cos⁡4θ +30cos⁡2θ+20\begin{align*} 64\cos^6\theta =&\,2\cos6\theta+12\cos4\theta\\[2mm] &\,\hspace{2pt}+30\cos2\theta+20 \end{align*}

Using the double-angle formula,

cos⁡2θ= 2cos⁡2θ−1cos⁡4θ= 2cos⁡22θ−1= 2(2cos⁡2θ−1)2−1= 8cos⁡4θ−8cos⁡2θ+1\begin{align*} \cos2\theta=&\,2\cos^2\theta-1\\[2mm] \cos4\theta=&\,2\cos^2 2\theta-1\\[2mm] =&\,2(2\cos^2\theta-1)^2-1\\[2mm] =&\,8\cos^4\theta-8\cos^2\theta+1 \end{align*}

Substitute these expressions:

64cos⁡6θ= 2cos⁡6θ +12(8cos⁡4θ−8cos⁡2θ+1) +30(2cos⁡2θ−1)+20= 2cos⁡6θ+96cos⁡4θ −36cos⁡2θ+2\begin{align*} 64\cos^6\theta =&\,2\cos6\theta\\[2mm] &\,\hspace{2pt}+12(8\cos^4\theta-8\cos^2\theta+1)\\[2mm] &\,\hspace{4pt}+30(2\cos^2\theta-1)+20\\[4mm] =&\,2\cos6\theta+96\cos^4\theta\\[2mm] &\,\hspace{2pt}-36\cos^2\theta+2 \end{align*}

Rearranging and dividing by 22,

cos⁡6θ≡ 32cos⁡6θ−48cos⁡4θ +18cos⁡2θ−1\begin{align*} \cos6\theta \equiv&\,32\cos^6\theta-48\cos^4\theta\\[2mm] &\,\hspace{2pt}+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*}

To find positive roots with 0<x<10<x<1, take 0<θ<π20<\theta<\frac\pi2, so 0<6θ<3π0<6\theta<3\pi. Let α=arccos⁡(−13)\alpha=\arccos(-\frac13). The three possibilities in this range are

6θ= α,2π−α,2π+α\begin{align*} 6\theta=&\,\alpha,\quad2\pi-\alpha,\quad2\pi+\alpha \end{align*}

Since cosine decreases on 0<θ<π20<\theta<\frac\pi2, the largest of these angles gives the smallest positive root. Thus

θ= 2π+arccos⁡(−13)6x= cos⁡(2π+arccos⁡(−13)6)= 0.203723…\begin{align*} \theta =&\, \frac{2\pi+\arccos\bigg(-\frac13\bigg)}{6}\\[4mm] x =&\, \cos\bigg( \frac{2\pi+\arccos\bigg(-\frac13\bigg)}{6} \bigg)\\[4mm] =&\, 0.203723\ldots \end{align*}

The other two angles give x≈0.950x\approx0.950 and x≈0.746x\approx0.746.

Therefore the smallest positive root is

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