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CIE 9231 2026 June Paper 24 Q7

A Level / CIE / FP2

CIE 9231 2026 June Paper 24 Paper · Question 7

题目

Problem

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

sec⁡7θ=sec⁡7θ64−112sec⁡2θ+56sec⁡4θ−7sec⁡6θ.\sec 7\theta=\frac{\sec^7\theta}{64-112\sec^2\theta+56\sec^4\theta-7\sec^6\theta}.
(6)

(b) Hence obtain the roots of the equation

x7−14x6+112x4−224x2+128=0x^7-14x^6+112x^4-224x^2+128=0

in the form sec⁡qπ\sec q\pi, where qq is rational.

(5)
题目中文翻译

(a) 用棣莫弗定理证明

sec⁡7θ=sec⁡7θ64−112sec⁡2θ+56sec⁡4θ−7sec⁡6θ.\sec 7\theta=\frac{\sec^7\theta}{64-112\sec^2\theta+56\sec^4\theta-7\sec^6\theta}.
(6)

(b) 由此求方程

x7−14x6+112x4−224x2+128=0x^7-14x^6+112x^4-224x^2+128=0

的根，并写成 sec⁡qπ\sec q\pi 的形式，其中 qq 为有理数。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

对七次方取实部。将正弦的偶次幂换成余弦，取倒数，再将分子分母同除以 cos⁡7θ\cos^7\theta。

答题过程

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Let c=cos⁡θc=\cos\theta and s=sin⁡θs=\sin\theta. De Moivre’s theorem gives (c+is)7=cos⁡7θ+isin⁡7θ(c+\mathrm{i}s)^7=\cos7\theta+\mathrm{i}\sin7\theta. The real terms of its binomial expansion are c7+21c5(is)2+35c3(is)4+7c(is)6c^7+21c^5(\mathrm{i}s)^2+35c^3(\mathrm{i}s)^4+7c(\mathrm{i}s)^6. Hence

cos⁡7θ= c7−21c5s2+35c3s4−7cs6= c7−21c5(1−c2)+35c3(1−2c2+c4) −7c(1−3c2+3c4−c6)= 64c7−112c5+56c3−7c.\begin{align*} \cos7\theta =&\,c^7-21c^5s^2+35c^3s^4-7cs^6\\ =&\,c^7-21c^5(1-c^2)+35c^3(1-2c^2+c^4)\\ &\,\hspace{2pt}-7c(1-3c^2+3c^4-c^6)\\ =&\,64c^7-112c^5+56c^3-7c. \end{align*}

Where both secants are defined, take the reciprocal and divide the numerator and denominator by c7c^7:

164c7−112c5+56c3−7c=c−764−112c−2+56c−4−7c−6.\frac1{64c^7-112c^5+56c^3-7c} =\frac{c^{-7}}{64-112c^{-2}+56c^{-4}-7c^{-6}}.

This gives

sec⁡7θ=sec⁡7θ64−112sec⁡2θ+56sec⁡4θ−7sec⁡6θ.\sec7\theta=\frac{\sec^7\theta}{64-112\sec^2\theta+56\sec^4\theta-7\sec^6\theta}.

(b)

解法一

思路

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把多项式写成 x7x^7 加上 (a) 分母的两倍。令 x=sec⁡θx=\sec\theta 后，留意得到的是 sec⁡7θ=−2\sec7\theta=-2。

答题过程

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The equation is

x7+2(64−112x2+56x4−7x6)=0.x^7+2(64-112x^2+56x^4-7x^6)=0.

The denominator 64−112x2+56x4−7x664-112x^2+56x^4-7x^6 cannot be zero at a root, because the equation would then force x=0x=0, where that denominator is 6464. With x=sec⁡θx=\sec\theta, part (a) gives sec⁡7θ=−2\sec7\theta=-2, hence cos⁡7θ=−12\cos7\theta=-\frac12. Taking 0≤θ≤π0\le\theta\le\pi, we obtain

7θ∈{ 2π3,4π3,8π3,10π3, 14π3,16π3,20π3}.\begin{align*} 7\theta\in\bigg\{&\,\frac{2\pi}{3},\frac{4\pi}{3},\frac{8\pi}{3},\frac{10\pi}{3},\\ &\,\hspace{2pt}\frac{14\pi}{3},\frac{16\pi}{3},\frac{20\pi}{3}\bigg\}. \end{align*}

Therefore the roots are

x1= sec⁡2π21,x2=sec⁡4π21,x3= sec⁡8π21,x4=sec⁡10π21,x5= sec⁡14π21,x6=sec⁡16π21,x7= sec⁡20π21.\begin{align*} x_1=&\,\sec\frac{2\pi}{21},\qquad x_2=\sec\frac{4\pi}{21},\\ x_3=&\,\sec\frac{8\pi}{21},\qquad x_4=\sec\frac{10\pi}{21},\\ x_5=&\,\sec\frac{14\pi}{21},\qquad x_6=\sec\frac{16\pi}{21},\\ x_7=&\,\sec\frac{20\pi}{21}. \end{align*}

None of the angles has zero cosine, and the cosine values are distinct on [0,π][0,\pi]. These seven values therefore account for every root of the polynomial.