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CIE 9231 2022 June Paper 21 Q7

A Level / CIE / FP2

CIE 9231 2022 June Paper 21 Paper · Question 7

题目

Problem

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

cosec⁡7θ=cosec⁡7θ7cosec⁡6θ−56cosec⁡4θ+112cosec⁡2θ−64.\operatorname{cosec}7\theta=\frac{\operatorname{cosec}^7\theta}{7\operatorname{cosec}^6\theta-56\operatorname{cosec}^4\theta+112\operatorname{cosec}^2\theta-64}.
(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 cosec⁡qπ\operatorname{cosec}q\pi, where qq is rational.

(5)
题目中文翻译

(a) 用棣莫弗定理证明

cosec⁡7θ=cosec⁡7θ7cosec⁡6θ−56cosec⁡4θ+112cosec⁡2θ−64.\operatorname{cosec}7\theta=\frac{\operatorname{cosec}^7\theta}{7\operatorname{cosec}^6\theta-56\operatorname{cosec}^4\theta+112\operatorname{cosec}^2\theta-64}.
(6)

(b) 由此求方程

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

的根，将答案写成 cosec⁡qπ\operatorname{cosec}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

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对七次方取虚部，把余弦的偶次幂改成正弦。然后取倒数，并将分子分母同除以 sin⁡7θ\sin^7\theta。

答题过程

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Let c=cos⁡θc=\cos\theta and s=sin⁡θs=\sin\theta. By de Moivre’s theorem,

cos⁡7θ+isin⁡7θ= (c+is)7= c7+7ic6s−21c5s2−35ic4s3 +35c3s4+21ic2s5−7cs6−is7.\begin{align*} \cos7\theta+\mathrm{i}\sin7\theta =&\,(c+\mathrm{i}s)^7\\ =&\,c^7+7\mathrm{i}c^6s-21c^5s^2-35\mathrm{i}c^4s^3\\ &\,\hspace{2pt}+35c^3s^4+21\mathrm{i}c^2s^5-7cs^6-\mathrm{i}s^7. \end{align*}

Taking imaginary parts and using c2=1−s2c^2=1-s^2 gives

sin⁡7θ= 7c6s−35c4s3+21c2s5−s7= 7s(1−3s2+3s4−s6) −35s3(1−2s2+s4)+21s5(1−s2)−s7= 7s−56s3+112s5−64s7.\begin{align*} \sin7\theta =&\,7c^6s-35c^4s^3+21c^2s^5-s^7\\ =&\,7s(1-3s^2+3s^4-s^6)\\ &\,\hspace{2pt}-35s^3(1-2s^2+s^4)+21s^5(1-s^2)-s^7\\ =&\,7s-56s^3+112s^5-64s^7. \end{align*}

Where the cosecants are defined, taking reciprocals and dividing numerator and denominator by s7s^7 gives

csc⁡7θ=csc⁡7θ7csc⁡6θ−56csc⁡4θ+112csc⁡2θ−64.\csc7\theta=\frac{\csc^7\theta}{7\csc^6\theta-56\csc^4\theta+112\csc^2\theta-64}.

(b)

解法一

思路

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将多项式整理为 x7x^7 等于 (a) 分母的两倍。令 x=csc⁡θx=\csc\theta，在 [−π/2,π/2][-\pi/2,\pi/2] 中寻找七个不同的根。

答题过程

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The equation is x7=2(7x6−56x4+112x2−64)x^7=2(7x^6-56x^4+112x^2-64). Substituting x=csc⁡θx=\csc\theta and using (a) gives csc⁡7θ=2\csc7\theta=2, hence sin⁡7θ=12\sin7\theta=\frac12. The angle 7θ7\theta is π/6+2kπ\pi/6+2k\pi or 5π/6+2kπ5\pi/6+2k\pi for integer kk. Restricting θ\theta to [−π/2,π/2][-\pi/2,\pi/2] gives

x1= csc⁡(−19π42),x2= csc⁡(−11π42),x3= csc⁡(−7π42),x4= csc⁡π42,x5= csc⁡5π42,x6= csc⁡13π42,x7= csc⁡17π42.\begin{align*} x_1=&\,\csc\bigg(-\frac{19\pi}{42}\bigg),\\ x_2=&\,\csc\bigg(-\frac{11\pi}{42}\bigg),\\ x_3=&\,\csc\bigg(-\frac{7\pi}{42}\bigg),\\ x_4=&\,\csc\frac{\pi}{42},\\ x_5=&\,\csc\frac{5\pi}{42},\\ x_6=&\,\csc\frac{13\pi}{42},\\ x_7=&\,\csc\frac{17\pi}{42}. \end{align*}

These seven angles have distinct non-zero sine values, giving seven distinct roots. The polynomial has degree seven, so there are no others.