题目
Problem
(a) Write down the fourth roots of unity.
(1)
(b) Use de Moivre’s theorem to show that
cos 4 θ = 8 cos 4 θ − 8 cos 2 θ + 1. \cos4\theta=8\cos^4\theta-8\cos^2\theta+1. cos 4 θ = 8 cos 4 θ − 8 cos 2 θ + 1. (4)
(c) Hence obtain the real roots of the equation
16 ( 8 x 4 − 8 x 2 + 1 ) 4 − 9 = 0 16(8x^4-8x^2+1)^4-9=0 16 ( 8 x 4 − 8 x 2 + 1 ) 4 − 9 = 0 in the form cos ( q π ) \cos(q\pi) cos ( q π ) q q q
(5)
题目中文翻译
(a) 写出四次单位根。
(1)
(b) 用棣莫弗定理证明
cos 4 θ = 8 cos 4 θ − 8 cos 2 θ + 1. \cos4\theta=8\cos^4\theta-8\cos^2\theta+1. cos 4 θ = 8 cos 4 θ − 8 cos 2 θ + 1. (4)
(c) 由此求方程
16 ( 8 x 4 − 8 x 2 + 1 ) 4 − 9 = 0 16(8x^4-8x^2+1)^4-9=0 16 ( 8 x 4 − 8 x 2 + 1 ) 4 − 9 = 0 的实根，将答案写成 cos ( q π ) \cos(q\pi) cos ( q π ) q q q
(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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四次单位根在单位圆上，相邻辐角相差 π / 2 \pi/2 π /2
答题过程
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They have modulus 1 1 1 0 , π / 2 , π , 3 π / 2 0,\pi/2,\pi,3\pi/2 0 , π /2 , π , 3 π /2 π / 2 \pi/2 π /2 1 , i , − 1 , − i \boxed{1,\ \mathrm{i},\ -1,\ -\mathrm{i}} 1 , i , − 1 , − i
(b)
解法一
思路
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对 ( cos θ + i sin θ ) 4 (\cos\theta+\mathrm{i}\sin\theta)^4 ( cos θ + i sin θ ) 4 sin 2 θ = 1 − cos 2 θ \sin^2\theta=1-\cos^2\theta sin 2 θ = 1 − cos 2 θ
答题过程
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Let c = cos θ c=\cos\theta c = cos θ s = sin θ s=\sin\theta s = sin θ
( c + i s ) 4 = c 4 + 4 i c 3 s − 6 c 2 s 2 − 4 i c s 3 + s 4 . (c+\mathrm{i}s)^4=c^4+4\mathrm{i}c^3s-6c^2s^2-4\mathrm{i}cs^3+s^4. ( c + i s ) 4 = c 4 + 4 i c 3 s − 6 c 2 s 2 − 4 i c s 3 + s 4 . Taking real parts,
cos 4 θ = c 4 − 6 c 2 s 2 + s 4 = c 4 − 6 c 2 ( 1 − c 2 ) + ( 1 − c 2 ) 2 = 8 c 4 − 8 c 2 + 1. \begin{align*}
\cos4\theta=&\,c^4-6c^2s^2+s^4\\
=&\,c^4-6c^2(1-c^2)+(1-c^2)^2\\
=&\,8c^4-8c^2+1.
\end{align*} cos 4 θ = = = c 4 − 6 c 2 s 2 + s 4 c 4 − 6 c 2 ( 1 − c 2 ) + ( 1 − c 2 ) 2 8 c 4 − 8 c 2 + 1.
(c)
解法一
思路
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先检查 ∣ x ∣ > 1 |x|>1 ∣ x ∣ > 1 9 / 16 9/16 9/16 x = cos θ x=\cos\theta x = cos θ
答题过程
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The equation gives ( 8 x 4 − 8 x 2 + 1 ) 4 = 9 / 16 (8x^4-8x^2+1)^4=9/16 ( 8 x 4 − 8 x 2 + 1 ) 4 = 9/16 ∣ x ∣ > 1 |x|>1 ∣ x ∣ > 1 8 x 4 − 8 x 2 + 1 = 1 + 8 x 2 ( x 2 − 1 ) > 1 8x^4-8x^2+1=1+8x^2(x^2-1)>1 8 x 4 − 8 x 2 + 1 = 1 + 8 x 2 ( x 2 − 1 ) > 1 x = cos θ x=\cos\theta x = cos θ 0 ≤ θ ≤ π 0\le\theta\le\pi 0 ≤ θ ≤ π
cos 4 θ = ± 3 2 . \cos4\theta=\pm\frac{\sqrt3}{2}. cos 4 θ = ± 2 3 . On 0 ≤ 4 θ ≤ 4 π 0\le4\theta\le4\pi 0 ≤ 4 θ ≤ 4 π 4 θ 4\theta 4 θ q 0 π / 6 q_0\pi/6 q 0 π /6 q 0 = 1 , 5 , 7 , 11 , 13 , 17 , 19 , 23 q_0=1,5,7,11,13,17,19,23 q 0 = 1 , 5 , 7 , 11 , 13 , 17 , 19 , 23
x 1 = cos π 24 , x 2 = cos 5 π 24 , x 3 = cos 7 π 24 , x 4 = cos 11 π 24 , x 5 = cos 13 π 24 , x 6 = cos 17 π 24 , x 7 = cos 19 π 24 , x 8 = cos 23 π 24 . \begin{align*}
x_1=&\,\cos\frac\pi{24},& x_2=&\,\cos\frac{5\pi}{24},\\
x_3=&\,\cos\frac{7\pi}{24},& x_4=&\,\cos\frac{11\pi}{24},\\
x_5=&\,\cos\frac{13\pi}{24},& x_6=&\,\cos\frac{17\pi}{24},\\
x_7=&\,\cos\frac{19\pi}{24},& x_8=&\,\cos\frac{23\pi}{24}.
\end{align*} x 1 = x 3 = x 5 = x 7 = cos 24 π , cos 24 7 π , cos 24 13 π , cos 24 19 π , x 2 = x 4 = x 6 = x 8 = cos 24 5 π , cos 24 11 π , cos 24 17 π , cos 24 23 π . 