题目
Problem
(a) Use de Moivre’s theorem to show that
cos 5 θ = 16 cos 5 θ − 20 cos 3 θ + 5 cos θ . \cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta. cos 5 θ = 16 cos 5 θ − 20 cos 3 θ + 5 cos θ . [4]
(b) Hence obtain the roots of the equation
32 x 5 − 40 x 3 + 10 x − 2 = 0 32x^5-40x^3+10x-\sqrt{2}=0 32 x 5 − 40 x 3 + 10 x − 2 = 0 in the form cos ( q π ) \cos(q\pi) cos ( q π ) q q q
[4]
题目中文翻译
(a) 使用棣莫弗定理证明
cos 5 θ = 16 cos 5 θ − 20 cos 3 θ + 5 cos θ \cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta cos 5 θ = 16 cos 5 θ − 20 cos 3 θ + 5 cos θ (4)
(b) 由此求方程
32 x 5 − 40 x 3 + 10 x − 2 = 0 32x^5-40x^3+10x-\sqrt{2}=0 32 x 5 − 40 x 3 + 10 x − 2 = 0 的根，并将其表示为 cos ( q π ) \cos(q\pi) cos ( q π ) q q q
(4)
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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从五次方的实部取出 cos 5 θ \cos5\theta cos 5 θ 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 ) 5 = cos 5 θ + i sin 5 θ (c+\mathrm{i}s)^5=\cos5\theta+\mathrm{i}\sin5\theta ( c + i s ) 5 = cos 5 θ + i sin 5 θ
( c + i s ) 5 = c 5 + 5 c 4 ( i s ) + 10 c 3 ( i s ) 2 + 10 c 2 ( i s ) 3 + 5 c ( i s ) 4 + ( i s ) 5 . \begin{align*}
(c+\mathrm{i}s)^5=&\,c^5+5c^4(\mathrm{i}s)+10c^3(\mathrm{i}s)^2\\
&\,\hspace{2pt}+10c^2(\mathrm{i}s)^3+5c(\mathrm{i}s)^4+(\mathrm{i}s)^5.
\end{align*} ( c + i s ) 5 = c 5 + 5 c 4 ( i s ) + 10 c 3 ( i s ) 2 + 10 c 2 ( i s ) 3 + 5 c ( i s ) 4 + ( i s ) 5 . Taking real parts gives
cos 5 θ = c 5 − 10 c 3 s 2 + 5 c s 4 = c 5 − 10 c 3 ( 1 − c 2 ) + 5 c ( 1 − 2 c 2 + c 4 ) = 16 c 5 − 20 c 3 + 5 c . \begin{align*}
\cos5\theta
=&\,c^5-10c^3s^2+5cs^4\\
=&\,c^5-10c^3(1-c^2)+5c(1-2c^2+c^4)\\
=&\,16c^5-20c^3+5c.
\end{align*} cos 5 θ = = = c 5 − 10 c 3 s 2 + 5 c s 4 c 5 − 10 c 3 ( 1 − c 2 ) + 5 c ( 1 − 2 c 2 + c 4 ) 16 c 5 − 20 c 3 + 5 c .
解法二
思路
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也可从余弦多项式逆推。令 w = z + z − 1 = 2 cos θ w=z+z^{-1}=2\cos\theta w = z + z − 1 = 2 cos θ
答题过程
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With z = cos θ + i sin θ z=\cos\theta+\mathrm{i}\sin\theta z = cos θ + i sin θ w = z + z − 1 w=z+z^{-1} w = z + z − 1
w 3 = z 3 + 3 z + 3 z − 1 + z − 3 = z 3 + z − 3 + 3 w , w 5 = z 5 + 5 z 3 + 10 z + 10 z − 1 + 5 z − 3 + z − 5 = z 5 + z − 5 + 5 ( z 3 + z − 3 ) + 10 w . \begin{align*}
w^3=&\,z^3+3z+3z^{-1}+z^{-3}\\
=&\,z^3+z^{-3}+3w,\\
w^5=&\,z^5+5z^3+10z+10z^{-1}+5z^{-3}+z^{-5}\\
=&\,z^5+z^{-5}+5(z^3+z^{-3})+10w.
\end{align*} w 3 = = w 5 = = z 3 + 3 z + 3 z − 1 + z − 3 z 3 + z − 3 + 3 w , z 5 + 5 z 3 + 10 z + 10 z − 1 + 5 z − 3 + z − 5 z 5 + z − 5 + 5 ( z 3 + z − 3 ) + 10 w . Therefore w 5 − 5 w 3 + 5 w = z 5 + z − 5 = 2 cos 5 θ w^5-5w^3+5w=z^5+z^{-5}=2\cos5\theta w 5 − 5 w 3 + 5 w = z 5 + z − 5 = 2 cos 5 θ w = 2 cos θ w=2\cos\theta w = 2 cos θ 2 2 2
(b)
解法一
思路
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令 x = cos θ x=\cos\theta x = cos θ 5 θ 5\theta 5 θ 0 ≤ θ ≤ π 0\le\theta\le\pi 0 ≤ θ ≤ π
答题过程
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With x = cos θ x=\cos\theta x = cos θ cos 5 θ = 2 2 \cos5\theta=\frac{\sqrt2}{2} cos 5 θ = 2 2 5 θ = 2 k π ± π 4 5\theta=2k\pi\pm\frac\pi4 5 θ = 2 k π ± 4 π 0 ≤ 5 θ ≤ 5 π 0\le5\theta\le5\pi 0 ≤ 5 θ ≤ 5 π π 4 , 7 π 4 , 9 π 4 , 15 π 4 , 17 π 4 \frac\pi4,\frac{7\pi}4,\frac{9\pi}4,\frac{15\pi}4,\frac{17\pi}4 4 π , 4 7 π , 4 9 π , 4 15 π , 4 17 π
x 1 = cos π 20 , x 2 = cos 7 π 20 , x 3 = cos 9 π 20 , x 4 = cos 15 π 20 , x 5 = cos 17 π 20 . \begin{align*}
x_1=&\,\cos\frac\pi{20},\qquad x_2=\cos\frac{7\pi}{20},\\
x_3=&\,\cos\frac{9\pi}{20},\qquad x_4=\cos\frac{15\pi}{20},\\
x_5=&\,\cos\frac{17\pi}{20}.
\end{align*} x 1 = x 3 = x 5 = cos 20 π , x 2 = cos 20 7 π , cos 20 9 π , x 4 = cos 20 15 π , cos 20 17 π . They are distinct because cosine is strictly decreasing on [ 0 , π ] [0,\pi] [ 0 , π ]
