题目
Problem
(a) Use de Moivre’s theorem to show that
sin 5 θ = 16 sin 5 θ − 20 sin 3 θ + 5 sin θ . \sin 5\theta
=
16\sin^5\theta - 20\sin^3\theta + 5\sin\theta. sin 5 θ = 16 sin 5 θ − 20 sin 3 θ + 5 sin θ . [4]
(b) Hence obtain the roots of the equation
32 x 5 − 40 x 3 + 10 x − 3 = 0 32x^5 - 40x^3 + 10x - \sqrt{3} = 0 32 x 5 − 40 x 3 + 10 x − 3 = 0 in the form sin q π \sin q\pi sin q π q q q
[4]
题目中文翻译
(a) 使用 de Moivre 定理证明
sin 5 θ = 16 sin 5 θ − 20 sin 3 θ + 5 sin θ . \sin 5\theta
=
16\sin^5\theta - 20\sin^3\theta + 5\sin\theta. sin 5 θ = 16 sin 5 θ − 20 sin 3 θ + 5 sin θ . (4)
(b) 由此求方程
32 x 5 − 40 x 3 + 10 x − 3 = 0 32x^5 - 40x^3 + 10x - \sqrt{3} = 0 32 x 5 − 40 x 3 + 10 x − 3 = 0 的根，答案写成 sin q π \sin q\pi sin 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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展开 ( cos θ + i sin θ ) 5 (\cos\theta+\mathrm{i}\sin\theta)^5 ( cos θ + i sin θ ) 5 i \mathrm{i} i cos 2 θ = 1 − sin 2 θ \cos^2\theta=1-\sin^2\theta cos 2 θ = 1 − sin 2 θ
答题过程
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By de Moivre’s theorem, sin 5 θ \sin5\theta sin 5 θ ( cos θ + i sin θ ) 5 (\cos\theta+\mathrm{i}\sin\theta)^5 ( cos θ + i sin θ ) 5 s = sin θ s=\sin\theta s = sin θ c = cos θ c=\cos\theta c = cos θ
sin 5 θ = 5 c 4 s − 10 c 2 s 3 + s 5 = 5 s ( 1 − 2 s 2 + s 4 ) − 10 s 3 ( 1 − s 2 ) + s 5 = 16 s 5 − 20 s 3 + 5 s . \begin{align*}
\sin5\theta
=&\,5c^4s-10c^2s^3+s^5\\
=&\,5s(1-2s^2+s^4)-10s^3(1-s^2)+s^5\\
=&\,16s^5-20s^3+5s.
\end{align*} sin 5 θ = = = 5 c 4 s − 10 c 2 s 3 + s 5 5 s ( 1 − 2 s 2 + s 4 ) − 10 s 3 ( 1 − s 2 ) + s 5 16 s 5 − 20 s 3 + 5 s .
解法二
思路
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也可从右边的正弦多项式逆推。先把 2 i sin θ 2\mathrm{i}\sin\theta 2 i sin θ z − z − 1 z-z^{-1} z − z − 1
答题过程
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Let z = cos θ + i sin θ z=\cos\theta+\mathrm{i}\sin\theta z = cos θ + i sin θ z m − z − m = 2 i sin ( m θ ) z^m-z^{-m}=2\mathrm{i}\sin(m\theta) z m − z − m = 2 i sin ( m θ ) w = z − z − 1 = 2 i sin θ w=z-z^{-1}=2\mathrm{i}\sin\theta w = z − z − 1 = 2 i sin θ
w 5 = z 5 − 5 z 3 + 10 z − 10 z − 1 + 5 z − 3 − z − 5 , 5 w 3 = 5 z 3 − 15 z + 15 z − 1 − 5 z − 3 , 5 w = 5 z − 5 z − 1 . \begin{align*}
w^5=&\,z^5-5z^3+10z-10z^{-1}+5z^{-3}-z^{-5},\\
5w^3=&\,5z^3-15z+15z^{-1}-5z^{-3},\\
5w=&\,5z-5z^{-1}.
\end{align*} w 5 = 5 w 3 = 5 w = z 5 − 5 z 3 + 10 z − 10 z − 1 + 5 z − 3 − z − 5 , 5 z 3 − 15 z + 15 z − 1 − 5 z − 3 , 5 z − 5 z − 1 . Adding cancels every term except z 5 − z − 5 z^5-z^{-5} z 5 − z − 5
w 5 + 5 w 3 + 5 w = z 5 − z − 5 . w^5+5w^3+5w=z^5-z^{-5}. w 5 + 5 w 3 + 5 w = z 5 − z − 5 . Dividing by 2 i 2\mathrm{i} 2 i 16 sin 5 θ − 20 sin 3 θ + 5 sin θ = sin 5 θ 16\sin^5\theta-20\sin^3\theta+5\sin\theta=\sin5\theta 16 sin 5 θ − 20 sin 3 θ + 5 sin θ = sin 5 θ
(b)
解法一
思路
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令 x = sin θ x=\sin\theta x = sin θ [ − π 2 , π 2 ] [-\frac{\pi}{2},\frac{\pi}{2}] [ − 2 π , 2 π ] θ \theta θ
答题过程
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By (a), the substitution x = sin θ x=\sin\theta x = sin θ sin 5 θ = 3 2 \sin5\theta=\frac{\sqrt3}{2} sin 5 θ = 2 3
5 θ = π 3 + 2 k π or 5 θ = 2 π 3 + 2 k π , k ∈ Z . 5\theta=\frac{\pi}{3}+2k\pi \quad\text{or}\quad 5\theta=\frac{2\pi}{3}+2k\pi, \qquad k\in\mathbb Z. 5 θ = 3 π + 2 k π or 5 θ = 3 2 π + 2 k π , k ∈ Z . Taking θ ∈ [ − π 2 , π 2 ] \theta\in[-\frac{\pi}{2},\frac{\pi}{2}] θ ∈ [ − 2 π , 2 π ]
x = sin ( − π 3 ) , sin ( − 4 π 15 ) , sin π 15 , sin 2 π 15 , sin 7 π 15 . \begin{align*}
x=&\,\sin\bigg(-\frac\pi3\bigg),\quad\sin\bigg(-\frac{4\pi}{15}\bigg),\\
&\,\sin\frac\pi{15},\quad\sin\frac{2\pi}{15},\quad\sin\frac{7\pi}{15}.
\end{align*} x = sin ( − 3 π ) , sin ( − 15 4 π ) , sin 15 π , sin 15 2 π , sin 15 7 π . These are distinct because sine is strictly increasing on the chosen interval. The polynomial has degree five, so these are all its roots.
