题目
Problem
(a) Use de Moivre’s theorem to show that
sec 5 θ = sec 5 θ 5 sec 4 θ − 20 sec 2 θ + 16 . \begin{align*}
\sec 5\theta
=&\, \frac{\sec^5\theta}{5\sec^4\theta - 20\sec^2\theta + 16}.
\end{align*} sec 5 θ = 5 sec 4 θ − 20 sec 2 θ + 16 sec 5 θ . [6]
(b) Hence, obtain the roots of the equation
3 x 5 − 10 x 4 + 40 x 2 − 32 = 0 \sqrt{3}x^5 - 10x^4 + 40x^2 - 32 = 0 3 x 5 − 10 x 4 + 40 x 2 − 32 = 0 in the form sec ( q π ) \sec(q\pi) sec ( q π ) q q q
[4]
题目中文翻译
(a) 使用 de Moivre 定理证明
sec 5 θ = sec 5 θ 5 sec 4 θ − 20 sec 2 θ + 16 . \begin{align*}
\sec 5\theta
=&\, \frac{\sec^5\theta}{5\sec^4\theta - 20\sec^2\theta + 16}.
\end{align*} sec 5 θ = 5 sec 4 θ − 20 sec 2 θ + 16 sec 5 θ . (6)
(b) 由此，求方程
3 x 5 − 10 x 4 + 40 x 2 − 32 = 0 \sqrt{3}x^5 - 10x^4 + 40x^2 - 32 = 0 3 x 5 − 10 x 4 + 40 x 2 − 32 = 0 的根，答案写成 sec ( q π ) \sec(q\pi) sec ( 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
展开
用 de Moivre 定理取五次方的实部，再把所有正弦的偶次幂改写成余弦。最后取倒数并除以 cos 5 θ \cos^5\theta cos 5 θ
答题过程
展开
Write c = cos θ c=\cos\theta c = cos θ s = sin θ s=\sin\theta s = sin θ
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 − c 2 ) 2 = 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-c^2)^2\\
=&\,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 − c 2 ) 2 16 c 5 − 20 c 3 + 5 c . Where the secants are defined, taking reciprocals and dividing the numerator and denominator by c 5 c^5 c 5
sec 5 θ = c − 5 16 − 20 c − 2 + 5 c − 4 = sec 5 θ 5 sec 4 θ − 20 sec 2 θ + 16 . \begin{align*}
\sec5\theta=&\,\frac{c^{-5}}{16-20c^{-2}+5c^{-4}}\\[4mm]
=&\,\frac{\sec^5\theta}{5\sec^4\theta-20\sec^2\theta+16}.
\end{align*} sec 5 θ = = 16 − 20 c − 2 + 5 c − 4 c − 5 5 sec 4 θ − 20 sec 2 θ + 16 sec 5 θ .
(b)
解法一
思路
展开
令 x = sec θ x=\sec\theta x = sec θ cos 5 θ = 3 / 2 \cos5\theta=\sqrt3/2 cos 5 θ = 3 /2
答题过程
展开
The equation can be written as
3 x 5 = 2 ( 5 x 4 − 20 x 2 + 16 ) . \sqrt3 x^5=2(5x^4-20x^2+16). 3 x 5 = 2 ( 5 x 4 − 20 x 2 + 16 ) . Its roots cannot make 5 x 4 − 20 x 2 + 16 = 0 5x^4-20x^2+16=0 5 x 4 − 20 x 2 + 16 = 0 x = 0 x=0 x = 0 x = sec θ x=\sec\theta x = sec θ
sec 5 θ = x 5 5 x 4 − 20 x 2 + 16 = 2 3 , cos 5 θ = 3 2 . \sec5\theta=\frac{x^5}{5x^4-20x^2+16}=\frac2{\sqrt3},
\qquad \cos5\theta=\frac{\sqrt3}{2}. sec 5 θ = 5 x 4 − 20 x 2 + 16 x 5 = 3 2 , cos 5 θ = 2 3 . For 0 ≤ θ ≤ π 0\le\theta\le\pi 0 ≤ θ ≤ π
θ = π 30 , 11 π 30 , 13 π 30 , 23 π 30 , 25 π 30 . \theta=\frac\pi{30},\quad\frac{11\pi}{30},\quad\frac{13\pi}{30},\quad\frac{23\pi}{30},\quad\frac{25\pi}{30}. θ = 30 π , 30 11 π , 30 13 π , 30 23 π , 30 25 π . Hence the roots are
x = sec π 30 , sec 11 π 30 , sec 13 π 30 , sec 23 π 30 , sec 25 π 30 . \begin{align*}
x=&\,\sec\frac\pi{30},\quad\sec\frac{11\pi}{30},\quad\sec\frac{13\pi}{30},\\[4mm]
&\,\hspace{2pt}\sec\frac{23\pi}{30},\quad\sec\frac{25\pi}{30}.
\end{align*} x = sec 30 π , sec 30 11 π , sec 30 13 π , sec 30 23 π , sec 30 25 π . These values are finite and distinct. Substituting the corresponding angles into the identity in (a) verifies that each satisfies the polynomial. Since it has degree five, these are all its roots.
