题目
Problem
The equation of a curve is y = x cos 2 x y=\dfrac{x}{\cos^2x} y = cos 2 x x 0 ≤ x < 1 2 π 0\le x<\frac12\pi 0 ≤ x < 2 1 π x = a x=a x = a 12 12 12
(a) Show that
a = cos − 1 ( cos a + 2 a sin a 12 3 ) . a=\cos^{-1}\Bigg(\sqrt[3]{\frac{\cos a+2a\sin a}{12}}\Bigg). a = cos − 1 ( 3 12 cos a + 2 a sin a ) . (3)
(b) Verify by calculation that a a a 0.9 0.9 0.9 1 1 1
(2)
(c) Use an iterative formula based on the equation in part (a) to determine a a a 2 2 2 4 4 4
(3)
题目中文翻译
一条曲线的方程为 y = x cos 2 x y=\dfrac{x}{\cos^2x} y = cos 2 x x 0 ≤ x < 1 2 π 0\le x<\frac12\pi 0 ≤ x < 2 1 π x = a x=a x = a 12 12 12
(a) 证明
a = cos − 1 ( cos a + 2 a sin a 12 3 ) . a=\cos^{-1}\Bigg(\sqrt[3]{\frac{\cos a+2a\sin a}{12}}\Bigg). a = cos − 1 ( 3 12 cos a + 2 a sin a ) . (3)
(b) 通过计算验证 a a a 0.9 0.9 0.9 1 1 1
(2)
(c) 使用基于 (a) 问方程的迭代公式求 a a a 2 2 2 4 4 4
(3)
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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先用商法则求导，再把分子中的一个 cos x \cos x cos x cos 3 x \cos^3x cos 3 x 12 12 12
答题过程
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For y = x / cos 2 x y=x/\cos^2x y = x / cos 2 x
d y d x = cos 2 x + 2 x sin x cos x cos 4 x = cos x + 2 x sin x cos 3 x . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,\frac{\cos^2x+2x\sin x\cos x}{\cos^4x}\\[4mm]
=&\,\frac{\cos x+2x\sin x}{\cos^3x}.
\end{align*} d x d y = = cos 4 x cos 2 x + 2 x sin x cos x cos 3 x cos x + 2 x sin x . At x = a x=a x = a 12 12 12
12 cos 3 a = cos a + 2 a sin a . 12\cos^3a=\cos a+2a\sin a. 12 cos 3 a = cos a + 2 a sin a . Thus
cos a = cos a + 2 a sin a 12 3 . \cos a=\sqrt[3]{\frac{\cos a+2a\sin a}{12}}. cos a = 3 12 cos a + 2 a sin a . Since 0 ≤ a < π / 2 0\le a<\pi/2 0 ≤ a < π /2
a = cos − 1 ( cos a + 2 a sin a 12 3 ) . \boxed{a=\cos^{-1}\Bigg(\sqrt[3]{\frac{\cos a+2a\sin a}{12}}\Bigg)}. a = cos − 1 ( 3 12 cos a + 2 a sin a ) .
(b)
解法一
思路
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使用(a)未取反函数前的连续方程检验符号，避免多层反函数造成计算混乱。
答题过程
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Let
f ( x ) = cos 3 x − cos x + 2 x sin x 12 . f(x)=\cos^3x-\frac{\cos x+2x\sin x}{12}. f ( x ) = cos 3 x − 12 cos x + 2 x sin x . Working in radians,
f ( 0.9 ) = 0.0708895 … > 0 , f ( 1 ) = − 0.0275417 … < 0. \begin{align*}
f(0.9)=&\,0.0708895\ldots>0,\\[4mm]
f(1)=&\,-0.0275417\ldots<0.
\end{align*} f ( 0.9 ) = f ( 1 ) = 0.0708895 … > 0 , − 0.0275417 … < 0. Since f f f 0.9 0.9 0.9 1 1 1 12 12 12 0.9 < a < 1 \boxed{0.9<a<1} 0.9 < a < 1
(c)
解法一
思路
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把(a)右侧作为迭代函数，使用弧度，保留内部精度并列到四位小数。
答题过程
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Use
a n + 1 = cos − 1 ( cos a n + 2 a n sin a n 12 3 ) , a 0 = 0.95. \begin{align*}
a_{n+1}=&\,\cos^{-1}\Bigg(
\sqrt[3]{\frac{\cos a_n+2a_n\sin a_n}{12}}
\Bigg),\\[4mm]
a_0=&\,0.95.
\end{align*} a n + 1 = a 0 = cos − 1 ( 3 12 cos a n + 2 a n sin a n ) , 0.95. The iterates are:
n n n a n a_n a n 0 0.9500 1 0.9743 2 0.9694 3 0.9704 4 0.9702 5 0.9702
Therefore a = 0.97 \boxed{a=0.97} a = 0.97
