题目
Problem
(a) Show that the equation 5 sec x + tan x = 4 \sqrt5\sec x+\tan x=4 5 sec x + tan x = 4 R cos ( x + α ) = 5 R\cos(x+\alpha)=\sqrt5 R cos ( x + α ) = 5 R > 0 R>0 R > 0 0 ∘ < α < 90 ∘ 0^\circ<\alpha<90^\circ 0 ∘ < α < 9 0 ∘ R R R α \alpha α 2 2 2
(4)
(b) Hence solve the equation 5 sec 2 x + tan 2 x = 4 \sqrt5\sec2x+\tan2x=4 5 sec 2 x + tan 2 x = 4 0 ∘ < x < 180 ∘ 0^\circ<x<180^\circ 0 ∘ < x < 18 0 ∘
(4)
题目中文翻译
(a) 证明方程 5 sec x + tan x = 4 \sqrt5\sec x+\tan x=4 5 sec x + tan x = 4 R cos ( x + α ) = 5 R\cos(x+\alpha)=\sqrt5 R cos ( x + α ) = 5 R > 0 R>0 R > 0 0 ∘ < α < 90 ∘ 0^\circ<\alpha<90^\circ 0 ∘ < α < 9 0 ∘ R R R α \alpha α 2 2 2
(4)
(b) 由此解方程 5 sec 2 x + tan 2 x = 4 \sqrt5\sec2x+\tan2x=4 5 sec 2 x + tan 2 x = 4 0 ∘ < x < 180 ∘ 0^\circ<x<180^\circ 0 ∘ < x < 18 0 ∘
(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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先乘以非零的 cos x \cos x cos x 4 cos x − sin x = 5 4\cos x-\sin x=\sqrt5 4 cos x − sin x = 5
答题过程
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Where the original equation is defined, cos x ≠ 0 \cos x\ne0 cos x = 0 cos x \cos x cos x
5 + sin x = 4 cos x , 4 cos x − sin x = 5 . \begin{align*}
\sqrt5+\sin x=&\,4\cos x,\\[4mm]
4\cos x-\sin x=&\,\sqrt5.
\end{align*} 5 + sin x = 4 cos x − sin x = 4 cos x , 5 . Expanding R cos ( x + α ) R\cos(x+\alpha) R cos ( x + α )
R cos α = 4 , R sin α = 1. R\cos\alpha=4,\qquad R\sin\alpha=1. R cos α = 4 , R sin α = 1. Hence
R = 17 , α = tan − 1 1 4 = 14.036243 … ∘ . \begin{align*}
R=&\,\sqrt{17},\\[4mm]
\alpha=&\,\tan^{-1}\frac14\\[4mm]
=&\,14.036243\ldots^\circ.
\end{align*} R = α = = 17 , tan − 1 4 1 14.036243 … ∘ . Therefore the equation can be written
17 cos ( x + α ) = 5 , α = 14.04 ∘ . \boxed{\sqrt{17}\cos(x+\alpha)=\sqrt5,\quad
\alpha=14.04^\circ}. 17 cos ( x + α ) = 5 , α = 14.0 4 ∘ .
(b)
解法一
思路
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承接(a)把角换成 2 x 2x 2 x
答题过程
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Using part (a), with unrounded α \alpha α
17 cos ( 2 x + α ) = 5 . \sqrt{17}\cos(2x+\alpha)=\sqrt5. 17 cos ( 2 x + α ) = 5 . Let β = cos − 1 5 / 17 \beta=\cos^{-1}\sqrt{5/17} β = cos − 1 5/17 0 ∘ < x < 180 ∘ 0^\circ<x<180^\circ 0 ∘ < x < 18 0 ∘
2 x = β − α or 2 x = 360 ∘ − β − α . \begin{align*}
2x=&\,\beta-\alpha\\[4mm]
\text{or}\quad 2x=&\,360^\circ-\beta-\alpha.
\end{align*} 2 x = or 2 x = β − α 36 0 ∘ − β − α . These give x = 21.5608 … ∘ x=21.5608\ldots^\circ x = 21.5608 … ∘ 144.4029 … ∘ 144.4029\ldots^\circ 144.4029 … ∘ cos 2 x ≠ 0 \cos2x\ne0 cos 2 x = 0
Thus
x = 21.6 ∘ or 144.4 ∘ . \boxed{x=21.6^\circ\quad\text{or}\quad144.4^\circ}. x = 21. 6 ∘ or 144. 4 ∘ . 