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CIE 9709 2022 November Paper 13 Q6

A Level / CIE / P1

CIE 9709 2022 Nov Paper 13 Paper · Question 6

题目

Problem

It is given that α=cos⁡−1(817)\alpha=\cos^{-1}\big(\tfrac{8}{17}\big).

Find, without using the trigonometric functions on your calculator, the exact value of 1sin⁡α+1tan⁡α\dfrac{1}{\sin\alpha}+\dfrac{1}{\tan\alpha}.

(5)
题目中文翻译

已知 α=cos⁡−1(817)\alpha=\cos^{-1}\big(\tfrac{8}{17}\big)。

不使用计算器的三角函数功能,求 1sin⁡α+1tan⁡α\dfrac{1}{\sin\alpha}+\dfrac{1}{\tan\alpha} 的精确值。

(5)
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

解法一

思路

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反余弦的主值在第一象限。由余弦求出正的正弦,再用 1/tan⁡α=cos⁡α/sin⁡α1/\tan\alpha=\cos\alpha/\sin\alpha 合并两项。

答题过程

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Because cos⁡α=8/17>0\cos\alpha=8/17>0 and α∈[0,π]\alpha\in[0,\pi], we have sin⁡α=1−(8/17)2=15/17\sin\alpha=\sqrt{1-(8/17)^2}=15/17. Also,

tan⁡α=sin⁡αcos⁡α=15/178/17=158.\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{15/17}{8/17}=\frac{15}{8}.

Thus

 1sin⁡α+1tan⁡α= 1+cos⁡αsin⁡α= 17+815=53.\begin{align*} &\,\frac1{\sin\alpha}+\frac1{\tan\alpha}\\ =&\,\frac{1+\cos\alpha}{\sin\alpha}\\ =&\,\frac{17+8}{15}=\frac53. \end{align*}