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P3 2023 May Q9

A Level / Edexcel / P3

IAL 2023 May Paper · Question 9

题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

(a) Show that

cos⁡2xsin⁡x+sin⁡2xcos⁡x≡cosec⁡x\frac{\cos 2x}{\sin x}+\frac{\sin 2x}{\cos x}\equiv \cosec x x≠nπ2, n∈Zx\ne \frac{n\pi}{2},\ n\in\mathbb{Z}
(3)

(b) Hence solve, for 0<θ<π20<\theta<\dfrac{\pi}{2},

 (cos⁡2θsin⁡θ +sin⁡2θcos⁡θ)2= 6cot⁡θ−4\begin{align*} &\,\biggl(\frac{\cos 2\theta}{\sin\theta} \\[2mm] &\,\hspace{14pt}+\frac{\sin 2\theta}{\cos\theta}\biggr)^2 \\[2mm] =&\,6\cot\theta-4 \end{align*}

giving your answers to 3 significant figures as appropriate.

(5)

(c) Using the result from part (a), or otherwise, find the exact value of

 ∫π/6π/4(cos⁡2xsin⁡x +sin⁡2xcos⁡x)cot⁡x dx\begin{align*} &\,\int_{\pi/6}^{\pi/4}\biggl(\frac{\cos 2x}{\sin x} \\[2mm] &\,\hspace{28pt}+\frac{\sin 2x}{\cos x}\biggr) \cot x\,\mathrm{d}x \end{align*}
(2)
题目中文翻译

本题中你必须写出解题过程的所有步骤。

不接受完全依赖计算器技术的解法。

(a) 证明

cos⁡2xsin⁡x+sin⁡2xcos⁡x≡cosec⁡x\frac{\cos 2x}{\sin x}+\frac{\sin 2x}{\cos x}\equiv \cosec x

其中

x≠nπ2, n∈Zx\ne \frac{n\pi}{2},\ n\in\mathbb{Z}

(b) 由此在 0<θ<π20<\theta<\dfrac{\pi}{2} 内解方程

 (cos⁡2θsin⁡θ +sin⁡2θcos⁡θ)2= 6cot⁡θ−4\begin{align*} &\,\biggl(\frac{\cos 2\theta}{\sin\theta} \\[2mm] &\,\hspace{14pt}+\frac{\sin 2\theta}{\cos\theta}\biggr)^2 \\[2mm] =&\,6\cot\theta-4 \end{align*}

答案按需要保留 3 位有效数字。

(c) 利用 (a) 的结果，或用其他方法，求

 ∫π/6π/4(cos⁡2xsin⁡x +sin⁡2xcos⁡x)cot⁡x dx\begin{align*} &\,\int_{\pi/6}^{\pi/4}\biggl(\frac{\cos 2x}{\sin x} \\[2mm] &\,\hspace{28pt}+\frac{\sin 2x}{\cos x}\biggr) \cot x\,\mathrm{d}x \end{align*}

的精确值。

解答

(a)

解法一

思路

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目标是把左边化成 cosec⁡x=1sin⁡x\cosec x=\frac1{\sin x}。先用 sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x，再用 cos⁡2x=1−2sin⁡2x\cos 2x=1-2\sin^2x，这样两个含 sin⁡x\sin x 的项会抵消。

答题过程

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Start with the left-hand side:

cos⁡2xsin⁡x+sin⁡2xcos⁡x.\frac{\cos 2x}{\sin x}+\frac{\sin 2x}{\cos x}.

Use

sin⁡2x=2sin⁡xcos⁡x.\sin 2x=2\sin x\cos x.

Then

cos⁡2xsin⁡x+sin⁡2xcos⁡x= cos⁡2xsin⁡x+2sin⁡xcos⁡xcos⁡x= cos⁡2xsin⁡x+2sin⁡x.\begin{align*} \frac{\cos 2x}{\sin x}+\frac{\sin 2x}{\cos x} =&\,\frac{\cos 2x}{\sin x} +\frac{2\sin x\cos x}{\cos x} \\[2mm] =&\,\frac{\cos 2x}{\sin x}+2\sin x. \end{align*}

Now use

cos⁡2x=1−2sin⁡2x.\cos 2x=1-2\sin^2x.

So

cos⁡2xsin⁡x+2sin⁡x= 1−2sin⁡2xsin⁡x+2sin⁡x= 1sin⁡x−2sin⁡x+2sin⁡x= 1sin⁡x= cosec⁡x.\begin{align*} \frac{\cos 2x}{\sin x}+2\sin x =&\,\frac{1-2\sin^2x}{\sin x}+2\sin x \\[2mm] =&\,\frac1{\sin x}-2\sin x+2\sin x \\[2mm] =&\,\frac1{\sin x} \\[2mm] =&\,\cosec x. \end{align*}

Therefore

cos⁡2xsin⁡x+sin⁡2xcos⁡x≡cosec⁡x.\boxed{\frac{\cos 2x}{\sin x}+\frac{\sin 2x}{\cos x}\equiv\cosec x}.

(b)

解法一

思路

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题目说 “Hence”，所以直接用 (a) 的恒等式，把括号内整体换成 cosec⁡θ\cosec\theta。然后用 cosec⁡2θ=1+cot⁡2θ\cosec^2\theta=1+\cot^2\theta，把方程化成关于 cot⁡θ\cot\theta 的二次方程。由于 0<θ<π20<\theta<\frac{\pi}{2}，两个解都在第一象限。

答题过程

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Using part (a),

cos⁡2θsin⁡θ+sin⁡2θcos⁡θ=cosec⁡θ.\frac{\cos 2\theta}{\sin\theta} +\frac{\sin 2\theta}{\cos\theta} =\cosec\theta.

Therefore the equation becomes

cosec⁡2θ=6cot⁡θ−4.\cosec^2\theta=6\cot\theta-4.

Use

cosec⁡2θ=1+cot⁡2θ.\cosec^2\theta=1+\cot^2\theta.

Then

1+cot⁡2θ=6cot⁡θ−4.1+\cot^2\theta=6\cot\theta-4.

Rearrange:

cot⁡2θ−6cot⁡θ+5=0.\cot^2\theta-6\cot\theta+5=0.

Factorise:

(cot⁡θ−1)(cot⁡θ−5)=0.(\cot\theta-1)(\cot\theta-5)=0.

So

cot⁡θ=1orcot⁡θ=5.\cot\theta=1 \qquad\text{or}\qquad \cot\theta=5.

Hence

tan⁡θ=1ortan⁡θ=15.\tan\theta=1 \qquad\text{or}\qquad \tan\theta=\frac15.

Since 0<θ<π20<\theta<\dfrac{\pi}{2},

θ=π4orθ=arctan⁡15.\theta=\frac{\pi}{4} \qquad\text{or}\qquad \theta=\arctan\frac15.

Therefore

θ=0.197orθ=π4.\boxed{\theta=0.197} \qquad\text{or}\qquad \boxed{\theta=\frac{\pi}{4}}.

(c)

解法一

思路

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继续使用 (a)，积分中的括号就是 cosec⁡x\cosec x。于是被积函数变成 cosec⁡xcot⁡x\cosec x\cot x，它的积分是 −cosec⁡x-\cosec x，再代入上下限即可。

答题过程

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Using part (a),

cos⁡2xsin⁡x+sin⁡2xcos⁡x=cosec⁡x.\frac{\cos 2x}{\sin x}+\frac{\sin 2x}{\cos x} =\cosec x.

So the integral is

∫π/6π/4cosec⁡xcot⁡x dx.\int_{\pi/6}^{\pi/4}\cosec x\cot x\,\mathrm{d}x.

Since

ddx(cosec⁡x)=−cosec⁡xcot⁡x,\frac{\mathrm{d}}{\mathrm{d}x}(\cosec x) =-\cosec x\cot x,

we have

∫cosec⁡xcot⁡x dx=−cosec⁡x+C.\int \cosec x\cot x\,\mathrm{d}x=-\cosec x+C.

Therefore

∫π/6π/4cosec⁡xcot⁡x dx= [−cosec⁡x]π/6π/4= −cosec⁡π4+cosec⁡π6= −2+2= 2−2.\begin{align*} \int_{\pi/6}^{\pi/4}\cosec x\cot x\,\mathrm{d}x =&\,\Bigl[-\cosec x\Bigr]_{\pi/6}^{\pi/4} \\[2mm] =&\,-\cosec\frac{\pi}{4} +\cosec\frac{\pi}{6} \\[2mm] =&\,-\sqrt2+2 \\[2mm] =&\,2-\sqrt2. \end{align*}

Hence the exact value is

2−2.\boxed{2-\sqrt2}.