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IAL 2021 Jan Q7

A Level / Edexcel / P3

IAL 2021 Jan Paper · Question 7

题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

(a) Prove that

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

(b) Hence solve, for −π2<θ<π2-\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2},

7+sin⁡4θcos⁡2θ+cos⁡4θsin⁡2θ=3cot⁡22θ7+\frac{\sin 4\theta}{\cos 2\theta}+\frac{\cos 4\theta}{\sin 2\theta}=3\cot^2 2\theta

giving your answers in radians to 3 significant figures where appropriate.

(6)
题目中文翻译

本题必须写出全部解题步骤。

不接受完全依赖计算器技术的解法。

(a) 证明

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

(b) 由此在 −π2<θ<π2-\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2} 内解方程

7+sin⁡4θcos⁡2θ+cos⁡4θsin⁡2θ=3cot⁡22θ7+\frac{\sin 4\theta}{\cos 2\theta}+\frac{\cos 4\theta}{\sin 2\theta}=3\cot^2 2\theta

答案用弧度表示，并在适当处保留 3 位有效数字。

解答

(a)

Start with the left hand side:

sin⁡2xcos⁡x+cos⁡2xsin⁡x\frac{\sin2x}{\cos x}+\frac{\cos2x}{\sin x}

Use

sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x

and

cos⁡2x=1−2sin⁡2x\cos2x=1-2\sin^2x

Then

sin⁡2xcos⁡x+cos⁡2xsin⁡x=2sin⁡xcos⁡xcos⁡x+1−2sin⁡2xsin⁡x=2sin⁡x+1sin⁡x−2sin⁡x=1sin⁡x=cosec⁡x\begin{aligned} \frac{\sin2x}{\cos x}+\frac{\cos2x}{\sin x} &=\frac{2\sin x\cos x}{\cos x}+\frac{1-2\sin^2x}{\sin x}\\ &=2\sin x+\frac{1}{\sin x}-2\sin x\\ &=\frac{1}{\sin x}\\ &=\cosec x \end{aligned}

Therefore

sin⁡2xcos⁡x+cos⁡2xsin⁡x≡cosec⁡x\boxed{\frac{\sin2x}{\cos x}+\frac{\cos2x}{\sin x}\equiv \cosec x}

(b)

Using part (a), with x=2θx=2\theta,

sin⁡4θcos⁡2θ+cos⁡4θsin⁡2θ=cosec⁡2θ\frac{\sin4\theta}{\cos2\theta}+\frac{\cos4\theta}{\sin2\theta} =\cosec2\theta

So the equation becomes

7+cosec⁡2θ=3cot⁡22θ7+\cosec2\theta=3\cot^2 2\theta

Use

cot⁡22θ=cosec⁡22θ−1\cot^2 2\theta=\cosec^2 2\theta-1

Then

7+cosec⁡2θ=3(cosec⁡22θ−1)7+\cosec2\theta=3(\cosec^2 2\theta-1)

Expand:

7+cosec⁡2θ=3cosec⁡22θ−37+\cosec2\theta=3\cosec^2 2\theta-3

Rearrange:

3cosec⁡22θ−cosec⁡2θ−10=03\cosec^2 2\theta-\cosec2\theta-10=0

Let

u=cosec⁡2θu=\cosec2\theta

Then

3u2−u−10=03u^2-u-10=0

Factorise:

(3u+5)(u−2)=0(3u+5)(u-2)=0

So

u=−53oru=2u=-\frac53\quad \text{or}\quad u=2

Therefore

cosec⁡2θ=−53orcosec⁡2θ=2\cosec2\theta=-\frac53 \quad \text{or}\quad \cosec2\theta=2

This gives

sin⁡2θ=−35orsin⁡2θ=12\sin2\theta=-\frac35 \quad \text{or}\quad \sin2\theta=\frac12

Since

−π2<θ<π2-\frac{\pi}{2}<\theta<\frac{\pi}{2}

we have

−π<2θ<π-\pi<2\theta<\pi

For

sin⁡2θ=12\sin2\theta=\frac12

we get

2θ=π6,5π62\theta=\frac{\pi}{6},\quad \frac{5\pi}{6}

so

θ=π12,5π12\theta=\frac{\pi}{12},\quad \frac{5\pi}{12}

For

sin⁡2θ=−35\sin2\theta=-\frac35

the solutions in −π<2θ<π-\pi<2\theta<\pi are

2θ=−0.6435…,−2.4981…2\theta=-0.6435\ldots,\quad -2.4981\ldots

so

θ=−0.32175…,−1.2490…\theta=-0.32175\ldots,\quad -1.2490\ldots

Therefore

θ=π12, 5π12, −0.322, −1.25\boxed{\theta=\frac{\pi}{12},\ \frac{5\pi}{12},\ -0.322,\ -1.25}

where the decimal answers are given to 3 significant figures.