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IAL 2024 Jan Q9

A Level / Edexcel / P3

IAL 2024 Jan 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 the equation

3sin⁡θcos⁡θcos⁡θ+sin⁡θ=(2+sec⁡2θ)(cos⁡θ−sin⁡θ)\frac{3\sin\theta\cos\theta}{\cos\theta+\sin\theta} =(2+\sec 2\theta)(\cos\theta-\sin\theta)

can be written in the form

3sin⁡2θ−4cos⁡2θ=23\sin 2\theta-4\cos 2\theta=2
(3)

(b) Hence solve for π<x<3π2\pi<x<\dfrac{3\pi}{2}

3sin⁡xcos⁡xcos⁡x+sin⁡x=(2+sec⁡2x)(cos⁡x−sin⁡x)\frac{3\sin x\cos x}{\cos x+\sin x} =(2+\sec 2x)(\cos x-\sin x)

giving the answer to 3 significant figures.

(5)
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) 证明方程

3sin⁡θcos⁡θcos⁡θ+sin⁡θ=(2+sec⁡2θ)(cos⁡θ−sin⁡θ)\frac{3\sin\theta\cos\theta}{\cos\theta+\sin\theta} =(2+\sec 2\theta)(\cos\theta-\sin\theta)

可写成

3sin⁡2θ−4cos⁡2θ=23\sin 2\theta-4\cos 2\theta=2

的形式。

(b) 由此解方程

3sin⁡xcos⁡xcos⁡x+sin⁡x=(2+sec⁡2x)(cos⁡x−sin⁡x)\frac{3\sin x\cos x}{\cos x+\sin x} =(2+\sec 2x)(\cos x-\sin x)

其中 π<x<3π2\pi<x<\dfrac{3\pi}{2}，答案精确到 3 位有效数字。

解答

(a)

解法一

思路

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先乘去分母，再把 (cos⁡θ−sin⁡θ)(cos⁡θ+sin⁡θ)(\cos\theta-\sin\theta)(\cos\theta+\sin\theta) 化为 cos⁡2θ\cos 2\theta，同时使用 sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta。最后由 sec⁡2θcos⁡2θ=1\sec2\theta\cos2\theta=1 得到目标式。

答题过程

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Starting from

3sin⁡θcos⁡θcos⁡θ+sin⁡θ=(2+sec⁡2θ)(cos⁡θ−sin⁡θ),\frac{3\sin\theta\cos\theta} {\cos\theta+\sin\theta} =(2+\sec2\theta)(\cos\theta-\sin\theta),

multiplying by cos⁡θ+sin⁡θ\cos\theta+\sin\theta gives

3sin⁡θcos⁡θ= (2+sec⁡2θ) ×(cos⁡θ−sin⁡θ)(cos⁡θ+sin⁡θ)= (2+sec⁡2θ)(cos⁡2θ−sin⁡2θ)= (2+sec⁡2θ)cos⁡2θ.\begin{align*} 3\sin\theta\cos\theta =&\,(2+\sec2\theta)\\[2mm] &\,\hspace{2pt}\times (\cos\theta-\sin\theta) (\cos\theta+\sin\theta)\\[4mm] =&\,(2+\sec2\theta) (\cos^2\theta-\sin^2\theta)\\[4mm] =&\,(2+\sec2\theta)\cos2\theta. \end{align*}

Using 2sin⁡θcos⁡θ=sin⁡2θ2\sin\theta\cos\theta=\sin2\theta,

32sin⁡2θ=2cos⁡2θ+1.\frac32\sin2\theta =2\cos2\theta+1.

Multiplying by 22 and rearranging,

3sin⁡2θ−4cos⁡2θ=2.\boxed{3\sin2\theta-4\cos2\theta=2}.

(b)

解法一

思路

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这是官方评分资料的 Way 1。把 3sin⁡2x−4cos⁡2x3\sin2x-4\cos2x 合并为 Rsin⁡(2x−α)R\sin(2x-\alpha)，再依据 xx 的严格范围筛选唯一合法角。

答题过程

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From part (a),

3sin⁡2x−4cos⁡2x=2.3\sin2x-4\cos2x=2.

Write

3sin⁡2x−4cos⁡2x=Rsin⁡(2x−α).3\sin2x-4\cos2x =R\sin(2x-\alpha).

Then,

R=32+42=5,R=\sqrt{3^2+4^2}=5,

with

cos⁡α=35,sin⁡α=45.\cos\alpha=\frac35, \qquad \sin\alpha=\frac45.

Thus,

α=tan⁡−1(43)=0.927295…\alpha=\tan^{-1}\bigg(\frac43\bigg) =0.927295\ldots

and the equation becomes

5sin⁡(2x−α)=2.5\sin(2x-\alpha)=2.

Since π<x<3π2\pi<x<\dfrac{3\pi}{2},

2π−α<2x−α<3π−α.2\pi-\alpha<2x-\alpha<3\pi-\alpha.

The only solution of

sin⁡(2x−α)=25\sin(2x-\alpha)=\frac25

in this interval is

2x−α=2π+sin⁡−1(25).2x-\alpha =2\pi+\sin^{-1}\bigg(\frac25\bigg).

Therefore,

x= 2π+sin⁡−1(2/5)+α2= 3.810998…\begin{align*} x =&\,\frac{ 2\pi+\sin^{-1}(2/5)+\alpha }{2}\\[4mm] =&\,3.810998\ldots \end{align*}

Hence, to 33 significant figures,

x=3.81.\boxed{x=3.81}.

解法二

思路

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这是官方评分资料的 Way 2。把正弦项单独放在一边后平方，再以 u=cos⁡2xu=\cos2x 化成二次方程。平方会引入增根，因此所得候选值必须代回平方前的方程。

答题过程

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Rearranging the equation from part (a),

3sin⁡2x=2+4cos⁡2x.3\sin2x=2+4\cos2x.

Squaring both sides,

9sin⁡22x=4+16cos⁡2x+16cos⁡22x.9\sin^2 2x =4+16\cos2x+16\cos^2 2x.

Using sin⁡22x=1−cos⁡22x\sin^2 2x=1-\cos^2 2x,

9(1−cos⁡22x)= 4+16cos⁡2x+16cos⁡22x25cos⁡22x+16cos⁡2x−5= 0.\begin{align*} 9(1-\cos^2 2x) =&\,4+16\cos2x+16\cos^2 2x\\[4mm] 25\cos^2 2x+16\cos2x-5 =&\,0. \end{align*}

Let u=cos⁡2xu=\cos2x. Then

25u2+16u−5=0,25u^2+16u-5=0,

so

u= −16±162+4(25)(5)50= −8±32125.\begin{align*} u =&\,\frac{-16\pm\sqrt{16^2+4(25)(5)}}{50}\\[4mm] =&\,\frac{-8\pm3\sqrt{21}}{25}. \end{align*}

For 2π<2x<3π2\pi<2x<3\pi, the corresponding candidates are

x= 12[2π+cos⁡−1(−8+32125)]= 3.810998…\begin{align*} x =&\,\frac12\Bigg[ 2\pi+\cos^{-1} \bigg(\frac{-8+3\sqrt{21}}{25}\bigg) \Bigg]\\[4mm] =&\,3.810998\ldots \end{align*}

and

x= 12[2π+cos⁡−1(−8−32125)]= 4.454500…\begin{align*} x =&\,\frac12\Bigg[ 2\pi+\cos^{-1} \bigg(\frac{-8-3\sqrt{21}}{25}\bigg) \Bigg]\\[4mm] =&\,4.454500\ldots \end{align*}

Substitution into the equation before squaring,

3sin⁡2x=2+4cos⁡2x,3\sin2x=2+4\cos2x,

shows that x=4.454500…x=4.454500\ldots is extraneous: for this candidate, 3sin⁡2x>03\sin2x>0 but 2+4cos⁡2x<02+4\cos2x<0. Hence, to 33 significant figures,

x=3.81.\boxed{x=3.81}.

解法三

思路

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这是官方评分资料的 Way 3。把 sin⁡2x\sin2x 与 cos⁡2x\cos2x 展开成单角形式，再除以 cos⁡2x\cos^2x，便可得到关于 tan⁡x\tan x 的二次方程。由于给定范围位于第三象限，只保留正的正切值。

答题过程

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Using the double-angle identities,

3sin⁡2x−4cos⁡2x= 26sin⁡xcos⁡x−4(cos⁡2x−sin⁡2x)= 2.\begin{align*} 3\sin2x-4\cos2x=&\,2\\[4mm] 6\sin x\cos x -4(\cos^2x-\sin^2x)=&\,2. \end{align*}

Since cos⁡x≠0\cos x\neq0 for π<x<3π2\pi<x<\dfrac{3\pi}{2}, divide by cos⁡2x\cos^2x:

6tan⁡x−4+4tan⁡2x=2sec⁡2x.6\tan x-4+4\tan^2x=2\sec^2x.

Using sec⁡2x=1+tan⁡2x\sec^2x=1+\tan^2x,

6tan⁡x−4+4tan⁡2x= 2+2tan⁡2xtan⁡2x+3tan⁡x−3= 0.\begin{align*} 6\tan x-4+4\tan^2x =&\,2+2\tan^2x\\[4mm] \tan^2x+3\tan x-3 =&\,0. \end{align*}

Therefore,

tan⁡x=−3±212.\tan x=\frac{-3\pm\sqrt{21}}{2}.

The interval π<x<3π2\pi<x<\dfrac{3\pi}{2} lies in quadrant III, where tan⁡x>0\tan x>0. Hence,

tan⁡x=−3+212.\tan x=\frac{-3+\sqrt{21}}{2}.

Thus,

x= π+tan⁡−1(−3+212)= 3.810998…\begin{align*} x =&\,\pi+\tan^{-1} \bigg(\frac{-3+\sqrt{21}}{2}\bigg)\\[4mm] =&\,3.810998\ldots \end{align*}

Therefore, to 33 significant figures,

x=3.81.\boxed{x=3.81}.

解法四

思路

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这是官方评分资料的 Way 4。由于原方程含有 sec⁡2x\sec2x，所以 cos⁡2x≠0\cos2x\neq0；可将 (a) 的方程除以 cos⁡2x\cos2x，再平方化成关于 tan⁡2x\tan2x 的二次方程。最后必须回代平方前的方程排除增根。

答题过程

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Dividing

3sin⁡2x−4cos⁡2x=23\sin2x-4\cos2x=2

by cos⁡2x\cos2x gives

3tan⁡2x−4=2sec⁡2x.3\tan2x-4=2\sec2x.

Squaring both sides,

9tan⁡22x−24tan⁡2x+16=4sec⁡22x.9\tan^2 2x-24\tan2x+16 =4\sec^2 2x.

Using sec⁡22x=1+tan⁡22x\sec^2 2x=1+\tan^2 2x,

9tan⁡22x−24tan⁡2x+16= 4+4tan⁡22x5tan⁡22x−24tan⁡2x+12= 0.\begin{align*} 9\tan^2 2x-24\tan2x+16 =&\,4+4\tan^2 2x\\[4mm] 5\tan^2 2x-24\tan2x+12 =&\,0. \end{align*}

Therefore,

tan⁡2x=12±2215.\tan2x=\frac{12\pm2\sqrt{21}}{5}.

Since 2π<2x<3π2\pi<2x<3\pi, the two positive values give

x= 12[2π+tan⁡−1(12+2215)]= 3.810998…\begin{align*} x =&\,\frac12\Bigg[ 2\pi+\tan^{-1} \bigg(\frac{12+2\sqrt{21}}{5}\bigg) \Bigg]\\[4mm] =&\,3.810998\ldots \end{align*}

and

x= 12[2π+tan⁡−1(12−2215)]= 3.399482…\begin{align*} x =&\,\frac12\Bigg[ 2\pi+\tan^{-1} \bigg(\frac{12-2\sqrt{21}}{5}\bigg) \Bigg]\\[4mm] =&\,3.399482\ldots \end{align*}

For x=3.399482…x=3.399482\ldots, the left-hand side of the equation before squaring,

3tan⁡2x−4=2sec⁡2x,3\tan2x-4=2\sec2x,

is negative while the right-hand side is positive. This candidate is therefore extraneous.

Hence, to 33 significant figures,

x=3.81.\boxed{x=3.81}.