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IAL 2023 Jan Q5

A Level / Edexcel / P3

IAL 2023 Jan Paper · Question 5

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

cot⁡2x−tan⁡2x≡4cot⁡2xcosec⁡2x\cot^2x-\tan^2x\equiv 4\cot 2x\cosec 2x
(4)

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

4cot⁡2θcosec⁡2θ=2tan⁡2θ4\cot 2\theta\cosec 2\theta=2\tan^2\theta

giving your answers to 2 decimal places.

(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

cot⁡2x−tan⁡2x≡4cot⁡2xcosec⁡2x\cot^2x-\tan^2x\equiv 4\cot 2x\cosec 2x

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

4cot⁡2θcosec⁡2θ=2tan⁡2θ4\cot 2\theta\cosec 2\theta=2\tan^2\theta

答案精确到小数点后 2 位。

解答

(a)

解法一

思路

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从左边出发，把 cot⁡x\cot x 和 tan⁡x\tan x 都写成 sin⁡x\sin x 与 cos⁡x\cos x 的形式，通分后用平方差公式。最后用 sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x，把分母改成含 sin⁡22x\sin^22x 的形式。

答题过程

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Start with the left-hand side:

cot⁡2x−tan⁡2x= cos⁡2xsin⁡2x−sin⁡2xcos⁡2x= cos⁡4x−sin⁡4xsin⁡2xcos⁡2x.\begin{align*} \cot^2x-\tan^2x =&\,\frac{\cos^2x}{\sin^2x} -\frac{\sin^2x}{\cos^2x} \\[2mm] =&\,\frac{\cos^4x-\sin^4x}{\sin^2x\cos^2x}. \end{align*}

Use the difference of two squares:

cos⁡4x−sin⁡4x=(cos⁡2x−sin⁡2x)(cos⁡2x+sin⁡2x).\cos^4x-\sin^4x =(\cos^2x-\sin^2x)(\cos^2x+\sin^2x).

Since

cos⁡2x+sin⁡2x=1\cos^2x+\sin^2x=1

and

cos⁡2x−sin⁡2x=cos⁡2x,\cos^2x-\sin^2x=\cos2x,

we get

cot⁡2x−tan⁡2x=cos⁡2xsin⁡2xcos⁡2x.\cot^2x-\tan^2x =\frac{\cos2x}{\sin^2x\cos^2x}.

Now

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

Therefore

sin⁡22x=4sin⁡2xcos⁡2x,\sin^22x=4\sin^2x\cos^2x,

so

sin⁡2xcos⁡2x=14sin⁡22x.\sin^2x\cos^2x=\frac14\sin^22x.

Hence

cot⁡2x−tan⁡2x= cos⁡2x14sin⁡22x= 4cos⁡2xsin⁡22x= 4(cos⁡2xsin⁡2x)(1sin⁡2x)= 4cot⁡2xcosec⁡2x.\begin{align*} \cot^2x-\tan^2x =&\,\frac{\cos2x}{\frac14\sin^22x} \\[2mm] =&\,\frac{4\cos2x}{\sin^22x} \\[2mm] =&\,4\left(\frac{\cos2x}{\sin2x}\right) \left(\frac{1}{\sin2x}\right) \\[2mm] =&\,4\cot2x\cosec2x. \end{align*}

Therefore

cot⁡2x−tan⁡2x≡4cot⁡2xcosec⁡2x.\boxed{\cot^2x-\tan^2x\equiv4\cot2x\cosec2x}.

(b)

解法一

思路

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题目说 Hence，所以要用 (a) 的恒等式。把 4cot⁡2θcosec⁡2θ4\cot2\theta\cosec2\theta 换成 cot⁡2θ−tan⁡2θ\cot^2\theta-\tan^2\theta，然后把方程化成只含 tan⁡θ\tan\theta 的方程。

答题过程

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Using part (a),

4cot⁡2θcosec⁡2θ=cot⁡2θ−tan⁡2θ.4\cot2\theta\cosec2\theta =\cot^2\theta-\tan^2\theta.

So the equation becomes

cot⁡2θ−tan⁡2θ=2tan⁡2θ.\cot^2\theta-\tan^2\theta=2\tan^2\theta.

Hence

cot⁡2θ=3tan⁡2θ.\cot^2\theta=3\tan^2\theta.

Since

cot⁡2θ=1tan⁡2θ,\cot^2\theta=\frac{1}{\tan^2\theta},

we have

1tan⁡2θ=3tan⁡2θ.\frac{1}{\tan^2\theta}=3\tan^2\theta.

Thus

3tan⁡4θ=1.3\tan^4\theta=1.

So

tan⁡4θ=13.\tan^4\theta=\frac13.

Taking fourth roots,

tan⁡θ=±134.\tan\theta=\pm\sqrt[4]{\frac13}.

Therefore

θ=arctan⁡(134)orθ=−arctan⁡(134).\theta=\arctan\left(\sqrt[4]{\frac13}\right) \quad\text{or}\quad \theta=-\arctan\left(\sqrt[4]{\frac13}\right).

These lie in the required interval

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

Numerically,

arctan⁡(134)=0.6493….\arctan\left(\sqrt[4]{\frac13}\right)=0.6493\ldots.

Hence, to 2 decimal places,

θ=−0.65, 0.65.\boxed{\theta=-0.65,\ 0.65}.