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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⁡x,x≠nπ2,n∈Z\begin{align*} \frac{\sin 2x}{\cos x} +&\,\frac{\cos 2x}{\sin x}\\[4mm] \equiv&\,\cosec x, \qquad x\ne\frac{n\pi}{2},\quad n\in\mathbb{Z} \end{align*}
(3)

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

7+sin⁡4θcos⁡2θ+ cos⁡4θsin⁡2θ= 3cot⁡22θ\begin{align*} 7+\frac{\sin 4\theta}{\cos 2\theta} +&\,\frac{\cos 4\theta}{\sin 2\theta}\\[4mm] =&\,3\cot^2 2\theta \end{align*}

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⁡x,x≠nπ2,n∈Z\begin{align*} \frac{\sin 2x}{\cos x} +&\,\frac{\cos 2x}{\sin x}\\[4mm] \equiv&\,\cosec x, \qquad x\ne\frac{n\pi}{2},\quad n\in\mathbb{Z} \end{align*}

(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⁡22θ\begin{align*} 7+\frac{\sin 4\theta}{\cos 2\theta} +&\,\frac{\cos 4\theta}{\sin 2\theta}\\[4mm] =&\,3\cot^2 2\theta \end{align*}

答案用弧度表示，并在适当处保留 3 位有效数字。

解答

(a)

解法一

思路

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把 sin⁡2x\sin2x 与 cos⁡2x\cos2x 都改写成关于 sin⁡x\sin x、cos⁡x\cos x 的式子，约分后即可化成 1/sin⁡x1/\sin x。

答题过程

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Using sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x and cos⁡2x=1−2sin⁡2x\cos2x=1-2\sin^2x,

 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{align*} &\,\frac{\sin2x}{\cos x}+\frac{\cos2x}{\sin x}\\[4mm] =&\,\frac{2\sin x\cos x}{\cos x} +\frac{1-2\sin^2x}{\sin x}\\[4mm] =&\,2\sin x+\frac{1}{\sin x}-2\sin x\\[4mm] =&\,\frac{1}{\sin x}\\[4mm] =&\,\cosec x. \end{align*}

Hence

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

解法二

思路

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按照官方替代路线先通分，分子再用复合角公式识别为 cos⁡(2x−x)\cos(2x-x)。

答题过程

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Combining the fractions gives

 sin⁡2xcos⁡x+cos⁡2xsin⁡x= sin⁡xsin⁡2x+cos⁡xcos⁡2xsin⁡xcos⁡x= cos⁡(2x−x)sin⁡xcos⁡x= cos⁡xsin⁡xcos⁡x= cosec⁡x.\begin{align*} &\,\frac{\sin2x}{\cos x}+\frac{\cos2x}{\sin x}\\[4mm] =&\,\frac{\sin x\sin2x+\cos x\cos2x} {\sin x\cos x}\\[4mm] =&\,\frac{\cos(2x-x)}{\sin x\cos x}\\[4mm] =&\,\frac{\cos x}{\sin x\cos x}\\[4mm] =&\,\cosec x. \end{align*}

Therefore the identity is proved.

(b)

解法一

思路

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先使用 (a) 把左侧两个分式合成 cosec⁡2θ\cosec2\theta，再用 cot⁡2u=cosec⁡2u−1\cot^2u=\cosec^2u-1 得到关于 cosec⁡2θ\cosec2\theta 的二次方程。

答题过程

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Using part (a) with x=2θx=2\theta, the equation becomes

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

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

3cosec⁡22θ−cosec⁡2θ−10=0.3\cosec^2 2\theta-\cosec2\theta-10=0.

Hence

(3cosec⁡2θ+5)(cosec⁡2θ−2)=0,(3\cosec2\theta+5)(\cosec2\theta-2)=0,

so

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

As −π<2θ<π-\pi<2\theta<\pi,

sin⁡2θ=12  ⟹   2θ=π6,5π6,sin⁡2θ=−35  ⟹   2θ=−0.6435…,−2.4981….\begin{align*} \sin2\theta=\frac12 \implies&\,2\theta=\frac{\pi}{6},\frac{5\pi}{6},\\[4mm] \sin2\theta=-\frac35 \implies&\,2\theta=-0.6435\ldots,-2.4981\ldots. \end{align*}

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.

解法二

思路

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官方评分资料也接受把 cosec⁡2θ\cosec2\theta 与 cot⁡22θ\cot^22\theta 直接写成正弦、余弦分式，清除分母后得到关于 sin⁡2θ\sin2\theta 的二次方程。

答题过程

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Using part (a), write

7+1sin⁡2θ=3cos⁡22θsin⁡22θ.7+\frac{1}{\sin2\theta} =3\frac{\cos^22\theta}{\sin^22\theta}.

Multiplying by sin⁡22θ\sin^22\theta and using cos⁡22θ=1−sin⁡22θ\cos^22\theta=1-\sin^22\theta gives

7sin⁡22θ+sin⁡2θ=3(1−sin⁡22θ).7\sin^22\theta+\sin2\theta =3(1-\sin^22\theta).

Therefore

10sin⁡22θ+sin⁡2θ−3=0,10\sin^22\theta+\sin2\theta-3=0,

and hence

(5sin⁡2θ+3)(2sin⁡2θ−1)=0.(5\sin2\theta+3)(2\sin2\theta-1)=0.

Thus

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

Using −π<2θ<π-\pi<2\theta<\pi gives

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