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FP2 2022 Jan Q6

A Level / Edexcel / FP2

IAL 2022 Jan Paper · Question 6

题目

Problem

6. Given that A>B>0A > B > 0, by letting x=arctan⁡Ax = \arctan A and y=arctan⁡By = \arctan B

(a) prove that

 arctan⁡A−arctan⁡B= arctan⁡(A−B1+AB)\begin{align*} &\,\arctan A-\arctan B\\[4mm] =&\,\arctan\Big(\frac{A-B}{1+AB}\Big) \end{align*}
(3)

(b) Show that when A=r+2A = r + 2 and B=rB = r

A−B1+AB=2(1+r)2\begin{align*} \frac{A - B}{1 + AB} = \frac{2}{(1 + r)^2} \end{align*}
(2)

(c) Hence, using the method of differences, show that

 ∑r=1narctan⁡(2(1+r)2)= arctan⁡(n+p)+arctan⁡(n+q) −arctan⁡2−π4\begin{align*} &\,\sum_{r=1}^n\arctan\Big(\frac{2}{(1+r)^2}\Big)\\[4mm] =&\,\arctan(n+p)+\arctan(n+q)\\[4mm] &\,\hspace{2pt}-\arctan2-\frac{\pi}{4} \end{align*}

where pp and qq are integers to be determined.

(4)

(d) Hence, making your reasoning clear, determine

∑r=1∞arctan⁡(2(1+r)2)\begin{align*} \sum_{r=1}^\infty\arctan\Big(\frac{2}{(1+r)^2}\Big) \end{align*}

giving the answer in the form kπ−arctan⁡2k\pi - \arctan 2, where kk is a constant.

(2)
题目中文翻译

6. 已知 A>B>0A>B>0，令 x=arctan⁡Ax=\arctan A、y=arctan⁡By=\arctan B，

(a) 证明

 arctan⁡A−arctan⁡B= arctan⁡(A−B1+AB).\begin{align*} &\,\arctan A-\arctan B\\[4mm] =&\,\arctan\Big(\frac{A-B}{1+AB}\Big). \end{align*}
(3)

(b) 证明，当 A=r+2A=r+2 且 B=rB=r 时，

A−B1+AB=2(1+r)2.\begin{align*} \frac{A-B}{1+AB}=\frac{2}{(1+r)^2}. \end{align*}
(2)

(c) 由此，使用差分法证明

 ∑r=1narctan⁡(2(1+r)2)= arctan⁡(n+p)+arctan⁡(n+q) −arctan⁡2−π4,\begin{align*} &\,\sum_{r=1}^n\arctan\Big(\frac{2}{(1+r)^2}\Big)\\[4mm] =&\,\arctan(n+p)+\arctan(n+q)\\[4mm] &\,\hspace{2pt}-\arctan2-\frac{\pi}{4}, \end{align*}

其中 pp 和 qq 是待求的整数。

(4)

(d) 由此，清楚说明理由，求

∑r=1∞arctan⁡(2(1+r)2),\begin{align*} \sum_{r=1}^\infty\arctan\Big(\frac{2}{(1+r)^2}\Big), \end{align*}

将答案写成 kπ−arctan⁡2k\pi-\arctan2 的形式，其中 kk 为常数。

(2)

解答

(a)

解法一

思路

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令 x=arctan⁡Ax=\arctan A 和 y=arctan⁡By=\arctan B，就有 tan⁡x=A\tan x=A、tan⁡y=B\tan y=B。使用

tan⁡(x−y)=tan⁡x−tan⁡y1+tan⁡xtan⁡y.\begin{align*} \tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}. \end{align*}

因为 A>B>0A>B>0，所以 x>y>0x>y>0，差角在主值范围内，可以对两边取 arctan⁡\arctan。

答题过程

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Let

x=arctan⁡A,y=arctan⁡B.\begin{align*} x=\arctan A,\qquad y=\arctan B. \end{align*}

Then

tan⁡x=A,tan⁡y=B.\begin{align*} \tan x=A,\qquad \tan y=B. \end{align*}

Using the compound angle formula,

tan⁡(x−y)= tan⁡x−tan⁡y1+tan⁡xtan⁡y= A−B1+AB.\begin{align*} \tan(x-y) =&\,\frac{\tan x-\tan y}{1+\tan x\tan y}\\[2mm] =&\,\frac{A-B}{1+AB}. \end{align*}

Since A>B>0A>B>0, we have x>y>0x>y>0, so x−yx-y is in the principal range for arctan⁡\arctan. Hence

x−y= arctan⁡(A−B1+AB).\begin{align*} x-y =&\,\arctan\Big(\frac{A-B}{1+AB}\Big). \end{align*}

Therefore

arctan⁡A−arctan⁡B= arctan⁡(A−B1+AB).\begin{align*} \arctan A-\arctan B =&\,\arctan\Big(\frac{A-B}{1+AB}\Big). \end{align*}

(b)

解法一

思路

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直接代入 A=r+2A=r+2、B=rB=r。关键是把分母展开后写成 (r+1)2(r+1)^2。

答题过程

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When

A=r+2,B=r,\begin{align*} A=r+2,\qquad B=r, \end{align*}

we have

A−B1+AB= (r+2)−r1+(r+2)r= 21+r2+2r= 2(r+1)2.\begin{align*} \frac{A-B}{1+AB} =&\,\frac{(r+2)-r}{1+(r+2)r}\\[2mm] =&\,\frac{2}{1+r^2+2r}\\[2mm] =&\,\frac{2}{(r+1)^2}. \end{align*}

(c)

解法一

思路

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由 (a)、(b) 可得

arctan⁡(2(r+1)2)=arctan⁡(r+2)−arctan⁡r.\begin{align*} \arctan\Big(\frac{2}{(r+1)^2}\Big) =\arctan(r+2)-\arctan r. \end{align*}

然后把 r=1r=1 到 nn 展开，中间项会抵消。

答题过程

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From parts (a) and (b),

arctan⁡(2(r+1)2)= arctan⁡(r+2)−arctan⁡r.\begin{align*} \arctan\Big(\frac{2}{(r+1)^2}\Big) =&\,\arctan(r+2)-\arctan r. \end{align*}

Therefore

 ∑r=1narctan⁡(2(r+1)2)= ∑r=1n[arctan⁡(r+2)−arctan⁡r].\begin{align*} &\,\sum_{r=1}^n \arctan\Big(\frac{2}{(r+1)^2}\Big)\\[4mm] =&\,\sum_{r=1}^n \Big[\arctan(r+2)-\arctan r\Big]. \end{align*}

Write out the terms:

 (arctan⁡3−arctan⁡1) +(arctan⁡4−arctan⁡2) +(arctan⁡5−arctan⁡3) +⋯ +(arctan⁡(n+1)−arctan⁡(n−1)) +(arctan⁡(n+2)−arctan⁡n).\begin{align*} &\,\Big(\arctan3-\arctan1\Big)\\[4mm] &\,\hspace{2pt}+\Big(\arctan4-\arctan2\Big)\\[4mm] &\,\hspace{4pt}+\Big(\arctan5-\arctan3\Big)\\[4mm] &\,\hspace{6pt}+\cdots\\[4mm] &\,\hspace{8pt}+\Big(\arctan(n+1)-\arctan(n-1)\Big)\\[4mm] &\,\hspace{10pt}+\Big(\arctan(n+2)-\arctan n\Big). \end{align*}

The middle terms cancel, leaving

 ∑r=1narctan⁡(2(r+1)2)= arctan⁡(n+2)+arctan⁡(n+1) −arctan⁡2−arctan⁡1.\begin{align*} &\,\sum_{r=1}^n \arctan\Big(\frac{2}{(r+1)^2}\Big)\\[4mm] =&\,\arctan(n+2)+\arctan(n+1)\\[2mm] &\,\hspace{2pt}-\arctan2-\arctan1. \end{align*}

Since

arctan⁡1=π4,\begin{align*} \arctan1=\frac{\pi}{4}, \end{align*}

we have

 ∑r=1narctan⁡(2(r+1)2)= arctan⁡(n+2)+arctan⁡(n+1) −arctan⁡2−π4.\begin{align*} &\,\sum_{r=1}^n \arctan\Big(\frac{2}{(r+1)^2}\Big)\\[4mm] =&\,\arctan(n+2)+\arctan(n+1)\\[2mm] &\,\hspace{2pt}-\arctan2-\frac{\pi}{4}. \end{align*}

Thus

p=2,q=1.\begin{align*} p=2,\qquad q=1. \end{align*}

(d)

解法一

思路

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当 n→∞n\to\infty 时，

arctan⁡(n+2)→π2,arctan⁡(n+1)→π2.\begin{align*} \arctan(n+2)\to\frac{\pi}{2}, \qquad \arctan(n+1)\to\frac{\pi}{2}. \end{align*}

把这个代入 (c) 的结果即可。

答题过程

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From part (c),

As n→∞n\to\infty,

arctan⁡(n+2)→π2,arctan⁡(n+1)→π2.\begin{align*} \arctan(n+2)\to\frac{\pi}{2}, \qquad \arctan(n+1)\to\frac{\pi}{2}. \end{align*}

Therefore

∑r=1∞arctan⁡(2(r+1)2)= π2+π2−arctan⁡2−π4= 3π4−arctan⁡2.\begin{align*} \sum_{r=1}^{\infty} \arctan\Big(\frac{2}{(r+1)^2}\Big) =&\,\frac{\pi}{2}+\frac{\pi}{2} -\arctan2-\frac{\pi}{4}\\[2mm] =&\,\frac{3\pi}{4}-\arctan2. \end{align*}

So

k=34.\begin{align*} k=\frac34. \end{align*}