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FP2 2025 Jan Q7

A Level / Edexcel / FP2

IAL 2025 Jan Paper · Question 7

题目

Problem

7.

(a) Use De Moivre’s theorem to

(i) show that

sin⁡5θ≡ 5cos⁡4θsin⁡θ−10cos⁡2θsin⁡3θ+sin⁡5θ\begin{align*} \sin5\theta \equiv&\, 5\cos^4\theta\sin\theta -10\cos^2\theta\sin^3\theta +\sin^5\theta \end{align*}

(ii) determine an expression for cos⁡5θ\cos5\theta in terms of sin⁡θ\sin\theta and cos⁡θ\cos\theta

(4)

(b) Hence show that, for cos⁡5θ≠0\cos5\theta\neq0

tan⁡5θ= 5tan⁡θ−10tan⁡3θ+tan⁡5θ1−10tan⁡2θ+5tan⁡4θ\begin{align*} \tan5\theta =&\, \frac{5\tan\theta-10\tan^3\theta+\tan^5\theta} {1-10\tan^2\theta+5\tan^4\theta} \end{align*}
(2)

(c) Using the result of part (b) and showing all stages of your working, determine the solutions of the equation

2x5−15x4−20x3+30x2+10x−3=0\begin{align*} 2x^5-15x^4-20x^3+30x^2+10x-3=0 \end{align*}

giving your answers to 33 decimal places.

(5)
题目中文翻译

(a) 使用棣莫弗定理：

(i) 证明

sin⁡5θ≡ 5cos⁡4θsin⁡θ −10cos⁡2θsin⁡3θ+sin⁡5θ;\begin{align*} \sin5\theta\equiv&\,5\cos^4\theta\sin\theta\\[3mm] &\,\hspace{2pt}-10\cos^2\theta\sin^3\theta+\sin^5\theta; \end{align*}

(ii) 求用 sin⁡θ\sin\theta 和 cos⁡θ\cos\theta 表示的 cos⁡5θ\cos5\theta 的表达式。

(4)

(b) 由此证明，当 cos⁡5θ≠0\cos5\theta\ne0 时，

tan⁡5θ= 5tan⁡θ−10tan⁡3θ+tan⁡5θ1−10tan⁡2θ+5tan⁡4θ.\begin{align*} \tan5\theta =&\,\frac{5\tan\theta-10\tan^3\theta+\tan^5\theta} {1-10\tan^2\theta+5\tan^4\theta}. \end{align*}
(2)

(c) 利用 (b) 的结果，并展示所有计算步骤，求方程

2x5−15x4−20x3+30x2+10x−3=0\begin{align*} 2x^5-15x^4-20x^3+30x^2+10x-3=0 \end{align*}

的所有解，答案保留 33 位小数。

(5)

解答

(a)

解法一

思路

展开

用 De Moivre’s theorem 展开 (cos⁡θ+isin⁡θ)5(\cos\theta+\mathrm{i}\sin\theta)^5。虚部给 sin⁡5θ\sin5\theta，实部给 cos⁡5θ\cos5\theta。

答题过程

展开

By De Moivre’s theorem,

(cos⁡θ+isin⁡θ)5=cos⁡5θ+isin⁡5θ\begin{align*} (\cos\theta+\mathrm{i}\sin\theta)^5 =\cos5\theta+\mathrm{i}\sin5\theta \end{align*}

Expand the left hand side:

(cos⁡θ+isin⁡θ)5= cos⁡5θ+5icos⁡4θsin⁡θ −10cos⁡3θsin⁡2θ −10icos⁡2θsin⁡3θ +5cos⁡θsin⁡4θ+isin⁡5θ\begin{align*} (\cos\theta+\mathrm{i}\sin\theta)^5 =&\, \cos^5\theta +5\mathrm{i}\cos^4\theta\sin\theta\\[2mm] &\,\hspace{2pt}-10\cos^3\theta\sin^2\theta\\[2mm] &\,\hspace{4pt}-10\mathrm{i}\cos^2\theta\sin^3\theta\\[2mm] &\,\hspace{6pt}+5\cos\theta\sin^4\theta +\mathrm{i}\sin^5\theta \end{align*}

(i) Equating imaginary parts gives

sin⁡5θ= 5cos⁡4θsin⁡θ−10cos⁡2θsin⁡3θ+sin⁡5θ\begin{align*} \sin5\theta =&\, 5\cos^4\theta\sin\theta -10\cos^2\theta\sin^3\theta +\sin^5\theta \end{align*}

(ii) Equating real parts gives

cos⁡5θ=cos⁡5θ−10cos⁡3θsin⁡2θ+5cos⁡θsin⁡4θ\begin{align*} \boxed{ \cos5\theta = \cos^5\theta -10\cos^3\theta\sin^2\theta +5\cos\theta\sin^4\theta } \end{align*}

(b)

解法一

思路

展开

由 tan⁡5θ=sin⁡5θcos⁡5θ\tan5\theta=\frac{\sin5\theta}{\cos5\theta}。把 (a) 的两个式子相除，再把分子分母同除以 cos⁡5θ\cos^5\theta，就会全部变成 tan⁡θ\tan\theta。

答题过程

展开

For cos⁡5θ≠0\cos5\theta\neq0,

tan⁡5θ= sin⁡5θcos⁡5θ= 5cos⁡4θsin⁡θ−10cos⁡2θsin⁡3θ+sin⁡5θcos⁡5θ−10cos⁡3θsin⁡2θ+5cos⁡θsin⁡4θ\begin{align*} \tan5\theta =&\, \frac{\sin5\theta}{\cos5\theta}\\[3mm] =&\, \frac{ 5\cos^4\theta\sin\theta -10\cos^2\theta\sin^3\theta +\sin^5\theta } { \cos^5\theta -10\cos^3\theta\sin^2\theta +5\cos\theta\sin^4\theta } \end{align*}

Divide numerator and denominator by cos⁡5θ\cos^5\theta:

tan⁡5θ= 5sin⁡θcos⁡θ−10sin⁡3θcos⁡3θ+sin⁡5θcos⁡5θ1−10sin⁡2θcos⁡2θ+5sin⁡4θcos⁡4θ= 5tan⁡θ−10tan⁡3θ+tan⁡5θ1−10tan⁡2θ+5tan⁡4θ\begin{align*} \tan5\theta =&\, \frac{ 5\frac{\sin\theta}{\cos\theta} -10\frac{\sin^3\theta}{\cos^3\theta} +\frac{\sin^5\theta}{\cos^5\theta} } { 1 -10\frac{\sin^2\theta}{\cos^2\theta} +5\frac{\sin^4\theta}{\cos^4\theta} }\\[4mm] =&\, \frac{5\tan\theta-10\tan^3\theta+\tan^5\theta} {1-10\tan^2\theta+5\tan^4\theta} \end{align*}

(c)

解法一

思路

展开

令 x=tan⁡θx=\tan\theta。题目的五次方程可以整理成

5x−10x3+x51−10x2+5x4=32\begin{align*} \frac{5x-10x^3+x^5}{1-10x^2+5x^4}=\frac32 \end{align*}

所以由 (b) 得 tan⁡5θ=32\tan5\theta=\frac32。在 tan⁡θ\tan\theta 的一个周期内，取 5θ5\theta 相差 π\pi 的五个角，可求出五个不同的实根。方程的次数为 55，因此这些就是全部根。

答题过程

展开

Let

x=tan⁡θ\begin{align*} x=\tan\theta \end{align*}

Using part (b),

tan⁡5θ= 5x−10x3+x51−10x2+5x4\begin{align*} \tan5\theta =&\, \frac{5x-10x^3+x^5}{1-10x^2+5x^4} \end{align*}

Now

2x5−15x4−20x3+30x2+10x−3= 02x5−20x3+10x= 15x4−30x2+32(x5−10x3+5x)= 3(5x4−10x2+1)\begin{align*} 2x^5-15x^4-20x^3+30x^2+10x-3 =&\, 0\\[2mm] 2x^5-20x^3+10x =&\, 15x^4-30x^2+3\\[2mm] 2(x^5-10x^3+5x) =&\, 3(5x^4-10x^2+1) \end{align*}

Hence

x5−10x3+5x5x4−10x2+1= 32\begin{align*} \frac{x^5-10x^3+5x}{5x^4-10x^2+1} =&\, \frac32 \end{align*}

Therefore

tan⁡5θ=32\begin{align*} \tan5\theta=\frac32 \end{align*}

So

5θ= arctan⁡32+kπ\begin{align*} 5\theta =&\, \arctan\frac32+k\pi \end{align*}

For five distinct values over one period of tan⁡θ\tan\theta, take k=0,1,2,3,4k=0,1,2,3,4:

θ= arctan⁡32+kπ5k=0,1,2,3,4\begin{align*} \theta =&\, \frac{\arctan\frac32+k\pi}{5} \qquad k=0,1,2,3,4 \end{align*}

Then

x= tan⁡(arctan⁡32+kπ5)\begin{align*} x =&\, \tan\bigg(\frac{\arctan\frac32+k\pi}{5}\bigg) \end{align*}

This gives

x= 0.1991…,1.0822…,8.464…, −1.7847…,−0.4607…\begin{align*} x =&\, 0.1991\ldots,\quad 1.0822\ldots,\quad 8.464\ldots,\\[2mm] &\,\hspace{2pt}-1.7847\ldots,\quad -0.4607\ldots \end{align*}

Therefore the solutions, to 33 decimal places, are

x=−1.785, −0.461, 0.199, 1.082, 8.464\begin{align*} \boxed{ x=-1.785,\ -0.461,\ 0.199,\ 1.082,\ 8.464 } \end{align*}