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FP2 2021 June Q7

A Level / Edexcel / FP2

IAL 2021 June Paper · Question 7

题目

Problem

7. (a) Use de Moivre’s theorem to show that

 tan⁡4θ≡ 4tan⁡θ−4tan⁡3θ1−6tan⁡2θ+tan⁡4θ\begin{align*} &\,\tan4\theta\\[4mm] \equiv&\,\frac{4\tan\theta-4\tan^3\theta}{1-6\tan^2\theta+\tan^4\theta} \end{align*}
(6)

(b) Use the identity given in part (a) to find the 22 positive roots of

 x4+2x3−6x2 −2x+1= 0.\begin{align*} &\,x^4+2x^3-6x^2\\[4mm] &\,-2x+1\\[4mm] =&\,0. \end{align*}

giving your answers to 33 significant figures.

(3)
题目中文翻译

7. (a) 利用 de Moivre 定理证明

 tan⁡4θ≡ 4tan⁡θ−4tan⁡3θ1−6tan⁡2θ+tan⁡4θ\begin{align*} &\,\tan4\theta\\[4mm] \equiv&\,\frac{4\tan\theta-4\tan^3\theta}{1-6\tan^2\theta+\tan^4\theta} \end{align*}
(6)

(b) 利用 (a) 中给出的恒等式，求方程

 x4+2x3−6x2 −2x+1= 0.\begin{align*} &\,x^4+2x^3-6x^2\\[4mm] &\,-2x+1\\[4mm] =&\,0. \end{align*}

的两个正根，答案保留 33 位有效数字。

(3)

解答

(a)

解法一

思路

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用 de Moivre 定理展开四次方，再比较实部和虚部，分别得到 cos⁡4θ\cos4\theta 与 sin⁡4θ\sin4\theta。最后取二者之比，并把分子、分母同时除以 cos⁡4θ\cos^4\theta，就能写成只含 tan⁡θ\tan\theta 的形式。

答题过程

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For compact notation, let

c= cos⁡θ,s= sin⁡θ.\begin{align*} c=&\,\cos\theta,\\[4mm] s=&\,\sin\theta. \end{align*}

By de Moivre’s theorem,

 (c+is)4= cos⁡4θ+isin⁡4θ.\begin{align*} &\,(c+\mathrm{i}s)^4\\[4mm] =&\,\cos4\theta+\mathrm{i}\sin4\theta. \end{align*}

Expanding all five terms,

 (c+is)4= c4+4ic3s −6c2s2 −4ics3+s4.\begin{align*} &\,(c+\mathrm{i}s)^4\\[4mm] =&\,c^4+4\mathrm{i}c^3s\\[4mm] &\,\hspace{2pt}-6c^2s^2\\[4mm] &\,\hspace{4pt}-4\mathrm{i}cs^3+s^4. \end{align*}

Comparing real and imaginary parts,

cos⁡4θ= c4−6c2s2+s4,sin⁡4θ= 4c3s−4cs3.\begin{align*} \cos4\theta=&\,c^4-6c^2s^2+s^4,\\[4mm] \sin4\theta=&\,4c^3s-4cs^3. \end{align*}

Therefore,

 tan⁡4θ= sin⁡4θcos⁡4θ= 4c3s−4cs3c4−6c2s2+s4.\begin{align*} &\,\tan4\theta\\[4mm] =&\,\frac{\sin4\theta}{\cos4\theta}\\[4mm] =&\,\frac{4c^3s-4cs^3}{c^4-6c^2s^2+s^4}. \end{align*}

Divide numerator and denominator by c4c^4. Since sc=tan⁡θ\dfrac{s}{c}=\tan\theta, this gives

 tan⁡4θ= 4tan⁡θ−4tan⁡3θ1−6tan⁡2θ+tan⁡4θ.\begin{align*} &\,\tan4\theta\\[4mm] =&\,\frac{4\tan\theta-4\tan^3\theta} {1-6\tan^2\theta+\tan^4\theta}. \end{align*}

This is the required identity.

解法二

思路

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也可以同时展开 z4z^4 与 z−4z^{-4}。两式相加时虚部抵消，相减时实部抵消，因而分别得到 cos⁡4θ\cos4\theta 与 sin⁡4θ\sin4\theta。最后同样取比值，并把分子、分母除以 cos⁡4θ\cos^4\theta。

答题过程

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Let

c= cos⁡θ,s= sin⁡θ,z= c+is.\begin{align*} c=&\,\cos\theta,\\[4mm] s=&\,\sin\theta,\\[4mm] z=&\,c+\mathrm{i}s. \end{align*}

Since c2+s2=1c^2+s^2=1,

z−1=c−is.z^{-1}=c-\mathrm{i}s.

By de Moivre’s theorem,

 z4= cos⁡4θ+isin⁡4θ, z−4= cos⁡4θ−isin⁡4θ.\begin{align*} &\,z^4\\[4mm] =&\,\cos4\theta+\mathrm{i}\sin4\theta,\\[4mm] &\,z^{-4}\\[4mm] =&\,\cos4\theta-\mathrm{i}\sin4\theta. \end{align*}

Hence

z4+z−4= 2cos⁡4θ,z4−z−4= 2isin⁡4θ.\begin{align*} z^4+z^{-4}=&\,2\cos4\theta,\\[4mm] z^4-z^{-4}=&\,2\mathrm{i}\sin4\theta. \end{align*}

Expanding both fourth powers completely,

z4= (c+is)4= c4+4ic3s −6c2s2 −4ics3+s4\begin{align*} z^4=&\,(c+\mathrm{i}s)^4\\[4mm] =&\,c^4+4\mathrm{i}c^3s\\[4mm] &\,\hspace{2pt}-6c^2s^2\\[4mm] &\,\hspace{4pt}-4\mathrm{i}cs^3+s^4 \end{align*}

and

z−4= (c−is)4= c4−4ic3s −6c2s2 +4ics3+s4.\begin{align*} z^{-4}=&\,(c-\mathrm{i}s)^4\\[4mm] =&\,c^4-4\mathrm{i}c^3s\\[4mm] &\,\hspace{2pt}-6c^2s^2\\[4mm] &\,\hspace{4pt}+4\mathrm{i}cs^3+s^4. \end{align*}

Adding gives

 2cos⁡4θ= 2c4−12c2s2+2s4,\begin{align*} &\,2\cos4\theta\\[4mm] =&\,2c^4-12c^2s^2+2s^4, \end{align*}

so

cos⁡4θ=c4−6c2s2+s4.\cos4\theta=c^4-6c^2s^2+s^4.

Subtracting gives

 2isin⁡4θ= 8ic3s−8ics3,\begin{align*} &\,2\mathrm{i}\sin4\theta\\[4mm] =&\,8\mathrm{i}c^3s-8\mathrm{i}cs^3, \end{align*}

so

sin⁡4θ=4c3s−4cs3.\sin4\theta=4c^3s-4cs^3.

Therefore,

 tan⁡4θ= 4c3s−4cs3c4−6c2s2+s4.\begin{align*} &\,\tan4\theta\\[4mm] =&\,\frac{4c^3s-4cs^3}{c^4-6c^2s^2+s^4}. \end{align*}

Dividing numerator and denominator by c4c^4 and using sc=tan⁡θ\dfrac{s}{c}=\tan\theta,

 tan⁡4θ= 4tan⁡θ−4tan⁡3θ1−6tan⁡2θ+tan⁡4θ.\begin{align*} &\,\tan4\theta\\[4mm] =&\,\frac{4\tan\theta-4\tan^3\theta} {1-6\tan^2\theta+\tan^4\theta}. \end{align*}

This is the required identity.

(b)

解法一

思路

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令 x=tan⁡θx=\tan\theta，把四次方程重排后，分母恰好成为 2x−2x32x-2x^3，因此由 (a) 得到 tan⁡4θ=2\tan4\theta=2。题目只要正根，可取第一象限的 θ\theta，求出对应的两个角，再还原成 xx。

答题过程

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For positive xx, let x=tan⁡θx=\tan\theta with 0<θ<π20<\theta<\dfrac\pi2.

The given equation is

 x4+2x3−6x2 −2x+1= 0.\begin{align*} &\,x^4+2x^3-6x^2\\[4mm] &\,\hspace{2pt}-2x+1\\[4mm] =&\,0. \end{align*}

Rearranging,

1−6x2+x4=2x−2x3.1-6x^2+x^4=2x-2x^3.

Using part (a),

tan⁡4θ= 4x−4x31−6x2+x4= 4x−4x32x−2x3= 2.\begin{align*} \tan4\theta =&\,\frac{4x-4x^3}{1-6x^2+x^4}\\[4mm] =&\,\frac{4x-4x^3}{2x-2x^3}\\[4mm] =&\,2. \end{align*}

Thus

4θ=arctan⁡2+kπ.4\theta=\arctan2+k\pi.

Since 0<4θ<2π0<4\theta<2\pi, the two suitable values are

θ= 14arctan⁡2or θ= 14(arctan⁡2+π).\begin{align*} \theta=&\,\frac14\arctan2\\[4mm] \text{or }\theta=&\,\frac14(\arctan2+\pi). \end{align*}

Hence the positive roots are

x= tan⁡(arctan⁡24).\begin{align*} x=&\,\tan\bigg(\frac{\arctan2}{4}\bigg). \end{align*}

or

x= tan⁡(arctan⁡2+π4).\begin{align*} x=&\,\tan\bigg(\frac{\arctan2+\pi}{4}\bigg). \end{align*}

Therefore, to 33 significant figures,

x=0.284,1.79.x=0.284,\qquad1.79.