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IAL 2021 June Q7

A Level / Edexcel / FP2

IAL 2021 June Paper · Question 7

题目

Problem

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

tan⁡4θ=4tan⁡θ−4tan⁡3θ1−6tan⁡2θ+tan⁡4θ\begin{align*} \tan 4\theta = \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 - 2x + 1 = 0 \end{align*}

giving your answers to 33 significant figures.

(3)

解答

(a)

解法一

思路

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用 de Moivre’s theorem 展开

(cos⁡θ+isin⁡θ)4=cos⁡4θ+isin⁡4θ.\begin{align*} (\cos\theta+\mathrm{i}\sin\theta)^4 =\cos4\theta+\mathrm{i}\sin4\theta. \end{align*}

然后比较实部和虚部，得到 cos⁡4θ\cos4\theta 与 sin⁡4θ\sin4\theta。最后用

tan⁡4θ=sin⁡4θcos⁡4θ\begin{align*} \tan4\theta=\frac{\sin4\theta}{\cos4\theta} \end{align*}

并把分子分母同时除以 cos⁡4θ\cos^4\theta。

答题过程

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By de Moivre’s theorem,

(cos⁡θ+isin⁡θ)4=cos⁡4θ+isin⁡4θ.\begin{align*} (\cos\theta+\mathrm{i}\sin\theta)^4 =\cos4\theta+\mathrm{i}\sin4\theta. \end{align*}

Expand the left hand side:

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

Comparing real and imaginary parts,

cos⁡4θ= cos⁡4θ−6cos⁡2θsin⁡2θ+sin⁡4θ,\begin{align*} \cos4\theta =&\,\cos^4\theta -6\cos^2\theta\sin^2\theta +\sin^4\theta, \end{align*}

and

sin⁡4θ= 4cos⁡3θsin⁡θ−4cos⁡θsin⁡3θ.\begin{align*} \sin4\theta =&\,4\cos^3\theta\sin\theta -4\cos\theta\sin^3\theta. \end{align*}

Therefore,

tan⁡4θ= sin⁡4θcos⁡4θ= 4cos⁡3θsin⁡θ−4cos⁡θsin⁡3θcos⁡4θ−6cos⁡2θsin⁡2θ+sin⁡4θ.\begin{align*} \tan4\theta =&\,\frac{\sin4\theta}{\cos4\theta}\\[4mm] =&\,\frac{ 4\cos^3\theta\sin\theta -4\cos\theta\sin^3\theta }{ \cos^4\theta -6\cos^2\theta\sin^2\theta +\sin^4\theta }. \end{align*}

Divide the numerator and denominator by cos⁡4θ\cos^4\theta:

tan⁡4θ= 4tan⁡θ−4tan⁡3θ1−6tan⁡2θ+tan⁡4θ.\begin{align*} \tan4\theta =&\,\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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除了直接展开 (cos⁡θ+isin⁡θ)4(\cos\theta+\mathrm{i}\sin\theta)^4，也可以同时使用

z=cos⁡θ+isin⁡θ\begin{align*} z=\cos\theta+\mathrm{i}\sin\theta \end{align*}

和

z−1=cos⁡θ−isin⁡θ.\begin{align*} z^{-1}=\cos\theta-\mathrm{i}\sin\theta. \end{align*}

这样 z4+z−4z^4+z^{-4} 会给出 cos⁡4θ\cos4\theta，而 z4−z−4z^4-z^{-4} 会给出 sin⁡4θ\sin4\theta。这个方法的好处是实部和虚部分开得很干净。

答题过程

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Let

z=cos⁡θ+isin⁡θ.\begin{align*} z=\cos\theta+\mathrm{i}\sin\theta. \end{align*}

Then

z−1=cos⁡θ−isin⁡θ.\begin{align*} z^{-1}=\cos\theta-\mathrm{i}\sin\theta. \end{align*}

By de Moivre’s theorem,

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

So

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

Now expand:

z4= (cos⁡θ+isin⁡θ)4= cos⁡4θ+4icos⁡3θsin⁡θ −6cos⁡2θsin⁡2θ −4icos⁡θsin⁡3θ+sin⁡4θ,\begin{align*} z^4 =&\,(\cos\theta+\mathrm{i}\sin\theta)^4\\[2mm] =&\,\cos^4\theta +4\mathrm{i}\cos^3\theta\sin\theta\\[2mm] &\,\hspace{2pt}-6\cos^2\theta\sin^2\theta\\[2mm] &\,\hspace{4pt}-4\mathrm{i}\cos\theta\sin^3\theta +\sin^4\theta, \end{align*}

and

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

Adding gives

2cos⁡4θ= 2cos⁡4θ−12cos⁡2θsin⁡2θ +2sin⁡4θ.\begin{align*} 2\cos4\theta =&\,2\cos^4\theta -12\cos^2\theta\sin^2\theta\\[2mm] &\,\hspace{2pt}+2\sin^4\theta. \end{align*}

Hence

cos⁡4θ= cos⁡4θ−6cos⁡2θsin⁡2θ+sin⁡4θ.\begin{align*} \cos4\theta =&\,\cos^4\theta -6\cos^2\theta\sin^2\theta +\sin^4\theta. \end{align*}

Subtracting gives

2isin⁡4θ= 8icos⁡3θsin⁡θ −8icos⁡θsin⁡3θ.\begin{align*} 2\mathrm{i}\sin4\theta =&\,8\mathrm{i}\cos^3\theta\sin\theta\\[2mm] &\,\hspace{2pt}-8\mathrm{i}\cos\theta\sin^3\theta. \end{align*}

Hence

sin⁡4θ= 4cos⁡3θsin⁡θ−4cos⁡θsin⁡3θ.\begin{align*} \sin4\theta =&\,4\cos^3\theta\sin\theta -4\cos\theta\sin^3\theta. \end{align*}

Therefore,

tan⁡4θ= 4cos⁡3θsin⁡θ−4cos⁡θsin⁡3θcos⁡4θ−6cos⁡2θsin⁡2θ+sin⁡4θ.\begin{align*} \tan4\theta =&\,\frac{ 4\cos^3\theta\sin\theta -4\cos\theta\sin^3\theta }{ \cos^4\theta -6\cos^2\theta\sin^2\theta +\sin^4\theta }. \end{align*}

Divide the numerator and denominator by cos⁡4θ\cos^4\theta:

tan⁡4θ= 4tan⁡θ−4tan⁡3θ1−6tan⁡2θ+tan⁡4θ.\begin{align*} \tan4\theta =&\,\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。题目中的 quartic 可以重排成

1−6x2+x4=2x−2x3.\begin{align*} 1-6x^2+x^4=2x-2x^3. \end{align*}

这刚好能和 (a) 的公式配合，得到 tan⁡4θ=2\tan4\theta=2。

答题过程

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Let

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

The equation

x4+2x3−6x2−2x+1=0\begin{align*} x^4+2x^3-6x^2-2x+1=0 \end{align*}

can be rearranged as

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

So

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

Using the identity from 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.\begin{align*} 4\theta=\arctan2+k\pi. \end{align*}

The two positive roots come from

θ= 14arctan⁡2andθ=14(arctan⁡2+π).\begin{align*} \theta =&\,\frac14\arctan2 \quad \text{and} \quad \theta=\frac14(\arctan2+\pi). \end{align*}

Hence

x= tan⁡(14arctan⁡2)orx=tan⁡(14(arctan⁡2+π)).\begin{align*} x =&\,\tan\left(\frac14\arctan2\right) \quad \text{or} \quad x=\tan\left(\frac14(\arctan2+\pi)\right). \end{align*}

Therefore, to 33 significant figures,

x=0.284,1.79.\begin{align*} x=0.284,\qquad 1.79. \end{align*}