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IAL 2021 Jan Q8

A Level / Edexcel / FP2

IAL 2021 Jan Paper · Question 8

题目

Problem

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Given that z=eiθz = \mathrm{e}^{\mathrm{i}\theta}

(a) show that zn+1zn=2cos⁡nθz^n + \dfrac{1}{z^n} = 2\cos n\theta

where nn is a positive integer.

(2)

(b) Show that

cos⁡6θ=132(cos⁡6θ+6cos⁡4θ+15cos⁡2θ+10)\begin{align*} \cos^6\theta = \frac{1}{32}\left(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10\right) \end{align*}
(5)

(c) Hence solve the equation

cos⁡6θ+6cos⁡4θ+15cos⁡2θ=00⩽θ⩽π\begin{align*} \cos 6\theta + 6\cos 4\theta + 15\cos 2\theta = 0 \qquad 0 \leqslant \theta \leqslant \pi \end{align*}

Give your answers to 33 significant figures.

(4)

(d) Use calculus to determine the exact value of

∫0π3(32cos⁡6θ−4cos⁡2θ) dθ\begin{align*} \int_{0}^{\frac{\pi}{3}} \left(32\cos^6\theta - 4\cos^2\theta\right)\,\mathrm{d}\theta \end{align*}
Solutions relying entirely on calculator technology are not acceptable.
(5)

解答

(a)

解法一

思路

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用 Euler form：zn=einθz^n=\mathrm{e}^{\mathrm{i}n\theta}，而 1zn=e−inθ\dfrac1{z^n}=\mathrm{e}^{-\mathrm{i}n\theta}。两者相加时虚部抵消，只剩 2cos⁡nθ2\cos n\theta。

答题过程

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Since z=eiθz=\mathrm{e}^{\mathrm{i}\theta},

zn= einθ=cos⁡nθ+isin⁡nθ.\begin{align*} z^n =&\,\mathrm{e}^{\mathrm{i}n\theta} =\cos n\theta+\mathrm{i}\sin n\theta. \end{align*}

Also,

1zn= e−inθ= cos⁡(−nθ)+isin⁡(−nθ)= cos⁡nθ−isin⁡nθ.\begin{align*} \frac1{z^n} =&\,\mathrm{e}^{-\mathrm{i}n\theta}\\[4mm] =&\,\cos(-n\theta)+\mathrm{i}\sin(-n\theta)\\[4mm] =&\,\cos n\theta-\mathrm{i}\sin n\theta. \end{align*}

Therefore,

zn+1zn= cos⁡nθ+isin⁡nθ+cos⁡nθ−isin⁡nθ= 2cos⁡nθ.\begin{align*} z^n+\frac1{z^n} =&\,\cos n\theta+\mathrm{i}\sin n\theta+\cos n\theta-\mathrm{i}\sin n\theta\\[4mm] =&\,2\cos n\theta. \end{align*}

(b)

解法一

思路

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因为 z+1z=2cos⁡θz+\dfrac1z=2\cos\theta，所以展开 (z+1z)6\left(z+\dfrac1z\right)^6，再把 zk+1zkz^k+\dfrac1{z^k} 换成 2cos⁡kθ2\cos k\theta。

答题过程

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From z=eiθz=\mathrm{e}^{\mathrm{i}\theta},

z+1z=2cos⁡θ.\begin{align*} z+\frac1z=2\cos\theta. \end{align*}

Hence,

64cos⁡6θ= (z+1z)6= z6+6z5(1z) +15z4(1z2)+20z3(1z3) +15z2(1z4)+6z(1z5) +1z6= z6+6z4+15z2+20 +15z2+6z4+1z6= (z6+1z6)+6(z4+1z4) +15(z2+1z2)+20.\begin{align*} 64\cos^6\theta =&\,\left(z+\frac1z\right)^6\\[4mm] =&\,z^6+6z^5\left(\frac1z\right)\\[4mm] &\,+15z^4\left(\frac1{z^2}\right) +20z^3\left(\frac1{z^3}\right)\\[4mm] &\,+15z^2\left(\frac1{z^4}\right) +6z\left(\frac1{z^5}\right)\\[4mm] &\,+\frac1{z^6}\\[4mm] =&\,z^6+6z^4+15z^2+20\\[4mm] &\,+\frac{15}{z^2}+\frac6{z^4}+\frac1{z^6}\\[4mm] =&\,\left(z^6+\frac1{z^6}\right) +6\left(z^4+\frac1{z^4}\right)\\[4mm] &\,+15\left(z^2+\frac1{z^2}\right)+20. \end{align*}

Using part (a),

64cos⁡6θ= 2cos⁡6θ+12cos⁡4θ +30cos⁡2θ+20.\begin{align*} 64\cos^6\theta =&\,2\cos6\theta+12\cos4\theta\\[4mm] &\,+30\cos2\theta+20. \end{align*}

Therefore,

cos⁡6θ= 132(cos⁡6θ+6cos⁡4θ +15cos⁡2θ+10).\begin{align*} \cos^6\theta =&\,\frac1{32} \left(\cos6\theta+6\cos4\theta\right.\\[4mm] &\,\hspace{24pt}\left.+15\cos2\theta+10\right). \end{align*}

(c)

解法一

思路

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用 (b) 把题目左边换成 32cos⁡6θ−1032\cos^6\theta-10。然后解 cos⁡6θ=516\cos^6\theta=\dfrac5{16}，在 0⩽θ⩽π0\leqslant\theta\leqslant\pi 内取两个角。

答题过程

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From part (b),

cos⁡6θ+6cos⁡4θ+15cos⁡2θ+10= 32cos⁡6θ.\begin{align*} \cos6\theta+6\cos4\theta+15\cos2\theta+10 =&\,32\cos^6\theta. \end{align*}

So the given equation becomes

32cos⁡6θ−10= 0cos⁡6θ= 516.\begin{align*} 32\cos^6\theta-10=&\,0\\[4mm] \cos^6\theta=&\,\frac5{16}. \end{align*}

Therefore,

cos⁡θ=±(516)1/6.\begin{align*} \cos\theta=\pm\left(\frac5{16}\right)^{1/6}. \end{align*}

For 0⩽θ⩽π0\leqslant\theta\leqslant\pi,

θ= 0.6027…,2.5388…\begin{align*} \theta=&\,0.6027\ldots,\quad 2.5388\ldots \end{align*}

Hence, to 33 significant figures,

θ=0.603,2.54.\begin{align*} \theta=0.603,\quad 2.54. \end{align*}

(d)

解法一

思路

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题目说用 calculus，且不能只靠计算器。先用 (b) 把 32cos⁡6θ32\cos^6\theta 改成多角表达式，再用

4cos⁡2θ=2+2cos⁡2θ.\begin{align*} 4\cos^2\theta=2+2\cos2\theta. \end{align*}

这样就能逐项积分。

答题过程

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Using part (b),

32cos⁡6θ= cos⁡6θ+6cos⁡4θ +15cos⁡2θ+10.\begin{align*} 32\cos^6\theta =&\,\cos6\theta+6\cos4\theta\\[4mm] &\,+15\cos2\theta+10. \end{align*}

Also,

4cos⁡2θ=2+2cos⁡2θ.\begin{align*} 4\cos^2\theta=2+2\cos2\theta. \end{align*}

Therefore,

 ∫0π3(32cos⁡6θ−4cos⁡2θ) dθ= ∫0π3(cos⁡6θ+6cos⁡4θ +13cos⁡2θ+8) dθ= [16sin⁡6θ+32sin⁡4θ+132sin⁡2θ+8θ]0π3.\begin{align*} &\,\int_0^{\frac{\pi}{3}} \left(32\cos^6\theta-4\cos^2\theta\right)\,\mathrm{d}\theta\\[4mm] =&\,\int_0^{\frac{\pi}{3}} \left(\cos6\theta+6\cos4\theta\right.\\[4mm] &\,\hspace{42pt}\left.+13\cos2\theta+8\right) \,\mathrm{d}\theta\\[4mm] =&\,\left[ \frac16\sin6\theta+\frac32\sin4\theta +\frac{13}{2}\sin2\theta+8\theta \right]_0^{\frac{\pi}{3}}. \end{align*}

Substituting the limits,

 16sin⁡2π+32sin⁡4π3 +132sin⁡2π3+8π3= 0+32(−32) +132(32)+8π3= 532+8π3.\begin{align*} &\,\frac16\sin2\pi+\frac32\sin\frac{4\pi}{3}\\[4mm] &\,+\frac{13}{2}\sin\frac{2\pi}{3} +\frac{8\pi}{3}\\[4mm] =&\,0+\frac32\left(-\frac{\sqrt3}{2}\right)\\[4mm] &\,+\frac{13}{2}\left(\frac{\sqrt3}{2}\right) +\frac{8\pi}{3}\\[4mm] =&\,\frac{5\sqrt3}{2}+\frac{8\pi}{3}. \end{align*}