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FP2 2026 Jan A Q7

A Level / Edexcel / FP2

IAL 2026 Jan A Paper · Question 7

题目

Problem

7.

(a) Show that

(z+1z)3(z−1z)3= z6−1z6−k(z2−1z2)\begin{align*} \Big(z+\frac1z\Big)^3 \Big(z-\frac1z\Big)^3 =&\, z^6-\frac{1}{z^6} -k\Big(z^2-\frac{1}{z^2}\Big) \end{align*}

where kk is a constant to be found.

(3)

Given that z=cos⁡θ+isin⁡θz=\cos\theta+\mathrm{i}\sin\theta, where θ\theta is real,

(b) show that

(i)

zn+1zn=2cos⁡nθ\begin{align*} z^n+\frac{1}{z^n}=2\cos n\theta \end{align*}

(ii)

zn−1zn=2isin⁡nθ\begin{align*} z^n-\frac{1}{z^n}=2\mathrm{i}\sin n\theta \end{align*}
(3)

(c) Hence show that

cos⁡3θsin⁡3θ= 132(3sin⁡2θ−sin⁡6θ)\begin{align*} \cos^3\theta\sin^3\theta =&\, \frac{1}{32}(3\sin2\theta-\sin6\theta) \end{align*}
(4)

(d) Use algebraic integration to find the exact value of

∫0π/8cos⁡3θsin⁡3θ dθ\begin{align*} \int_{0}^{\pi/8}\cos^3\theta\sin^3\theta\,\mathrm{d}\theta \end{align*}
(4)
题目中文翻译

(a) 证明

(z+1z)3(z−1z)3= z6−1z6−k(z2−1z2)\begin{align*} \Big(z+\frac1z\Big)^3 \Big(z-\frac1z\Big)^3 =&\, z^6-\frac{1}{z^6} -k\Big(z^2-\frac{1}{z^2}\Big) \end{align*}

其中 kk 为待求常数。

(3)

已知 z=cos⁡θ+isin⁡θz=\cos\theta+\mathrm{i}\sin\theta，其中 θ\theta 为实数，

(b) 证明

(i)

zn+1zn=2cos⁡nθ\begin{align*} z^n+\frac{1}{z^n}=2\cos n\theta \end{align*}

(ii)

zn−1zn=2isin⁡nθ\begin{align*} z^n-\frac{1}{z^n}=2\mathrm{i}\sin n\theta \end{align*}
(3)

(c) 由此证明

cos⁡3θsin⁡3θ= 132(3sin⁡2θ−sin⁡6θ)\begin{align*} \cos^3\theta\sin^3\theta =&\, \frac{1}{32}(3\sin2\theta-\sin6\theta) \end{align*}
(4)

(d) 用代数积分求下列积分的精确值：

∫0π/8cos⁡3θsin⁡3θ dθ\begin{align*} \int_{0}^{\pi/8}\cos^3\theta\sin^3\theta\,\mathrm{d}\theta \end{align*}
(4)

解答

(a)

解法一

思路

展开

先把两个括号相乘，再整体三次方：

(z+1z)(z−1z)=z2−1z2\begin{align*} \Big(z+\frac1z\Big)\Big(z-\frac1z\Big)=z^2-\frac1{z^2} \end{align*}

这样比直接展开两个三次方更短。

答题过程

展开 (z+1z)3(z−1z)3= [(z+1z)(z−1z)]3= (z2−1z2)3\begin{align*} \Big(z+\frac1z\Big)^3 \Big(z-\frac1z\Big)^3 =&\, \Big[ \Big(z+\frac1z\Big) \Big(z-\frac1z\Big) \Big]^3\\[3mm] =&\, \Big(z^2-\frac1{z^2}\Big)^3 \end{align*}

Now expand:

(z2−1z2)3= z6−3z2+31z2−1z6= z6−1z6−3(z2−1z2)\begin{align*} \Big(z^2-\frac1{z^2}\Big)^3 =&\, z^6 -3z^2 +3\frac1{z^2} -\frac1{z^6}\\[3mm] =&\, z^6-\frac1{z^6} -3\Big(z^2-\frac1{z^2}\Big) \end{align*}

Therefore

−k=−3⟹k=3\begin{align*} -k=-3\quad\Longrightarrow\quad k=3 \end{align*}

解法二

思路

展开

也可以分别展开两个三次方，再相乘。这个方法较长，但能训练二项式展开与合并同类项。

答题过程

展开

First,

(z+1z)3= z3+3z+3z+1z3\begin{align*} \Big(z+\frac1z\Big)^3 =&\, z^3+3z+\frac3z+\frac1{z^3} \end{align*}

and

(z−1z)3= z3−3z+3z−1z3\begin{align*} \Big(z-\frac1z\Big)^3 =&\, z^3-3z+\frac3z-\frac1{z^3} \end{align*}

Regrouping the two expansions and using the difference of two squares gives

(z+1z)3(z−1z)3= (z3+3z)2−(3z+1z3)2= (z6+6z2+9z2) −(9z2+6z2+1z6)= z6−3z2+31z2−1z6= z6−1z6−3(z2−1z2)\begin{align*} \Big(z+\frac1z\Big)^3 \Big(z-\frac1z\Big)^3 =&\, \Big(z^3+\frac3z\Big)^2 -\Big(3z+\frac1{z^3}\Big)^2\\[3mm] =&\, \Big(z^6+6z^2+\frac9{z^2}\Big) \\[2mm] &\,\hspace{2pt}-\Big(9z^2+\frac6{z^2}+\frac1{z^6}\Big)\\[3mm] =&\, z^6-3z^2+3\frac1{z^2}-\frac1{z^6}\\[3mm] =&\, z^6-\frac1{z^6} -3\Big(z^2-\frac1{z^2}\Big) \end{align*}

Hence

−k=−3⟹k=3\begin{align*} -k=-3\quad\Longrightarrow\quad k=3 \end{align*}

(b)

解法一

思路

展开

由 de Moivre 定理，zn=cos⁡nθ+isin⁡nθz^n=\cos n\theta+\mathrm{i}\sin n\theta。而 1zn=z−n\dfrac1{z^n}=z^{-n}，对应角度是 −nθ-n\theta。

答题过程

展开

Since

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

by de Moivre’s theorem,

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

Also,

1zn= z−n= cos⁡(−nθ)+isin⁡(−nθ)= cos⁡nθ−isin⁡nθ\begin{align*} \frac1{z^n} =&\, z^{-n}\\[3mm] =&\, \cos(-n\theta)+\mathrm{i}\sin(-n\theta)\\[3mm] =&\, \cos n\theta-\mathrm{i}\sin n\theta \end{align*}

Therefore

zn+1zn= cos⁡nθ+isin⁡nθ +cos⁡nθ−isin⁡nθ= 2cos⁡nθzn−1zn= cos⁡nθ+isin⁡nθ −(cos⁡nθ−isin⁡nθ)= 2isin⁡nθ\begin{align*} z^n+\frac1{z^n} =&\, \cos n\theta+\mathrm{i}\sin n\theta\\[2mm] &\,\hspace{2pt}+\cos n\theta-\mathrm{i}\sin n\theta\\[3mm] =&\, 2\cos n\theta\\[3mm] z^n-\frac1{z^n} =&\, \cos n\theta+\mathrm{i}\sin n\theta\\[2mm] &\,\hspace{2pt}-\big(\cos n\theta-\mathrm{i}\sin n\theta\big)\\[3mm] =&\, 2\mathrm{i}\sin n\theta \end{align*}

(c)

解法一

思路

展开

把 (a) 左边用 (b) 的 n=1n=1 结果表示成 cos⁡3θsin⁡3θ\cos^3\theta\sin^3\theta，右边用 n=6n=6 和 n=2n=2 结果表示成 sin⁡6θ\sin6\theta 与 sin⁡2θ\sin2\theta。

答题过程

展开

From part (b),

z+1z=2cos⁡θ,z−1z=2isin⁡θ\begin{align*} z+\frac1z=2\cos\theta, \qquad z-\frac1z=2\mathrm{i}\sin\theta \end{align*}

So the left hand side of part (a) is

(2cos⁡θ)3(2isin⁡θ)3= 8cos⁡3θ⋅8i3sin⁡3θ= −64icos⁡3θsin⁡3θ\begin{align*} \Big(2\cos\theta\Big)^3 \Big(2\mathrm{i}\sin\theta\Big)^3 =&\, 8\cos^3\theta\cdot 8\mathrm{i}^3\sin^3\theta\\[3mm] =&\, -64\mathrm{i}\cos^3\theta\sin^3\theta \end{align*}

Also, from part (b),

z6−1z6= 2isin⁡6θ,z2−1z2= 2isin⁡2θ\begin{align*} z^6-\frac1{z^6} =&\, 2\mathrm{i}\sin6\theta, \qquad z^2-\frac1{z^2} =&\, 2\mathrm{i}\sin2\theta \end{align*}

Using k=3k=3 from part (a),

z6−1z6−3(z2−1z2)= 2isin⁡6θ−6isin⁡2θ\begin{align*} z^6-\frac1{z^6} -3\Big(z^2-\frac1{z^2}\Big) =&\, 2\mathrm{i}\sin6\theta-6\mathrm{i}\sin2\theta \end{align*}

Equating both expressions:

−64icos⁡3θsin⁡3θ= 2isin⁡6θ−6isin⁡2θ\begin{align*} -64\mathrm{i}\cos^3\theta\sin^3\theta =&\, 2\mathrm{i}\sin6\theta-6\mathrm{i}\sin2\theta \end{align*}

Divide by −64i-64\mathrm{i}:

cos⁡3θsin⁡3θ= 132(3sin⁡2θ−sin⁡6θ)\begin{align*} \cos^3\theta\sin^3\theta =&\, \frac{1}{32}(3\sin2\theta-\sin6\theta) \end{align*}

(d)

解法一

思路

展开

直接用 (c) 的结果把被积函数换成正弦的线性组合，再逐项积分。上下限是 00 和 π8\dfrac{\pi}{8}，所以会用到 cos⁡π4\cos\dfrac{\pi}{4} 和 cos⁡3π4\cos\dfrac{3\pi}{4}。

答题过程

展开

Using part (c),

∫0π/8cos⁡3θsin⁡3θ dθ= 132∫0π/8(3sin⁡2θ−sin⁡6θ) dθ\begin{align*} \int_{0}^{\pi/8}\cos^3\theta\sin^3\theta\,\mathrm{d}\theta =&\, \frac1{32} \int_{0}^{\pi/8} (3\sin2\theta-\sin6\theta)\,\mathrm{d}\theta \end{align*}

Integrate:

 132[−32cos⁡2θ+16cos⁡6θ]0π/8= 132[(−32cos⁡π4+16cos⁡3π4)−(−32+16)]= 132[(−324−212)+43]= 132(43−526)\begin{align*} &\, \frac1{32} \Big[ -\frac32\cos2\theta+\frac16\cos6\theta \Big]_{0}^{\pi/8}\\[3mm] =&\, \frac1{32} \Big[ \Big( -\frac32\cos\frac{\pi}{4} +\frac16\cos\frac{3\pi}{4} \Big) - \Big( -\frac32+\frac16 \Big) \Big]\\[3mm] =&\, \frac1{32} \Big[ \Big( -\frac{3\sqrt2}{4} -\frac{\sqrt2}{12} \Big) +\frac43 \Big]\\[3mm] =&\, \frac1{32} \Big( \frac43-\frac{5\sqrt2}{6} \Big) \end{align*}

Therefore

∫0π/8cos⁡3θsin⁡3θ dθ= 124−52192\begin{align*} \int_{0}^{\pi/8}\cos^3\theta\sin^3\theta\,\mathrm{d}\theta =&\, \frac{1}{24}-\frac{5\sqrt2}{192} \end{align*}