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IAL 2026 Jan A Q7

A Level / Edexcel / FP2

IAL 2026 Jan A Paper · Question 7

题目

Problem

(a) Show that

(z+1z)3(z−1z)3=z6−1z6+k(z2−1z2)\begin{align*} \left(z+\frac1z\right)^3 \left(z-\frac1z\right)^3 = z^6-\frac{1}{z^6} +k\left(z^2-\frac{1}{z^2}\right) \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)

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*} \left(z+\frac1z\right)\left(z-\frac1z\right)=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*} \left(z+\frac1z\right)^3 \left(z-\frac1z\right)^3 =&\, \left[ \left(z+\frac1z\right) \left(z-\frac1z\right) \right]^3\\[3mm] =&\, \left(z^2-\frac1{z^2}\right)^3 \end{align*}

Now expand:

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

Therefore

k=−3\begin{align*} 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*} \left(z+\frac1z\right)^3 = z^3+3z+\frac3z+\frac1{z^3} \end{align*}

and

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

Multiplying and collecting terms gives

(z+1z)3(z−1z)3=z6−3z2+31z2−1z6=z6−1z6−3(z2−1z2)\begin{align*} \left(z+\frac1z\right)^3 \left(z-\frac1z\right)^3 ={}& z^6-3z^2+3\frac1{z^2}-\frac1{z^6}\\[3mm] ={}& z^6-\frac1{z^6} -3\left(z^2-\frac1{z^2}\right) \end{align*}

Hence

k=−3\begin{align*} 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} = \cos(-n\theta)+\mathrm{i}\sin(-n\theta) = \cos n\theta-\mathrm{i}\sin n\theta \end{align*}

Therefore

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

(c)

解法一

思路

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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*} \left(2\cos\theta\right)^3 \left(2\mathrm{i}\sin\theta\right)^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\left(z^2-\frac1{z^2}\right) = 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} \left[ -\frac32\cos2\theta+\frac16\cos6\theta \right]_{0}^{\pi/8}\\[3mm] ={}& \frac1{32} \left[ \left( -\frac32\cos\frac{\pi}{4} +\frac16\cos\frac{3\pi}{4} \right) - \left( -\frac32+\frac16 \right) \right]\\[3mm] ={}& \frac1{32} \left[ \left( -\frac{3\sqrt2}{4} -\frac{\sqrt2}{12} \right) +\frac43 \right]\\[3mm] ={}& \frac1{32} \left( \frac43-\frac{5\sqrt2}{6} \right) \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*}