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IAL 2025 June A Q8

A Level / Edexcel / FP2

IAL 2025 June A 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=cos⁡θ+isin⁡θz=\cos\theta+\mathrm{i}\sin\theta

(a) show that, for n∈Zn\in\mathbb{Z}

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

(b) Hence show that

cos⁡4θ=18(cos⁡4θ+acos⁡2θ+b)\begin{align*} \cos^4\theta=\frac18(\cos4\theta+a\cos2\theta+b) \end{align*}

where aa and bb are integers to be determined.

(4)

Figure 4

Figure 4 shows a sketch of the curve with equation

y=cos⁡2x1+sin⁡x−π2⩽x⩽π2\begin{align*} y=\cos^2x\sqrt{1+\sin x} \qquad -\frac{\pi}{2}\leqslant x\leqslant \frac{\pi}{2} \end{align*}

The region RR, shown shaded in Figure 4, is bounded by the curve, the xx-axis and the line with equation x=π4x=\dfrac{\pi}{4}

The region RR is rotated through 2π2\pi radians about the xx-axis to form a solid of revolution.

(c) Use the answer to part (b) and algebraic integration to determine the exact volume of this solid.

Give the answer in the form π160(p+qπ+r2)\dfrac{\pi}{160}(p+q\pi+r\sqrt2) where pp, qq and rr are integers.

(6)

解答

(a)

解法一

思路

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用 de Moivre 定理分别写出 znz^n 和 z−nz^{-n}。相加时虚部会抵消，留下 2cos⁡nθ2\cos n\theta。

#### 答题过程
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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}\\[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\\[2mm] &\,\hspace{2pt}+\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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由 (a)，z+1z=2cos⁡θz+\dfrac1z=2\cos\theta。把两边四次方展开，再把 z4+1z4z^4+\dfrac1{z^4} 和 z2+1z2z^2+\dfrac1{z^2} 用 (a) 换成余弦。

答题过程

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From part (a),

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

Raise both sides to the fourth power:

(z+1z)4= z4+4z3(1z)+6z2(1z)2 +4z(1z)3+(1z)4= z4+4z2+6+4z2+1z4= (z4+1z4)+4(z2+1z2)+6\begin{align*} \left(z+\frac1z\right)^4 =&\,z^4+4z^3\left(\frac1z\right) +6z^2\left(\frac1z\right)^2\\[2mm] &\,\hspace{2pt}+4z\left(\frac1z\right)^3 +\left(\frac1z\right)^4\\[4mm] =&\,z^4+4z^2+6+\frac4{z^2}+\frac1{z^4}\\[4mm] =&\,\left(z^4+\frac1{z^4}\right) +4\left(z^2+\frac1{z^2}\right)+6 \end{align*}

Using part (a) again,

(2cos⁡θ)4= 2cos⁡4θ+4(2cos⁡2θ)+616cos⁡4θ= 2cos⁡4θ+8cos⁡2θ+6cos⁡4θ= 18(cos⁡4θ+4cos⁡2θ+3)\begin{align*} (2\cos\theta)^4 =&\,2\cos4\theta+4(2\cos2\theta)+6\\[4mm] 16\cos^4\theta =&\,2\cos4\theta+8\cos2\theta+6\\[4mm] \cos^4\theta =&\,\frac18(\cos4\theta+4\cos2\theta+3) \end{align*}

Therefore

a=4,b=3\begin{align*} a=4,\qquad b=3 \end{align*}

解法二

思路

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也可以直接把 z4+z−4=2cos⁡4θz^4+z^{-4}=2\cos4\theta 展开。虚部相消后会得到含有 cos⁡4θ,cos⁡2θsin⁡2θ,sin⁡4θ\cos^4\theta,\cos^2\theta\sin^2\theta,\sin^4\theta 的式子，再用 sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta 化简。

答题过程

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Let c=cos⁡θc=\cos\theta and s=sin⁡θs=\sin\theta. From part (a) with n=4n=4,

2cos⁡4θ= z4+z−4= (c+is)4+(c−is)4\begin{align*} 2\cos4\theta =&\,z^4+z^{-4}\\[4mm] =&\,(c+\mathrm{i}s)^4+(c-\mathrm{i}s)^4 \end{align*}

Expanding,

(c+is)4+(c−is)4= 2c4−12c2s2+2s4\begin{align*} (c+\mathrm{i}s)^4+(c-\mathrm{i}s)^4 =&\,2c^4-12c^2s^2+2s^4 \end{align*}

So

cos⁡4θ= c4−6c2s2+s4= c4−6c2(1−c2)+(1−c2)2= 8c4−8c2+1\begin{align*} \cos4\theta =&\,c^4-6c^2s^2+s^4\\[4mm] =&\,c^4-6c^2(1-c^2)+(1-c^2)^2\\[4mm] =&\,8c^4-8c^2+1 \end{align*}

Hence

8cos⁡4θ= cos⁡4θ+8cos⁡2θ−1\begin{align*} 8\cos^4\theta =&\,\cos4\theta+8\cos^2\theta-1 \end{align*}

Using

cos⁡2θ=2cos⁡2θ−1⇒8cos⁡2θ=4cos⁡2θ+4\begin{align*} \cos2\theta=2\cos^2\theta-1 \quad \Rightarrow \quad 8\cos^2\theta=4\cos2\theta+4 \end{align*}

we get

8cos⁡4θ= cos⁡4θ+4cos⁡2θ+3cos⁡4θ= 18(cos⁡4θ+4cos⁡2θ+3)\begin{align*} 8\cos^4\theta =&\,\cos4\theta+4\cos2\theta+3\\[4mm] \cos^4\theta =&\,\frac18(\cos4\theta+4\cos2\theta+3) \end{align*}

Therefore

a=4,b=3\begin{align*} a=4,\qquad b=3 \end{align*}

(c)

解法一

思路

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绕 xx 轴旋转用体积公式 V=π∫y2 dxV=\pi\int y^2\,\mathrm{d}x。这里 y2=cos⁡4x(1+sin⁡x)y^2=\cos^4x(1+\sin x)，其中 cos⁡4x\cos^4x 用 (b) 的结果积分，sin⁡xcos⁡4x\sin x\cos^4x 用代换 u=cos⁡xu=\cos x 积分。

答题过程

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The volume is

V= π∫−π2π4y2 dx= π∫−π2π4cos⁡4x(1+sin⁡x) dx= π∫−π2π4(cos⁡4x+sin⁡xcos⁡4x) dx\begin{align*} V =&\,\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{4}}y^2\,\mathrm{d}x\\[4mm] =&\,\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{4}} \cos^4x(1+\sin x)\,\mathrm{d}x\\[4mm] =&\,\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{4}} (\cos^4x+\sin x\cos^4x)\,\mathrm{d}x \end{align*}

Using part (b),

∫cos⁡4x dx= ∫18(cos⁡4x+4cos⁡2x+3) dx= 18(14sin⁡4x+2sin⁡2x+3x)\begin{align*} \int \cos^4x\,\mathrm{d}x =&\,\int \frac18(\cos4x+4\cos2x+3)\,\mathrm{d}x\\[4mm] =&\,\frac18\left(\frac14\sin4x+2\sin2x+3x\right) \end{align*}

Also,

∫sin⁡xcos⁡4x dx= −15cos⁡5x\begin{align*} \int \sin x\cos^4x\,\mathrm{d}x =&\,-\frac15\cos^5x \end{align*}

Therefore

V= π[18(14sin⁡4x+2sin⁡2x+3x)−15cos⁡5x]−π2π4\begin{align*} V =&\,\pi\left[ \frac18\left(\frac14\sin4x+2\sin2x+3x\right) -\frac15\cos^5x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{4}} \end{align*}

At x=π4x=\dfrac{\pi}{4},

18(14sin⁡π+2sin⁡π2+3π4)−15cos⁡5π4= 18(2+3π4)−15(22)5= 14+3π32−240\begin{align*} \frac18\left(\frac14\sin\pi+2\sin\frac{\pi}{2}+\frac{3\pi}{4}\right) -\frac15\cos^5\frac{\pi}{4} =&\,\frac18\left(2+\frac{3\pi}{4}\right) -\frac15\left(\frac{\sqrt2}{2}\right)^5\\[4mm] =&\,\frac14+\frac{3\pi}{32}-\frac{\sqrt2}{40} \end{align*}

At x=−π2x=-\dfrac{\pi}{2},

18(14sin⁡(−2π)+2sin⁡(−π)−3π2)−15cos⁡5(−π2)= −3π16\begin{align*} \frac18\left(\frac14\sin(-2\pi)+2\sin(-\pi)-\frac{3\pi}{2}\right) -\frac15\cos^5\left(-\frac{\pi}{2}\right) =&\,-\frac{3\pi}{16} \end{align*}

Hence

V= π(14+3π32−240+3π16)= π(14+9π32−240)= π160(40+45π−42)\begin{align*} V =&\,\pi\left( \frac14+\frac{3\pi}{32}-\frac{\sqrt2}{40} +\frac{3\pi}{16} \right)\\[4mm] =&\,\pi\left( \frac14+\frac{9\pi}{32}-\frac{\sqrt2}{40} \right)\\[4mm] =&\,\frac{\pi}{160}(40+45\pi-4\sqrt2) \end{align*}