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IAL 2022 June Q8

A Level / Edexcel / FP2

IAL 2022 June Paper · Question 8

题目

Problem

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

sin⁡5θ≡ 16sin⁡5θ−20sin⁡3θ+5sin⁡θ\begin{align*} \sin 5\theta \equiv\,& 16 \sin^5 \theta - 20 \sin^3 \theta + 5 \sin \theta\\[2mm] \end{align*}
(5)

(b) Hence determine the five distinct solutions of the equation

16x5−20x3+5x+15= 0\begin{align*} 16x^5 - 20x^3 + 5x + \frac{1}{5} =\,& 0\\[2mm] \end{align*}

giving your answers to 3 decimal places.

(5)

(c) Use the identity given in part (a) to show that

∫0π4 (4sin⁡5θ−5sin⁡3θ −6sin⁡θ) dθ= a2+b\begin{align*} \int_{0}^{\frac{\pi}{4}} &\,(4 \sin^5 \theta - 5 \sin^3 \theta\\[2mm] &\,\hspace{2pt}- 6 \sin \theta)\,\mathrm{d}\theta\\[2mm] =&\, a\sqrt{2} + b\\[2mm] \end{align*}

where aa and bb are rational numbers to be determined.

(4)

解答

(a)

解法一

思路

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用 de Moivre’s theorem：

(cos⁡θ+isin⁡θ)5=cos⁡5θ+isin⁡5θ.\begin{align*} (\cos\theta+\mathrm{i}\sin\theta)^5 =\cos5\theta+\mathrm{i}\sin5\theta. \end{align*}

把左边二项展开后，取虚部。最后把所有 cos⁡2θ\cos^2\theta 换成 1−sin⁡2θ1-\sin^2\theta，就能得到只含 sin⁡θ\sin\theta 的恒等式。

答题过程

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By de Moivre’s theorem,

(cos⁡θ+isin⁡θ)5=cos⁡5θ+isin⁡5θ.\begin{align*} (\cos\theta+\mathrm{i}\sin\theta)^5 =\cos5\theta+\mathrm{i}\sin5\theta. \end{align*}

Expanding the left hand side,

(cos⁡θ+isin⁡θ)5= cos⁡5θ+5icos⁡4θsin⁡θ −10cos⁡3θsin⁡2θ−10icos⁡2θsin⁡3θ +5cos⁡θsin⁡4θ+isin⁡5θ.\begin{align*} (\cos\theta+\mathrm{i}\sin\theta)^5 =&\,\cos^5\theta +5\mathrm{i}\cos^4\theta\sin\theta\\[2mm] &\,\hspace{2pt}-10\cos^3\theta\sin^2\theta -10\mathrm{i}\cos^2\theta\sin^3\theta\\[2mm] &\,\hspace{4pt}+5\cos\theta\sin^4\theta +\mathrm{i}\sin^5\theta. \end{align*}

Taking imaginary parts,

sin⁡5θ= 5cos⁡4θsin⁡θ−10cos⁡2θsin⁡3θ +sin⁡5θ.\begin{align*} \sin5\theta =&\,5\cos^4\theta\sin\theta -10\cos^2\theta\sin^3\theta\\[2mm] &\,\hspace{2pt}+\sin^5\theta. \end{align*}

Now use cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta:

sin⁡5θ= 5(1−sin⁡2θ)2sin⁡θ −10(1−sin⁡2θ)sin⁡3θ+sin⁡5θ= 5(sin⁡θ−2sin⁡3θ+sin⁡5θ) −10(sin⁡3θ−sin⁡5θ)+sin⁡5θ= 5sin⁡θ−10sin⁡3θ+5sin⁡5θ −10sin⁡3θ+10sin⁡5θ+sin⁡5θ= 16sin⁡5θ−20sin⁡3θ+5sin⁡θ.\begin{align*} \sin5\theta =&\,5(1-\sin^2\theta)^2\sin\theta\\[2mm] &\,\hspace{2pt}-10(1-\sin^2\theta)\sin^3\theta +\sin^5\theta\\[2mm] =&\,5(\sin\theta-2\sin^3\theta+\sin^5\theta)\\[2mm] &\,\hspace{2pt}-10(\sin^3\theta-\sin^5\theta) +\sin^5\theta\\[2mm] =&\,5\sin\theta-10\sin^3\theta+5\sin^5\theta\\[2mm] &\,\hspace{2pt}-10\sin^3\theta+10\sin^5\theta +\sin^5\theta\\[2mm] =&\,16\sin^5\theta-20\sin^3\theta+5\sin\theta. \end{align*}

Therefore

sin⁡5θ≡16sin⁡5θ−20sin⁡3θ+5sin⁡θ.\begin{align*} \boxed{ \sin5\theta\equiv 16\sin^5\theta-20\sin^3\theta+5\sin\theta }. \end{align*}

(b)

解法一

思路

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由 (a)，如果令 x=sin⁡θx=\sin\theta，原方程会变成

sin⁡5θ+15=0.\begin{align*} \sin5\theta+\frac{1}{5}=0. \end{align*}

也就是 sin⁡5θ=−15\sin5\theta=-\frac{1}{5}。解出一整轮内的角，再取不同的 sin⁡θ\sin\theta 值即可。因为原方程是五次方程，所以最多有五个实根；找到五个不同的值后就完整了。

答题过程

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Let

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

Using the identity from part (a),

16x5−20x3+5x+15=0\begin{align*} 16x^5-20x^3+5x+\frac{1}{5}=0 \end{align*}

becomes

sin⁡5θ+15=0.\begin{align*} \sin5\theta+\frac{1}{5}=0. \end{align*}

So

sin⁡5θ=−15.\begin{align*} \sin5\theta=-\frac{1}{5}. \end{align*}

Solving this equation for one complete period of θ\theta gives the following distinct values of x=sin⁡θx=\sin\theta:

x= −0.962727…,−0.554737…,−0.040261…, 0.619880…,0.937844….\begin{align*} x=&\,-0.962727\ldots,\quad -0.554737\ldots,\quad -0.040261\ldots,\\[2mm] &\,\hspace{2pt}0.619880\ldots,\quad 0.937844\ldots. \end{align*}

Therefore the five distinct solutions, to 3 decimal places, are

x=−0.963, −0.555, −0.040, 0.620, 0.938.\begin{align*} \boxed{x=-0.963,\ -0.555,\ -0.040,\ 0.620,\ 0.938}. \end{align*}

(c)

解法一

思路

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先把 (a) 的恒等式除以 44：

4sin⁡5θ−5sin⁡3θ=14sin⁡5θ−54sin⁡θ.\begin{align*} 4\sin^5\theta-5\sin^3\theta =\frac{1}{4}\sin5\theta-\frac{5}{4}\sin\theta. \end{align*}

这样被积函数就会变成只含 sin⁡5θ\sin5\theta 和 sin⁡θ\sin\theta 的式子，积分会很直接。

答题过程

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From part (a),

sin⁡5θ=16sin⁡5θ−20sin⁡3θ+5sin⁡θ.\begin{align*} \sin5\theta =16\sin^5\theta-20\sin^3\theta+5\sin\theta. \end{align*}

Dividing by 44,

14sin⁡5θ=4sin⁡5θ−5sin⁡3θ+54sin⁡θ.\begin{align*} \frac{1}{4}\sin5\theta =4\sin^5\theta-5\sin^3\theta+\frac{5}{4}\sin\theta. \end{align*}

Therefore

4sin⁡5θ−5sin⁡3θ−6sin⁡θ= 14sin⁡5θ−54sin⁡θ−6sin⁡θ= 14sin⁡5θ−294sin⁡θ.\begin{align*} 4\sin^5\theta-5\sin^3\theta-6\sin\theta =&\,\frac{1}{4}\sin5\theta-\frac{5}{4}\sin\theta-6\sin\theta\\[2mm] =&\,\frac{1}{4}\sin5\theta-\frac{29}{4}\sin\theta. \end{align*}

Hence

∫0π4(4sin⁡5θ−5sin⁡3θ −6sin⁡θ) dθ= ∫0π4(14sin⁡5θ−294sin⁡θ) dθ= [−120cos⁡5θ+294cos⁡θ]0π4.\begin{align*} \int_{0}^{\frac{\pi}{4}} &(4\sin^5\theta-5\sin^3\theta\\[2mm] &\,\hspace{2pt}-6\sin\theta)\,\mathrm{d}\theta\\[2mm] =&\,\int_{0}^{\frac{\pi}{4}} \left(\frac{1}{4}\sin5\theta-\frac{29}{4}\sin\theta\right) \,\mathrm{d}\theta\\[2mm] =&\,\left[ -\frac{1}{20}\cos5\theta +\frac{29}{4}\cos\theta \right]_{0}^{\frac{\pi}{4}}. \end{align*}

At θ=π4\theta=\dfrac{\pi}{4},

cos⁡5θ=cos⁡5π4=−22,cos⁡θ=22.\begin{align*} \cos5\theta=\cos\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}, \qquad \cos\theta=\frac{\sqrt{2}}{2}. \end{align*}

Therefore

∫0π4(4sin⁡5θ−5sin⁡3θ −6sin⁡θ) dθ= (240+2928) −(−120+294)= 73220−365.\begin{align*} \int_{0}^{\frac{\pi}{4}} &(4\sin^5\theta-5\sin^3\theta\\[2mm] &\,\hspace{2pt}-6\sin\theta)\,\mathrm{d}\theta\\[2mm] =&\,\left(\frac{\sqrt{2}}{40} +\frac{29\sqrt{2}}{8}\right)\\[2mm] &\,\hspace{2pt}-\left(-\frac{1}{20}+\frac{29}{4}\right)\\[2mm] =&\,\frac{73\sqrt{2}}{20}-\frac{36}{5}. \end{align*}

Thus

a=7320,b=−365.\begin{align*} \boxed{a=\frac{73}{20},\qquad b=-\frac{36}{5}}. \end{align*}