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FP2 2022 June Q8

A Level / Edexcel / FP2

IAL 2022 June Paper · Question 8

题目

Problem

8. (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 \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 \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}} \big(4 \sin^5 \theta - 5 \sin^3 \theta - 6 \sin \theta\big)\,\mathrm{d}\theta\\[2mm] =&\, a\sqrt{2} + b \end{align*}

where aa and bb are rational numbers to be determined.

(4)
题目中文翻译
  1. (a) 用 de Moivre 定理证明
sin⁡5θ≡ 16sin⁡5θ−20sin⁡3θ+5sin⁡θ.\begin{align*} \sin5\theta\equiv&\,16\sin^5\theta-20\sin^3\theta+5\sin\theta. \end{align*}
(5)

(b) 由此求方程

16x5−20x3+5x+15= 0\begin{align*} 16x^5-20x^3+5x+\frac15=&\,0 \end{align*}

的五个不同的解，答案保留三位小数。

(5)

(c) 用 (a) 中的恒等式证明

 ∫0π4(4sin⁡5θ−5sin⁡3θ−6sin⁡θ) dθ= a2+b,\begin{align*} &\,\int_0^{\frac\pi4} \big(4\sin^5\theta-5\sin^3\theta-6\sin\theta\big)\,\mathrm{d}\theta\\[2mm] =&\,a\sqrt2+b, \end{align*}

其中 aa 和 bb 为待求的有理数，并求出它们的值。

(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*}

解法二

思路

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令 z=cos⁡θ+isin⁡θz=\cos\theta+\mathrm{i}\sin\theta，则 z−1=cos⁡θ−isin⁡θz^{-1}=\cos\theta-\mathrm{i}\sin\theta。由 de Moivre 定理，zm−z−m=2isin⁡mθz^m-z^{-m}=2\mathrm{i}\sin m\theta。

展开 (z−z−1)5(z-z^{-1})^5 后，把正负次幂配成三组，就会得到 sin⁡5θ\sin5\theta、sin⁡3θ\sin3\theta 和 sin⁡θ\sin\theta。最后用三倍角公式消去 sin⁡3θ\sin3\theta。

答题过程

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Let z=cos⁡θ+isin⁡θz=\cos\theta+\mathrm{i}\sin\theta. Since ∣z∣=1|z|=1, de Moivre’s theorem gives

zm= cos⁡mθ+isin⁡mθ,z−m= cos⁡mθ−isin⁡mθ.\begin{align*} z^m=&\,\cos m\theta+\mathrm{i}\sin m\theta,\\[2mm] z^{-m}=&\,\cos m\theta-\mathrm{i}\sin m\theta. \end{align*}

Hence zm−z−m=2isin⁡mθz^m-z^{-m}=2\mathrm{i}\sin m\theta, and in particular z−z−1=2isin⁡θz-z^{-1}=2\mathrm{i}\sin\theta.

Using the binomial theorem,

(z−z−1)5= z5−5z3+10z −10z−1+5z−3−z−5= (z5−z−5)−5(z3−z−3) +10(z−z−1).\begin{align*} (z-z^{-1})^5 =&\,z^5-5z^3+10z\\[2mm] &\,\hspace{2pt}-10z^{-1}+5z^{-3}-z^{-5}\\[2mm] =&\,(z^5-z^{-5})-5(z^3-z^{-3})\\[2mm] &\,\hspace{2pt}+10(z-z^{-1}). \end{align*}

Substituting the sine expressions and using (2i)5=32i(2\mathrm{i})^5=32\mathrm{i},

32isin⁡5θ= 2isin⁡5θ−10isin⁡3θ +20isin⁡θ.\begin{align*} 32\mathrm{i}\sin^5\theta =&\,2\mathrm{i}\sin5\theta-10\mathrm{i}\sin3\theta\\[2mm] &\,\hspace{2pt}+20\mathrm{i}\sin\theta. \end{align*}

Dividing by 2i2\mathrm{i} and rearranging,

sin⁡5θ= 16sin⁡5θ+5sin⁡3θ−10sin⁡θ.\begin{align*} \sin5\theta=&\,16\sin^5\theta+5\sin3\theta-10\sin\theta. \end{align*}

The angle addition formula gives

sin⁡3θ= sin⁡2θcos⁡θ+cos⁡2θsin⁡θ= 2sin⁡θcos⁡2θ+(1−2sin⁡2θ)sin⁡θ= 2sin⁡θ(1−sin⁡2θ)+sin⁡θ−2sin⁡3θ= 3sin⁡θ−4sin⁡3θ.\begin{align*} \sin3\theta =&\,\sin2\theta\cos\theta+\cos2\theta\sin\theta\\[2mm] =&\,2\sin\theta\cos^2\theta +(1-2\sin^2\theta)\sin\theta\\[2mm] =&\,2\sin\theta(1-\sin^2\theta) +\sin\theta-2\sin^3\theta\\[2mm] =&\,3\sin\theta-4\sin^3\theta. \end{align*}

Therefore

sin⁡5θ= 16sin⁡5θ+5(3sin⁡θ−4sin⁡3θ) −10sin⁡θ= 16sin⁡5θ−20sin⁡3θ+5sin⁡θ.\begin{align*} \sin5\theta =&\,16\sin^5\theta+5(3\sin\theta-4\sin^3\theta)\\[2mm] &\,\hspace{2pt}-10\sin\theta\\[2mm] =&\,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}。选取 −π2⩽θ⩽π2-\frac{\pi}{2}\leqslant\theta\leqslant\frac{\pi}{2}，在这个区间内 sin⁡θ\sin\theta 单调递增，不会重复取到相同的 xx。先列出两组角，再按照 5θ5\theta 的范围筛选，除以 55 后取正弦。因为原方程是五次方程，找到五个不同的实根后就完整了。

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*}

Choose −π2⩽θ⩽π2-\dfrac{\pi}{2}\leqslant\theta\leqslant\dfrac{\pi}{2}, so that

−5π2⩽5θ⩽5π2.\begin{align*} -\frac{5\pi}{2}\leqslant5\theta\leqslant\frac{5\pi}{2}. \end{align*}

Let α=arcsin⁡15≈0.201357921\alpha=\arcsin\dfrac15\approx0.201357921 radians. The two families of solutions are

5θ= −α+2kπ,or5θ= π+α+2kπ,k∈Z.\begin{align*} 5\theta=&\,-\alpha+2k\pi,\\[2mm] \text{or}\quad5\theta=&\,\pi+\alpha+2k\pi, \qquad k\in\mathbb Z. \end{align*}

Within the required interval, the first family has k=−1,0,1k=-1,0,1, and the second has k=−1,0k=-1,0. Thus

5θ= −2π−α, −π+α, −α, π+α, 2π−α.\begin{align*} 5\theta=&\,-2\pi-\alpha,\ -\pi+\alpha,\ -\alpha,\\[2mm] &\,\hspace{2pt}\pi+\alpha,\ 2\pi-\alpha. \end{align*}

Dividing each angle by 55 gives

θ≈ −1.296908646, −0.588046947, −0.040271584, 0.668590115, 1.216365477.\begin{align*} \theta\approx&\,-1.296908646,\ -0.588046947,\\[2mm] &\,\hspace{2pt}-0.040271584,\ 0.668590115,\ 1.216365477. \end{align*}

Taking x=sin⁡θx=\sin\theta for these five angles,

x≈ −0.962726649, −0.554737091, −0.040260700, 0.619880272, 0.937844168.\begin{align*} x\approx&\,-0.962726649,\ -0.554737091,\\[2mm] &\,\hspace{2pt}-0.040260700,\ 0.619880272,\ 0.937844168. \end{align*}

These are five distinct roots. A polynomial of degree 55 has at most five distinct roots, so this list is complete.

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}} \Big(\frac{1}{4}\sin5\theta-\frac{29}{4}\sin\theta\Big) \,\mathrm{d}\theta\\[2mm] =&\,\Big[ -\frac{1}{20}\cos5\theta +\frac{29}{4}\cos\theta \Big]_{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] =&\,\Big(\frac{\sqrt{2}}{40} +\frac{29\sqrt{2}}{8}\Big)\\[2mm] &\,\hspace{2pt}-\Big(-\frac{1}{20}+\frac{29}{4}\Big)\\[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*}