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IAL 2023 Jan Q7

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 7

题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

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

cos⁡5x≡ cos⁡x(asin⁡4x+bsin⁡2x+c)\begin{align*} \cos 5x \equiv\,& \cos x(a\sin^4 x + b\sin^2 x + c) \end{align*}

where aa, bb and cc are integers to be determined.

(4)

(b) Hence solve, for 0<θ<π20 < \theta < \frac{\pi}{2}

cos⁡5θ= sin⁡2θsin⁡θ−cos⁡θ\begin{align*} \cos 5\theta =\,& \sin 2\theta \sin \theta - \cos \theta \end{align*}

giving your answers to 33 decimal places.

(4)

解答

(a)

解法一

思路

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用 de Moivre’s theorem 展开

(cos⁡x+isin⁡x)5.\begin{align*} (\cos x+\mathrm{i}\sin x)^5. \end{align*}

取实部得到 cos⁡5x\cos5x，再把 cos⁡2x\cos^2x 换成 1−sin⁡2x1-\sin^2x，使括号里只含 sin⁡x\sin x。

答题过程

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By de Moivre’s theorem,

(cos⁡x+isin⁡x)5=cos⁡5x+isin⁡5x.\begin{align*} (\cos x+\mathrm{i}\sin x)^5 =\cos5x+\mathrm{i}\sin5x. \end{align*}

Taking the real terms in the binomial expansion,

cos⁡5x= cos⁡5x+10cos⁡3x(isin⁡x)2+5cos⁡x(isin⁡x)4= cos⁡5x−10cos⁡3xsin⁡2x+5cos⁡xsin⁡4x.\begin{align*} \cos5x =&\,\cos^5x +10\cos^3x(\mathrm{i}\sin x)^2 +5\cos x(\mathrm{i}\sin x)^4\\[2mm] =&\,\cos^5x-10\cos^3x\sin^2x +5\cos x\sin^4x. \end{align*}

Factor out cos⁡x\cos x:

cos⁡5x= cos⁡x(cos⁡4x−10cos⁡2xsin⁡2x+5sin⁡4x).\begin{align*} \cos5x =&\,\cos x(\cos^4x-10\cos^2x\sin^2x+5\sin^4x). \end{align*}

Using cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x,

cos⁡5x= cos⁡x{(1−sin⁡2x)2 −10(1−sin⁡2x)sin⁡2x+5sin⁡4x}= cos⁡x(1−2sin⁡2x+sin⁡4x −10sin⁡2x+10sin⁡4x+5sin⁡4x)= cos⁡x(16sin⁡4x−12sin⁡2x+1).\begin{align*} \cos5x =&\,\cos x\{(1-\sin^2x)^2\\[2mm] &\,\hspace{2pt}-10(1-\sin^2x)\sin^2x+5\sin^4x\}\\[2mm] =&\,\cos x(1-2\sin^2x+\sin^4x\\[2mm] &\,\hspace{2pt}-10\sin^2x+10\sin^4x+5\sin^4x)\\[2mm] =&\,\cos x(16\sin^4x-12\sin^2x+1). \end{align*}

Therefore

a=16,b=−12,c=1.\begin{align*} \boxed{a=16,\qquad b=-12,\qquad c=1}. \end{align*}

(b)

解法一

思路

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把 (a) 的结果代入，再用

sin⁡2θ=2sin⁡θcos⁡θ.\begin{align*} \sin2\theta=2\sin\theta\cos\theta. \end{align*}

由于 0<θ<π20<\theta<\frac{\pi}{2}，所以 cos⁡θ≠0\cos\theta\neq0，可以除以 cos⁡θ\cos\theta。

答题过程

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Using part (a),

cos⁡5θ=cos⁡θ(16sin⁡4θ−12sin⁡2θ+1).\begin{align*} \cos5\theta =\cos\theta(16\sin^4\theta-12\sin^2\theta+1). \end{align*}

The equation is

cos⁡5θ=sin⁡2θsin⁡θ−cos⁡θ.\begin{align*} \cos5\theta=\sin2\theta\sin\theta-\cos\theta. \end{align*}

So

cos⁡θ(16sin⁡4θ−12sin⁡2θ+1)= 2sin⁡2θcos⁡θ−cos⁡θ.\begin{align*} \cos\theta(16\sin^4\theta-12\sin^2\theta+1) =&\,2\sin^2\theta\cos\theta-\cos\theta. \end{align*}

Since 0<θ<π20<\theta<\dfrac{\pi}{2}, cos⁡θ≠0\cos\theta\neq0. Divide by cos⁡θ\cos\theta:

16sin⁡4θ−12sin⁡2θ+1= 2sin⁡2θ−116sin⁡4θ−14sin⁡2θ+2= 08sin⁡4θ−7sin⁡2θ+1= 0.\begin{align*} 16\sin^4\theta-12\sin^2\theta+1 =&\,2\sin^2\theta-1\\[2mm] 16\sin^4\theta-14\sin^2\theta+2=&\,0\\[2mm] 8\sin^4\theta-7\sin^2\theta+1=&\,0. \end{align*}

Let

u=sin⁡2θ.\begin{align*} u=\sin^2\theta. \end{align*}

Then

8u2−7u+1= 0u= 7±49−3216= 7±1716.\begin{align*} 8u^2-7u+1=&\,0\\[2mm] u=&\,\frac{7\pm\sqrt{49-32}}{16}\\[2mm] =&\,\frac{7\pm\sqrt{17}}{16}. \end{align*}

Since 0<θ<π20<\theta<\dfrac{\pi}{2},

sin⁡θ=7±1716.\begin{align*} \sin\theta=\sqrt{\frac{7\pm\sqrt{17}}{16}}. \end{align*}

Therefore

θ= sin⁡−1(7+1716)orsin⁡−1(7−1716)= 0.986…or0.438….\begin{align*} \theta =&\,\sin^{-1}\left(\sqrt{\frac{7+\sqrt{17}}{16}}\right) \quad\text{or}\quad \sin^{-1}\left(\sqrt{\frac{7-\sqrt{17}}{16}}\right)\\[2mm] =&\,0.986\ldots \quad\text{or}\quad 0.438\ldots. \end{align*}

Thus, to 33 decimal places,

θ=0.438, 0.986.\begin{align*} \boxed{\theta=0.438,\ 0.986}. \end{align*}