CalcGospel 國際數學圖譜
返回

FP2 2023 Jan Q7

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 7

题目

Problem

7.

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)
题目中文翻译

本题必须展示所有运算步骤。

完全依赖计算器技术的解答不予接受。

(a) 利用 de Moivre 定理证明

cos⁡5x≡ cos⁡x(asin⁡4x+bsin⁡2x+c), \begin{align*} \cos5x\equiv&\,\cos x(a\sin^4x+b\sin^2x+c), \end{align*}

其中 aa、bb、cc 为待求的整数。

(4)

(b) 由此解方程

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

其中 0<θ<π20<\theta<\frac{\pi}{2}。答案保留三位小数。

(4)

解答

(a)

解法一

思路

展开

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

展开

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

展开

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

展开

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

θ1= sin⁡−1(7+1716)=0.986…,θ2= sin⁡−1(7−1716)=0.438….\begin{align*} \theta_1 =&\,\sin^{-1}\bigg(\sqrt{\frac{7+\sqrt{17}}{16}}\bigg) =0.986\ldots,\\[2mm] \theta_2 =&\,\sin^{-1}\bigg(\sqrt{\frac{7-\sqrt{17}}{16}}\bigg) =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*}

解法二

思路

展开

仍从 (a) 的结果出发，但把 sin⁡2θ\sin^2\theta 换成 1−cos⁡2θ1-\cos^2\theta，得到关于 cos⁡2θ\cos^2\theta 的二次方程。第一象限内余弦为正，取正平方根后求角度。

答题过程

展开

Using part (a) and sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta,

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

Since 0<θ<π/20<\theta<\pi/2, we have cos⁡θ>0\cos\theta>0. Dividing by cos⁡θ\cos\theta and collecting terms gives

16sin⁡4θ−14sin⁡2θ+2= 0.\begin{align*} 16\sin^4\theta-14\sin^2\theta+2=&\,0. \end{align*}

Put v=cos⁡2θv=\cos^2\theta, so sin⁡2θ=1−v\sin^2\theta=1-v. Then

16(1−v)2−14(1−v)+2= 016−32v+16v2−14+14v+2= 016v2−18v+4= 0v= 18±182−4(16)(4)32= 9±1716.\begin{align*} 16(1-v)^2-14(1-v)+2=&\,0\\[2mm] 16-32v+16v^2-14+14v+2=&\,0\\[2mm] 16v^2-18v+4=&\,0\\[2mm] v=&\,\frac{18\pm\sqrt{18^2-4(16)(4)}}{32}\\[2mm] =&\,\frac{9\pm\sqrt{17}}{16}. \end{align*}

Taking the positive square root,

cos⁡θ= 9±1716.\begin{align*} \cos\theta=&\,\sqrt{\frac{9\pm\sqrt{17}}{16}}. \end{align*}

Therefore

θ1= cos⁡−1(9+1716)=0.438…,θ2= cos⁡−1(9−1716)=0.986….\begin{align*} \theta_1=&\,\cos^{-1}\bigg(\sqrt{\frac{9+\sqrt{17}}{16}}\bigg) =0.438\ldots,\\[2mm] \theta_2=&\,\cos^{-1}\bigg(\sqrt{\frac{9-\sqrt{17}}{16}}\bigg) =0.986\ldots. \end{align*}

Both angles lie in the required interval. To 33 decimal places,

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