CalcGospel 國際數學圖譜
返回

IAL 2025 Jan Q7

A Level / Edexcel / FP2

IAL 2025 Jan Paper · Question 7

(a) Use De Moivre’s theorem to

(i) show that

sin⁡5θ≡5cos⁡4θsin⁡θ−10cos⁡2θsin⁡3θ+sin⁡5θ\begin{align*} \sin 5\theta \equiv 5\cos^4 \theta\sin \theta - 10\cos^2 \theta\sin^3 \theta + \sin^5 \theta \end{align*}

(ii) determine an expression for cos⁡5θ\cos 5\theta in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta

(4)

(b) Hence show that, for cos⁡5θ≠0\cos 5\theta \neq 0

tan⁡5θ=5tan⁡θ−10tan⁡3θ+tan⁡5θ1−10tan⁡2θ+5tan⁡4θ\begin{align*} \tan 5\theta = \frac{5\tan \theta - 10\tan^3 \theta + \tan^5 \theta}{1 - 10\tan^2 \theta + 5\tan^4 \theta} \end{align*}
(2)

(c) Using the result of part (b) and showing all stages of your working, determine the solutions of the equation

2x5−15x4−20x3+30x2+10x−3=0\begin{align*} 2x^5 - 15x^4 - 20x^3 + 30x^2 + 10x - 3 = 0 \end{align*}

giving your answers to 33 decimal places.

(5)
解答

(a)

Using De Moivre’s Theorem and binomial expansion:

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

(i) Equating the imaginary parts gives:

sin⁡5θ= 5cos⁡4θsin⁡θ−10cos⁡2θsin⁡3θ+sin⁡5θ\begin{align*} \sin 5\theta = &\,5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta \end{align*}

(ii) Equating the real parts gives:

cos⁡5θ= cos⁡5θ−10cos⁡3θsin⁡2θ+5cos⁡θsin⁡4θ\begin{align*} \cos 5\theta = &\,\cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta \end{align*}

(b)

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

Divide the numerator and denominator by cos⁡5θ\cos^5\theta:

tan⁡5θ= 5cos⁡4θsin⁡θcos⁡5θ−10cos⁡2θsin⁡3θcos⁡5θ+sin⁡5θcos⁡5θcos⁡5θcos⁡5θ−10cos⁡3θsin⁡2θcos⁡5θ+5cos⁡θsin⁡4θcos⁡5θ= 5tan⁡θ−10tan⁡3θ+tan⁡5θ1−10tan⁡2θ+5tan⁡4θ\begin{align*} \tan 5\theta = &\,\frac{\frac{5\cos^4\theta\sin\theta}{\cos^5\theta} - \frac{10\cos^2\theta\sin^3\theta}{\cos^5\theta} + \frac{\sin^5\theta}{\cos^5\theta}}{\frac{\cos^5\theta}{\cos^5\theta} - \frac{10\cos^3\theta\sin^2\theta}{\cos^5\theta} + \frac{5\cos\theta\sin^4\theta}{\cos^5\theta}}\\[4mm] = &\,\frac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta} \end{align*}

(c)

We are given the equation:

2x5−15x4−20x3+30x2+10x−3= 0\begin{align*} 2x^5 - 15x^4 - 20x^3 + 30x^2 + 10x - 3 = &\,0 \end{align*}

Rearranging to isolate odd and even powers:

2x5−20x3+10x= 15x4−30x2+32(x5−10x3+5x)= 3(5x4−10x2+1)\begin{align*} 2x^5 - 20x^3 + 10x = &\,15x^4 - 30x^2 + 3\\[4mm] 2(x^5 - 10x^3 + 5x) = &\,3(5x^4 - 10x^2 + 1) \end{align*}

Letting x=tan⁡θx = \tan\theta, this becomes:

2(tan⁡5θ−10tan⁡3θ+5tan⁡θ)= 3(5tan⁡4θ−10tan⁡2θ+1)5tan⁡θ−10tan⁡3θ+tan⁡5θ1−10tan⁡2θ+5tan⁡4θ= 32\begin{align*} 2(\tan^5\theta - 10\tan^3\theta + 5\tan\theta) = &\,3(5\tan^4\theta - 10\tan^2\theta + 1)\\[4mm] \frac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta} = &\,\frac{3}{2} \end{align*}

Using the identity from (b):

tan⁡5θ= 325θ= arctan⁡(32)+kπ≈ 0.98279+kπ\begin{align*} \tan 5\theta = &\,\frac{3}{2}\\[4mm] 5\theta = &\,\arctan\left(\frac{3}{2}\right) + k\pi\\[4mm] \approx &\,0.98279 + k\pi \end{align*}

For principal solutions, θ=0.19656+kπ5\theta = 0.19656 + \frac{k\pi}{5} for k=0,1,2,3,4k = 0, 1, 2, 3, 4:

k=0  ⟹  x=tan⁡(0.19656)≈ 0.199k=1  ⟹  x=tan⁡(0.82488)≈ 1.082k=2  ⟹  x=tan⁡(1.4532)≈ 8.464k=3  ⟹  x=tan⁡(2.0815)≈ −1.785k=4  ⟹  x=tan⁡(2.7098)≈ −0.461\begin{align*} k=0 \implies x = \tan(0.19656) \approx &\,0.199\\[4mm] k=1 \implies x = \tan(0.82488) \approx &\,1.082\\[4mm] k=2 \implies x = \tan(1.4532) \approx &\,8.464\\[4mm] k=3 \implies x = \tan(2.0815) \approx &\,-1.785\\[4mm] k=4 \implies x = \tan(2.7098) \approx &\,-0.461 \end{align*}

Hence the solutions are 0.199,1.082,8.464,−1.785,−0.4610.199, 1.082, 8.464, -1.785, -0.461.