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IAL 2025 June Q9

A Level / Edexcel / FP2

IAL 2025 June Paper · Question 9

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

sin⁡5θ≡asin⁡5θ+bsin⁡3θ+csin⁡θ\begin{align*} \sin 5\theta \equiv a\sin^5\theta + b\sin^3\theta + c\sin\theta \end{align*}

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

(5)

(b) Hence determine the possible exact values of sin⁡2(kπ5)\sin^2\left(\dfrac{k\pi}{5}\right) where k∈Zk \in \mathbb{Z}

(4)
解答

(a)

Using de Moivre’s theorem:

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

Expanding the right-hand side using the binomial theorem, we get:

(cos⁡θ+isin⁡θ)5= cos⁡5θ+(51)cos⁡4θ(isin⁡θ)+(52)cos⁡3θ(isin⁡θ)2 +(53)cos⁡2θ(isin⁡θ)3+(54)cos⁡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 \theta + \mathrm{i}\sin \theta)^5 = &\,\cos^5 \theta + \binom{5}{1}\cos^4 \theta(\mathrm{i}\sin \theta) + \binom{5}{2}\cos^3 \theta(\mathrm{i}\sin \theta)^2\\[4mm] &\,\hspace{2pt}+ \binom{5}{3}\cos^2 \theta(\mathrm{i}\sin \theta)^3 + \binom{5}{4}\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*}

Equating the imaginary parts:

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*}

Using the identity 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(1−2sin⁡2θ+sin⁡4θ)sin⁡θ−10sin⁡3θ+10sin⁡5θ+sin⁡5θ= 5sin⁡θ−10sin⁡3θ+5sin⁡5θ−10sin⁡3θ+11sin⁡5θ= 16sin⁡5θ−20sin⁡3θ+5sin⁡θ\begin{align*} \sin 5\theta = &\,5(1 - \sin^2 \theta)^2\sin \theta - 10(1 - \sin^2 \theta)\sin^3 \theta + \sin^5 \theta\\[4mm] = &\,5(1 - 2\sin^2 \theta + \sin^4 \theta)\sin \theta - 10\sin^3 \theta + 10\sin^5 \theta + \sin^5 \theta\\[4mm] = &\,5\sin \theta - 10\sin^3 \theta + 5\sin^5 \theta - 10\sin^3 \theta + 11\sin^5 \theta\\[4mm] = &\,16\sin^5 \theta - 20\sin^3 \theta + 5\sin \theta \end{align*}

Thus, a=16a = 16, b=−20b = -20, and c=5c = 5.

(b)

Let θ=kπ5\theta = \frac{k\pi}{5} where k∈Zk \in \mathbb{Z}. Then sin⁡5θ=sin⁡(kπ)=0\sin 5\theta = \sin(k\pi) = 0.

Using the identity from part (a):

16sin⁡5θ−20sin⁡3θ+5sin⁡θ= 0sin⁡θ(16sin⁡4θ−20sin⁡2θ+5)= 0\begin{align*} 16\sin^5 \theta - 20\sin^3 \theta + 5\sin \theta = &\,0\\[4mm] \sin \theta(16\sin^4 \theta - 20\sin^2 \theta + 5) = &\,0 \end{align*}

This gives sin⁡θ=0\sin \theta = 0, which means sin⁡2θ=0\sin^2 \theta = 0. Or, solving the quadratic in sin⁡2θ\sin^2 \theta:

16(sin⁡2θ)2−20(sin⁡2θ)+5= 0sin⁡2θ= −(−20)±(−20)2−4(16)(5)2(16)= 20±400−32032= 20±8032= 20±4532= 5±58\begin{align*} 16(\sin^2 \theta)^2 - 20(\sin^2 \theta) + 5 = &\,0\\[4mm] \sin^2 \theta = &\,\frac{-(-20) \pm \sqrt{(-20)^2 - 4(16)(5)}}{2(16)}\\[4mm] = &\,\frac{20 \pm \sqrt{400 - 320}}{32}\\[4mm] = &\,\frac{20 \pm \sqrt{80}}{32}\\[4mm] = &\,\frac{20 \pm 4\sqrt{5}}{32}\\[4mm] = &\,\frac{5 \pm \sqrt{5}}{8} \end{align*}

So the possible exact values of sin⁡2(kπ5)\sin^2\left(\frac{k\pi}{5}\right) are:

0,5−58,and5+58\begin{align*} 0, \quad \frac{5 - \sqrt{5}}{8}, \quad \text{and} \quad \frac{5 + \sqrt{5}}{8} \end{align*}