IAL 2023 Jan Q7 | 國際數學站

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

\begin{align*} \cos 5x \equiv\,& \cos x(a\sin^4 x + b\sin^2 x + c)\\[2mm] \end{align*}

where $a$ , $b$ and $c$ are integers to be determined.

(4)

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

\begin{align*} \cos 5\theta =\,& \sin 2\theta \sin \theta - \cos \theta\\[2mm] \end{align*}

giving your answers to $3$ decimal places.

(4)

解答

(a)

By De Moivre’s theorem, we have:

\begin{align*} \cos 5x + \mathrm{i}\sin 5x = &\,(\cos x + \mathrm{i}\sin x)^5 \end{align*}

Expand $(\cos x + \mathrm{i}\sin x)^5$ using the binomial theorem, focusing on the real parts to find $\cos 5x$ :

\begin{align*} (\cos x + \mathrm{i}\sin x)^5 = &\,\cos^5 x + \binom{5}{1}\cos^4 x (\mathrm{i}\sin x) + \binom{5}{2}\cos^3 x (\mathrm{i}\sin x)^2\\[4mm] &\,\hspace{2pt}+ \binom{5}{3}\cos^2 x (\mathrm{i}\sin x)^3 + \binom{5}{4}\cos x (\mathrm{i}\sin x)^4 + (\mathrm{i}\sin x)^5 \end{align*}

The real terms are formed when the power of $\mathrm{i}$ is even (i.e., $\mathrm{i}^0$ , $\mathrm{i}^2$ , $\mathrm{i}^4$ ):

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

Factor out $\cos x$ and use the identity $\cos^2 x = 1 - \sin^2 x$ :

\begin{align*} \cos 5x = &\,\cos x \left( \cos^4 x - 10\cos^2 x \sin^2 x + 5\sin^4 x \right)\\[4mm] = &\,\cos x \left( (1 - \sin^2 x)^2 - 10(1 - \sin^2 x)\sin^2 x + 5\sin^4 x \right)\\[4mm] = &\,\cos x \left( 1 - 2\sin^2 x + \sin^4 x - 10\sin^2 x + 10\sin^4 x + 5\sin^4 x \right)\\[4mm] = &\,\cos x \left( 16\sin^4 x - 12\sin^2 x + 1 \right) \end{align*}

This is in the required form $\cos x (a\sin^4 x + b\sin^2 x + c)$ , where $a=16$ , $b=-12$ , and $c=1$ .

(b)

Given the equation:

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

Using the identity from part (a) and the double angle identity $\sin 2\theta = 2\sin\theta\cos\theta$ :

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

Since $0 < \theta < \frac{\pi}{2}$ , we know $\cos\theta \neq 0$ . Dividing the entire equation by $\cos\theta$ :

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

Let $u = \sin^2\theta$ , so $8u^2 - 7u + 1 = 0$ . Use the quadratic formula to solve for $u$ :

\begin{align*} u = &\,\frac{-(-7) \pm \sqrt{(-7)^2 - 4(8)(1)}}{2(8)}\\[4mm] = &\,\frac{7 \pm \sqrt{49 - 32}}{16}\\[4mm] = &\,\frac{7 \pm \sqrt{17}}{16} \end{align*}

Since $u = \sin^2\theta$ , we have $\sin\theta = \sqrt{\frac{7 \pm \sqrt{17}}{16}}$ (taking only the positive root as $0 < \theta < \frac{\pi}{2}$ ).

\begin{align*} \sin\theta = &\,\sqrt{\frac{7 + \sqrt{17}}{16}} \approx 0.83378 \quad \implies \quad \theta \approx \arcsin(0.83378) \approx 0.986\\[4mm] \sin\theta = &\,\sqrt{\frac{7 - \sqrt{17}}{16}} \approx 0.42403 \quad \implies \quad \theta \approx \arcsin(0.42403) \approx 0.438 \end{align*}

The solutions are $\theta = 0.438$ and $\theta = 0.986$ .