IAL 2024 June Q9 | 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 6\theta \equiv 32\cos^6 \theta - 48\cos^4 \theta + 18\cos^2 \theta - 1 \end{align*}

(4)

(b) Hence determine the smallest positive root of the equation

\begin{align*} 48x^6 - 72x^4 + 27x^2 - 1 = 0 \end{align*}

giving your answer to $3$ decimal places.

(4)

解答

(a)

By De Moivre’s Theorem, $(\cos\theta + \mathrm{i}\sin\theta)^6 = \cos 6\theta + \mathrm{i}\sin 6\theta$ . We expand $(\cos\theta + \mathrm{i}\sin\theta)^6$ using the binomial expansion and equate the real parts.

\begin{align*} (\cos\theta + \mathrm{i}\sin\theta)^6 = &\,\cos^6\theta + \binom{6}{1}\cos^5\theta(\mathrm{i}\sin\theta) + \binom{6}{2}\cos^4\theta(\mathrm{i}\sin\theta)^2\\[4mm] &\,\hspace{2pt}+ \binom{6}{3}\cos^3\theta(\mathrm{i}\sin\theta)^3 + \binom{6}{4}\cos^2\theta(\mathrm{i}\sin\theta)^4\\[4mm] &\,\hspace{4pt}+ \binom{6}{5}\cos\theta(\mathrm{i}\sin\theta)^5 + (\mathrm{i}\sin\theta)^6 \end{align*}

Extracting only the real terms (since $\mathrm{i}^2 = -1, \mathrm{i}^4 = 1, \mathrm{i}^6 = -1$ ):

\begin{align*} \cos 6\theta = &\,\cos^6\theta - 15\cos^4\theta \sin^2\theta + 15\cos^2\theta \sin^4\theta - \sin^6\theta \end{align*}

Use the identity $\sin^2\theta = 1 - \cos^2\theta$ :

\begin{align*} \cos 6\theta = &\,\cos^6\theta - 15\cos^4\theta(1 - \cos^2\theta) + 15\cos^2\theta(1 - \cos^2\theta)^2 - (1 - \cos^2\theta)^3\\[4mm] = &\,\cos^6\theta - 15\cos^4\theta + 15\cos^6\theta + 15\cos^2\theta(1 - 2\cos^2\theta + \cos^4\theta)\\[4mm] &\,\hspace{2pt}- (1 - 3\cos^2\theta + 3\cos^4\theta - \cos^6\theta)\\[4mm] = &\,\cos^6\theta - 15\cos^4\theta + 15\cos^6\theta + 15\cos^2\theta - 30\cos^4\theta + 15\cos^6\theta\\[4mm] &\,\hspace{2pt}- 1 + 3\cos^2\theta - 3\cos^4\theta + \cos^6\theta \end{align*}

Combine like terms:

\begin{align*} \cos 6\theta = &\,(1 + 15 + 15 + 1)\cos^6\theta + (-15 - 30 - 3)\cos^4\theta\\[4mm] &\,\hspace{2pt}+ (15 + 3)\cos^2\theta - 1\\[4mm] = &\,32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1 \end{align*}

This completes the proof.

(b)

We need to solve the equation $48x^6 - 72x^4 + 27x^2 - 1 = 0$ . Notice that this equation is closely related to our identity for $\cos 6\theta$ . Let $x = \cos\theta$ .

\begin{align*} \frac{3}{2}(32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1) = &\,48\cos^6\theta - 72\cos^4\theta + 27\cos^2\theta - \frac{3}{2} \end{align*}

Thus, the given equation can be rewritten as:

\begin{align*} \left( 48\cos^6\theta - 72\cos^4\theta + 27\cos^2\theta - \frac{3}{2} \right) + \frac{1}{2} = &\,0\\[4mm] \frac{3}{2} \cos 6\theta + \frac{1}{2} = &\,0\\[4mm] \cos 6\theta = &\,-\frac{1}{3} \end{align*}

We want to find the smallest positive root for $x$ . Since $x = \cos\theta$ , we want to find the value of $\theta$ that gives the smallest positive $\cos\theta$ . This occurs when $\theta$ is as close to $\frac{\pi}{2}$ as possible (but less than $\frac{\pi}{2}$ ).

The principal value is:

\begin{align*} 6\theta = &\,\arccos\left(-\frac{1}{3}\right) \approx 1.9106 \text{ rad} \end{align*}

The general solutions for $6\theta$ in the positive domain are $2\pi - 1.9106$ , $2\pi + 1.9106$ , $4\pi - 1.9106$ , etc. Let’s find the values of $\theta$ :

\begin{align*} 6\theta_1 \approx 1.9106 \quad &\Rightarrow \quad \theta_1 \approx 0.318 \quad \Rightarrow \quad x = \cos(0.318) \approx 0.950\\[4mm] 6\theta_2 = 2\pi - 1.9106 \approx 4.373 \quad &\Rightarrow \quad \theta_2 \approx 0.729 \quad \Rightarrow \quad x = \cos(0.729) \approx 0.746\\[4mm] 6\theta_3 = 2\pi + 1.9106 \approx 8.194 \quad &\Rightarrow \quad \theta_3 \approx 1.366 \quad \Rightarrow \quad x = \cos(1.366) \approx 0.204\\[4mm] 6\theta_4 = 4\pi - 1.9106 \approx 10.656 \quad &\Rightarrow \quad \theta_4 \approx 1.776 \quad \Rightarrow \quad x = \cos(1.776) \approx -0.204 \end{align*}

The value $x = -0.204$ is negative. Therefore, the smallest positive root occurs when $6\theta = 2\pi + \arccos\left(-\frac{1}{3}\right)$ .

\begin{align*} x = &\,\cos\left( \frac{2\pi + \arccos(-1/3)}{6} \right)\\[4mm] \approx &\,0.204 \end{align*}