IAL 2025 June A Q8 | 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.

Given that $z = \cos\theta + \mathrm{i}\sin\theta$

(a) show that, for $n \in \mathbb{Z}$

\begin{align*} z^n + \frac{1}{z^n} = 2\cos n\theta \end{align*}

(2)

(b) Hence show that

\begin{align*} \cos^4\theta = \frac{1}{8}(\cos 4\theta + a\cos 2\theta + b) \end{align*}

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

(4)

Figure 4

Figure 4 shows a sketch of the curve with equation

\begin{align*} y = \cos^2 x\sqrt{1 + \sin x} \qquad -\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2} \end{align*}

The region $R$ , shown shaded in Figure 4, is bounded by the curve, the $x$ -axis and the line with equation $x = \dfrac{\pi}{4}$

The region $R$ is rotated through $2\pi$ radians about the $x$ -axis to form a solid of revolution.

Give the answer in the form $\frac{\pi}{160}(p + q\pi + r\sqrt{2}\,)$ where $p$ , $q$ and $r$ are integers.

(6)

解答

(a)

By De Moivre’s Theorem, for $z = \cos\theta + i\sin\theta$ :

\begin{align*} z^n = &\,\cos(n\theta) + i\sin(n\theta) \end{align*}

And for the reciprocal:

\begin{align*} \frac{1}{z^n} = z^{-n} = &\,\cos(-n\theta) + i\sin(-n\theta)\\[4mm] = &\,\cos(n\theta) - i\sin(n\theta) \end{align*}

Adding these two equations:

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

(b)

Using the binomial expansion on $\left( z + \frac{1}{z} \right)^4$ :

\begin{align*} \left( z + \frac{1}{z} \right)^4 = &\,z^4 + 4z^3\left(\frac{1}{z}\right) + 6z^2\left(\frac{1}{z}\right)^2 + 4z\left(\frac{1}{z}\right)^3 + \left(\frac{1}{z}\right)^4\\[4mm] = &\,z^4 + 4z^2 + 6 + \frac{4}{z^2} + \frac{1}{z^4}\\[4mm] = &\,\left( z^4 + \frac{1}{z^4} \right) + 4\left( z^2 + \frac{1}{z^2} \right) + 6 \end{align*}

Using the result from (a), where $z^n + \frac{1}{z^n} = 2\cos(n\theta)$ , we can substitute:

\begin{align*} (2\cos\theta)^4 = &\,(2\cos 4\theta) + 4(2\cos 2\theta) + 6\\[4mm] 16\cos^4\theta = &\,2\cos 4\theta + 8\cos 2\theta + 6\\[4mm] \cos^4\theta = &\,\frac{1}{16}(2\cos 4\theta + 8\cos 2\theta + 6)\\[4mm] = &\,\frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3) \end{align*}

This gives the required form, with $a=4$ and $b=3$ .

(c)

The volume $V$ of the solid of revolution formed by rotating the curve about the $x$ -axis is:

\begin{align*} V = &\,\pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{4}} y^2 \,\mathrm{d}x\\[4mm] = &\,\pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{4}} \left( \cos^2 x \sqrt{1 + \sin x} \right)^2 \,\mathrm{d}x\\[4mm] = &\,\pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{4}} \cos^4 x (1 + \sin x) \,\mathrm{d}x\\[4mm] = &\,\pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{4}} (\cos^4 x + \cos^4 x \sin x) \,\mathrm{d}x \end{align*}

Using the identity from (b), we replace $\cos^4 x$ :

\begin{align*} \int \cos^4 x \,\mathrm{d}x = &\,\int \frac{1}{8}(\cos 4x + 4\cos 2x + 3) \,\mathrm{d}x\\[4mm] = &\,\frac{1}{8}\left( \frac{1}{4}\sin 4x + 2\sin 2x + 3x \right) \end{align*}

For the second part of the integral:

\begin{align*} \int \cos^4 x \sin x \,\mathrm{d}x = &\,-\frac{1}{5}\cos^5 x \end{align*}

Combining these, the indefinite integral is:

\begin{align*} I = &\,\pi \left[ \frac{1}{8}\left( \frac{1}{4}\sin 4x + 2\sin 2x + 3x \right) - \frac{1}{5}\cos^5 x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{4}} \end{align*}

Now, evaluate at the upper limit $x = \frac{\pi}{4}$ :

\begin{align*} &\,\pi \left( \frac{1}{8} \left( \frac{1}{4}\sin(\pi) + 2\sin\left(\frac{\pi}{2}\right) + 3\left(\frac{\pi}{4}\right) \right) - \frac{1}{5}\cos^5\left(\frac{\pi}{4}\right) \right)\\[4mm] = &\,\pi \left( \frac{1}{8} \left( 0 + 2 + \frac{3\pi}{4} \right) - \frac{1}{5}\left(\frac{\sqrt{2}}{2}\right)^5 \right)\\[4mm] = &\,\pi \left( \frac{1}{4} + \frac{3\pi}{32} - \frac{1}{5}\left(\frac{4\sqrt{2}}{32}\right) \right)\\[4mm] = &\,\pi \left( \frac{1}{4} + \frac{3\pi}{32} - \frac{\sqrt{2}}{40} \right) \end{align*}

Evaluate at the lower limit $x = -\frac{\pi}{2}$ :

\begin{align*} &\,\pi \left( \frac{1}{8} \left( \frac{1}{4}\sin(-2\pi) + 2\sin(-\pi) + 3\left(-\frac{\pi}{2}\right) \right) - \frac{1}{5}\cos^5\left(-\frac{\pi}{2}\right) \right)\\[4mm] = &\,\pi \left( \frac{1}{8} \left( 0 + 0 - \frac{3\pi}{2} \right) - 0 \right)\\[4mm] = &\,-\frac{3\pi}{16} \pi \end{align*}

Subtract the lower limit value from the upper limit value:

\begin{align*} V = &\,\pi \left( \frac{1}{4} + \frac{3\pi}{32} - \frac{\sqrt{2}}{40} - \left( -\frac{3\pi}{16} \right) \right)\\[4mm] = &\,\pi \left( \frac{1}{4} + \frac{3\pi}{32} + \frac{6\pi}{32} - \frac{\sqrt{2}}{40} \right)\\[4mm] = &\,\pi \left( \frac{1}{4} + \frac{9\pi}{32} - \frac{\sqrt{2}}{40} \right) \end{align*}

To write this over a common denominator of $160$ :

\begin{align*} \frac{1}{4} = \frac{40}{160}, \quad \frac{9\pi}{32} = \frac{45\pi}{160}, \quad \frac{\sqrt{2}}{40} = \frac{4\sqrt{2}}{160} \end{align*}

Thus, the volume is:

\begin{align*} V = &\,\frac{\pi}{160} \left( 40 + 45\pi - 4\sqrt{2} \right) \end{align*}