a) Express $\frac{3 - i}{2 + i}$ in the form $x + iy$, where $x$ and $y$ are real numbers - HSC - SSCE Mathematics Extension 2 - Question 11 - 2022 - Paper 1

Using the identity $\sin 2x = 2 \sin x \cos x$, we rewrite:

$\sin 2x \cos 2x = \frac{1}{2} \sin 4x$

Thus, we have:

$\int \sin 2x \cos 2x \,dx = \frac{1}{2} \int \sin 4x \,dx$.

Calculating the integral:

$\frac{1}{2} \cdot \left( -\frac{1}{4} \cos 4x \right) + C = -\frac{1}{8} \cos 4x + C$.

To write $-\sqrt{3} + i$ in exponential form, we first find its modulus and argument:

The modulus is:

$r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$

The argument is:

$\theta = \tan^{-1}\left(\frac{1}{-\sqrt{3}}\right) = \frac{5\pi}{6}$

Thus, $-\sqrt{3} + i$ can be expressed in exponential form as:

$2 e^{i\frac{5\pi}{6}}$.

Using De Moivre's theorem:

$(-\sqrt{3} + i)^{10} = (2 e^{i\frac{5\pi}{6}})^{10} = 2^{10} e^{i\frac{50\pi}{6}} = 1024 e^{i\frac{25\pi}{3}}$

Since $\frac{25\pi}{3}$ exceeds $2\pi$, we reduce it:

$\frac{25\pi}{3} = 8\pi + \frac{1\pi}{3}$

Thus:

$e^{i\frac{25\pi}{3}} = e^{i\frac{\pi}{3}}\Rightarrow -\sqrt{3} + i = 1024 \left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)$

This gives us:

$= 512 + 512i\Rightarrow 512 + 512i$.

To find $\angle ABC$, first calculate vectors $\vec{AB}$ and $\vec{AC}$. The coordinates are:

$\vec{AB} = B - A = (0 - 1, 2 + 1, -1 - 2) = (-1, 3, -3)$ $\vec{AC} = C - A = (2 - 1, 1 + 1, 1 - 2) = (1, 2, -1)$

Now, use the dot product to find the angle:

$\vec{AB} \cdot \vec{AC} = (-1)(1) + (3)(2) + (-3)(-1) = -1 + 6 + 3 = 8$

The magnitudes are:

$|\vec{AB}| = \sqrt{(-1)^2 + 3^2 + (-3)^2} = \sqrt{1 + 9 + 9} = \sqrt{19}$ $|\vec{AC}| = \sqrt{(1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$

Using the cosine rule:

$\cos \angle ABC = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}| |\vec{AC}|} = \frac{8}{\sqrt{19} \cdot \sqrt{6}}$

Calculating gives:

$\angle ABC = \cos^{-1}\left(\frac{8}{\sqrt{114}}\right) \approx 33^{\circ}$.

Since line $l_2$ is parallel to $l_1$, it has the same direction vector $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$. The point $A(-6, 5)$ gives:

The parametric form is:

$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -6 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 2 \end{pmatrix}$

This leads to:

$x = -6 + 3\mu, \, y = 5 + 2\mu$

To write in the form $y = mx + c$, solve for $\mu$:

From $x = -6 + 3\mu$:

$\mu = \frac{x + 6}{3}$

Substituting into $y$ gives:

$y = 5 + 2\left(\frac{x + 6}{3}\right) = 5 + \frac{2x + 12}{3} = \frac{2x + 12 + 15}{3} = \frac{2x + 27}{3}$

Thus, the equation is:

$y = \frac{2}{3}x + 9$.

Using the substitution $t = \tan \frac{x}{2}$, we have:

$dx = \frac{2}{1 + t^2} dt$

Expressing $\cos x$ and $\sin x$ in terms of $t$ gives:

$\cos x = \frac{1 - t^2}{1 + t^2}, \quad \sin x = \frac{2t}{1 + t^2}$

Thus, the integral becomes:

$\int \frac{dx}{1 + \frac{1 - t^2}{1 + t^2} - \frac{2t}{1 + t^2}} = \int \frac{2dt}{1 + t^2 + 1 - t^2 - 2t} = \int \frac{2dt}{2 - 2t} = -\ln|1 - t| + C$.

Substituting back gives:

$= -\ln\left|1 - \tan \frac{x}{2}\right| + C$.