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

To evaluate ( \int \sin^2 2x \cos 2x , dx ), we can use the substitution method.

Let ( u = \sin 2x ), therefore ( du = 2 \cos 2x , dx ) or ( dx = \frac{du}{2 \cos 2x} ).

The integral becomes:

[ \int \sin^2 2x \cos 2x , dx = \frac{1}{2} \int u^2 , du = \frac{1}{2} \cdot \frac{u^3}{3} + C = \frac{1}{6} \sin^3 2x + C ]

The complex number ( -\sqrt{3} + i ) can be expressed in exponential form as follows:

First, we find the modulus:

[ |z| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2 ]

Next, we determine the argument:

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

Thus, in exponential form, we have:

[ z = 2 e^{i \frac{5\pi}{6}}]

Using the exponential form we obtained, we can compute:

[ (-\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}} ]

To simplify, we reduce the argument:

[ \frac{25\pi}{3} - 8\pi = \frac{25\pi - 24\pi}{3} = \frac{\pi}{3} ]

Finally, [ (-\sqrt{3} + i)^{10} = 1024 e^{i \frac{\pi}{3}} = 1024 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = 1024 \left(\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = 512 + 512i ]

To find the size of ( \angle ABC ), we first need to calculate the vectors ( \overrightarrow{AB} ) and ( \overrightarrow{AC} ):

[\overrightarrow{AB} = B - A = \begin{pmatrix} 0 - 1 \ 2 - (-1) \ -1 - 2 \end{pmatrix} = \begin{pmatrix} -1 \ 3 \ -3 \end{pmatrix}]

[\overrightarrow{AC} = C - A = \begin{pmatrix} 2 - 1 \ 1 - (-1) \ 1 - 2 \end{pmatrix} = \begin{pmatrix} 1 \ 2 \ -1 \end{pmatrix}]

Now we calculate the dot product:\n [ \overrightarrow{AB} \cdot \overrightarrow{AC} = (-1)(1) + (3)(2) + (-3)(-1) = -1 + 6 + 3 = 8 ]

Next, we find the magnitudes of the vectors:

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

Now we can use the cosine formula:

[\cos \angle ABC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|} = \frac{8}{\sqrt{19} \sqrt{6}}]

Thus, [\angle ABC = \cos^{-1} \left( \frac{8}{\sqrt{114}} \right) \approx 33^\circ]

Given the line ( l_1 ):

[\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} -1 \ 2 \end{pmatrix} + \lambda \begin{pmatrix} 3 \ -7 \end{pmatrix}, \lambda \in \mathbb{R},]

The direction vector of ( l_1 ) is ( \begin{pmatrix} 3 \ -7 \end{pmatrix}). Since ( l_2 ) is parallel to ( l_1 ), it has the same slope. We also know that it passes through point ( A(-6, 5) ).

Using point-slope form:

[y - 5 = m(x + 6)]

Calculating the slope, we find:

Slope ( m = \frac{-7}{3} \Rightarrow y - 5 = \frac{-7}{3}(x + 6)]

Expanding this, we find: [y = \frac{-7}{3}x - 14 + 5 = \frac{-7}{3}x - 9]

Starting with the substitution:

[t = \tan \frac{x}{2} \Rightarrow dx = \frac{2}{1 + t^2} dt]

We also know that:

[\cos x = \frac{1 - t^2}{1 + t^2} \quad \text{and} \quad \sin x = \frac{2t}{1 + t^2}]

Therefore, [1 + \cos x - \sin x = 1 + \frac{1 - t^2}{1 + t^2} - \frac{2t}{1 + t^2} = \frac{(1 + t^2 + 1 - t^2 - 2t)}{1 + t^2} = \frac{2 - 2t}{1 + t^2}]

Now substituting back into the integral:

[\int \frac{dx}{1 + \cos x - \sin x} = \int \frac{2}{(2 - 2t)(1 + t^2)} dt = \int \frac{1}{-(t - 1)(1 + t^2)} dt]

This can then be integrated using partial fractions.