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

To evaluate ( \sin^2 2x \cos 2x , dx ), we can use the substitution ( u = \sin 2x ), then ( du = 2\cos 2x , dx ), hence:

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

To write ( -\sqrt{3} + i ) in exponential form, we first convert it to polar coordinates:

1. Calculate the modulus:

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

2. Calculate the argument:

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

Thus, the exponential form is:

$-\sqrt{3} + i = 2e^{i\frac{5\pi}{6}}.$

Using the exponential form, we can find:

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

Reducing the exponent:

$\frac{25\pi}{3} - 8\pi = \frac{25\pi - 24\pi}{3} = \frac{\pi}{3}.$

Thus:
$(-\sqrt{3} + i)^{10} = 1024 e^{i\frac{\pi}{3}} = 1024 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = 1024 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = 512 + 512\sqrt{3}i.$

To find the size of ( \angle ABC ):

1. Compute the vectors ( \overrightarrow{AB} ) and ( \overrightarrow{AC} ):
   ( \overrightarrow{AB} = B - A = (0 - 1, 2 - (-1), -1 - 2) = (-1, 3, -3) )
   ( \overrightarrow{AC} = C - A = (2 - 1, 1 - (-1), 1 - 2) = (1, 2, -1) )

2. Use the dot product to find the angle:
   $\cos \angle ABC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|}$
   where ( \overrightarrow{AB} \cdot \overrightarrow{AC} = (-1)(1) + (3)(2) + (-3)(-1) = -1 + 6 + 3 = 8 $$
   and ( |\overrightarrow{AB}| = \sqrt{(-1)^2 + (3)^2 + (-3)^2} = \sqrt{1 + 9 + 9} = \sqrt{19} )
   and ( |\overrightarrow{AC}| = \sqrt{(1)^2 + (2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} ).

3. Plug these into the cosine formula:
   $\cos \angle ABC = \frac{8}{\sqrt{19} \cdot \sqrt{6}}$
   Calculate the angle with ( \angle ABC = \cos^{-1}\left( \frac{8}{\sqrt{114}} \right) ), rounding to the nearest degree gives the size of ( \angle ABC ).

Since ( l_2 ) is parallel to ( l_1 ), it has the same slope. The slope of ( l_1 ) can be derived from its equation. We simplify the equation of ( l_1 ):

1. From the equation of ( l_1 ): $\frac{x}{-1} = \frac{y}{-7} + \alpha(3)$
   Rearranging gives us a slope of ( m = \frac{-7}{-1} = 7. )

2. Using point-slope form for point ( A(-6, 5) ): $y - 5 = 7(x + 6).$
   Simplifying gives: $y = 7x + 42 + 5 \Rightarrow y = 7x + 47.$

Using the substitution ( t = \tan \frac{x}{2} ), we know:

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

Substituting yields:

$\int \frac{dx}{1 + \cos x - \sin x} = \int \frac{\frac{2}{1 + t^2} dt}{1 + \frac{1 - t^2}{1 + t^2}} = \int \frac{2 dt}{2 + 2t^2} = \int \frac{dt}{1 + t^2} = \tan^{-1}(t) + C$

Thus: $= \tan^{-1}\left( \tan \frac{x}{2} \right) + C = \frac{x}{2} + C.$