P(−√7; 3) en S(a; b) is punte in die Cartesiese vlak soos in die diagram hieronder getoon - NSC Mathematics - Question 5 - 2016 - Paper 2

Using the property of sine, we find:

$sin(−θ) = −sin(θ)$

With $sin(θ) = \frac{3}{4}$ (from previous calculations), therefore:

$sin(−θ) = −\frac{3}{4}.$

To find α, we use the cosine relation:

$\frac{a}{6} = cos 2θ$

From the earlier calculations, we have: $a = 6(1 - 2 sin^2 θ) = 6(1 - 2(\frac{3}{4})^2)$

Calculating this gives:

a = 6(1 - 2(\frac{9}{16})) = 6(1 - \frac{18}{16}) = 6(−\frac{2}{16}) = -\frac{3\sqrt{7}}{4}.$$

We first recognize:

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

This simplifies to:

$2tan 2x.$

Substituting the values:

$\frac{4sin15^{\circ}cos15^{\circ}}{2sin15^{\circ}-1} = 2tan 2(15^{\circ})$

Since $tan(30^{\circ}) = \frac{1}{\sqrt{3}}$, we have:

$= 2 \cdot \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}.$