Determine, WITHOUT using a calculator, the value of: 5.1.1 tan θ 5.1.2 sin(−θ) 5.1.3 a 5.2 5.2.1 Simplify \( 4 \sin x \cos x \) to a single trigonometric ratio - NSC Mathematics - Question 5 - 2016 - Paper 2

Using the property of sine, we have:

$\sin(-\theta) = -\sin(\theta)$

From our previous calculation, since ( \sin \theta = \frac{3}{4} ), we find:

$\sin(-\theta) = -\frac{3}{4}$

Given that ( a = \frac{\cos 2\theta}{6} ), we can use the double angle formula for cosine:

$\cos(2\theta) = 2\cos^2(\theta) - 1$

We first need ( \cos \theta ):

Knowing ( \sin^2 \theta + \cos^2 \theta = 1 ), we can calculate:

$\cos^2 \theta = 1 - \left(\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{7}{16}$

Thus,

$\cos \theta = -\sqrt{\frac{7}{16}} = -\frac{\sqrt{7}}{4}$

Now substituting into the cosine double angle formula:

$\cos(2\theta) = 2\left(-\frac{\sqrt{7}}{4}\right)^2 - 1 = 2\times\frac{7}{16} - 1 = \frac{7}{8} - 1 = -\frac{1}{8}$

Therefore:

$a = 6 \times \left(-\frac{1}{8}\right) = -\frac{3}{4}$

Using the double angle identity:

$4 \sin x \cos x = 2 \sin(2x)$

Substituting our previous finding into this expression,

$\frac{4 \sin 15^{\circ} \cos 15^{\circ}}{2 \sin 15^{\circ} - 1} = \frac{2 \sin(30^{\circ})}{2 \sin 15^{\circ} - 1}$

We know that:

$\sin(30^{\circ}) = \frac{1}{2}$

Thus,

$= \frac{1}{2(2 \sin 15^{\circ} - 1)}$

Substituting the known value of ( \sin 15^{\circ} = \frac{\sqrt{3}}{4} ):

Hence, simplifying gives:

$= 2 \tan(2 \cdot 15^{\circ}) = 2 \tan(30^{\circ}) = 2 \times \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$