5.1 P(−√7; 3) and S(a; b) are points on the Cartesian plane, as shown in the diagram below - NSC Mathematics - Question 5 - 2016 - Paper 2

Using the property of the sine function that states ( \sin(-\theta) = -\sin(\theta) ):

Calculating ( \sin(\theta) ):

[ OP^2 = (-\sqrt{7})^2 + 3^2] [ OP^2 = 7 + 9 = 16 \implies OP = 4 ] [ \sin(\theta) = \frac{OP}{OR} = \frac{4}{6} = \frac{2}{3} ]

Therefore, ( \sin(-\theta) = -\frac{2}{3} ).

Using the relationship between angles:

[ a = \frac{\cos 2\theta}{6} ]

We can find ( a ) as follows:

[ \cos 2\theta = 1 - 2\sin^2\theta = 1 - 2\left(\frac{2}{3}\right)^2 = 1 - 2 \cdot \frac{4}{9} = 1 - \frac{8}{9} = \frac{1}{9} ]

Thus, substituting back, we get: [ a = \frac{1/9}{6} = \frac{1}{54} ].

To simplify ( \frac{4 \sin x \cos x}{2 \sin x - 1} ), we first notice that:

[ 4 \sin x \cos x = 2 \sin 2x \text{ and consider } 2 \sin x - 1. ]

This gives us: [ \frac{2 \sin 2x}{2 \sin x - 1} = 2 \tan 2x. ]

Using the result from the previous part:

[ 4 \sin 15^{\circ} \cos 15^{\circ} = 2 \sin 30^{\circ} = 2\left(\frac{1}{2}\right) = 1 ]

We substitute:

[ 2 \sin 15^{\circ} - 1 = 2 \left(\frac{\sqrt{6}-\sqrt{2}}{4}\right) - 1 = \frac{\sqrt{6}-\sqrt{2}}{2} - 1 ]

Hence: [ \frac{1}{\frac{\sqrt{6}-\sqrt{2}}{2} - 1} = -2 \tan 2(15^{\circ}) ].