5.1 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

To find \( ext{sin}(-θ) \, we can use the property of sine:

$$\sin(-θ) = -\sin(θ)$$

Using the triangle formed by the point P, we find:

$$\sin θ = \frac{opposite}{hypotenuse} = \frac{3}{\sqrt{16 + 7}} = \frac{3}{4}$$

Then:

$$\sin(-θ) = -\frac{3}{4}$$

To find \( α \, we can use the cosine relationship. Since \( cos(2θ) = 2cos^2(θ) - 1 \, we need to find \( cos(θ) \).

Using:

$$\sin^2(θ) + \cos^2(θ) = 1$$

Substituting:

$$\left(\frac{3}{4}\right)^2 + \cos^2(θ) = 1$$

Calculating gives:

$$\cos^2(θ) = 1 - \frac{9}{16} = \frac{7}{16}$$

Thus:

$$\cos(θ) = \frac{√7}{4}$$

Thus, \( \alpha = \frac{\cos(2θ)}{6} \) can be calculated as:

$$\alpha = \frac{6(2(\frac{√7}{4})^2 - 1)}{6} = 2(\frac{7}{16} - 1) = -\frac{9}{8}$$

To simplify:

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

By the identity of double angles, \( an(2x) = \frac{2tan x}{1 - tan^2 x}\.

$$$$

To find:

$$\frac{4sin(15°)cos(15°)}{2sin(15°)-1}$$

Using the identity for sine:

$$= 2tan(2(15°))$$

Thus:

$$= 2tan(30°) = 2√{3}$$