In the diagram, P(k ; 1) is a point in the 2nd quadrant and is \sqrt{5} units from the origin - NSC Mathematics - Question 5 - 2018 - Paper 2

(a) \tan \theta

To find \tan \theta, we use the reference angle in the 2nd quadrant:

$\tan \theta = -\frac{1}{\sqrt{5}}.$

(b) \cos(180^\circ + \theta)

Using the cosine identity for the addition of angles:

$\cos(180^\circ + \theta) = -\cos \theta = -\left(-\frac{2}{\sqrt{5}}\right) = \frac{2}{\sqrt{5}}.$

(c) \sin(\theta + 60^\circ)

Using the sine addition formula:

$\sin(\theta + 60^\circ) = \sin \theta \cos 60^\circ + \cos \theta \sin 60^\circ = \left(-\frac{1}{\sqrt{5}}\right) \left(\frac{1}{2}\right) + \left(-\frac{2}{\sqrt{5}}\right) \left(\frac{\sqrt{3}}{2}\right)$

Combining terms:

$= -\frac{1}{2\sqrt{5}} - \frac{\sqrt{3}}{\sqrt{5}} = \frac{-1 - \sqrt{3}}{2\sqrt{5}}.$

Calculating:

$\tan(20^\circ - 40^\circ) = \tan(-20^\circ) = -\tan(20^\circ)$

Using a calculator gives:

\tan(-20^\circ) \approx -0.364.

To prove the first part of the identity:

$LHS = \cos x + \sin x$

Transforming it to single fraction:

$= \frac{(\cos^2 x + \sin^2 x)(\cos x + \sin x)}{(\cos x + \sin x)(\cos x - \sin x)}$

Simplifying:

$= 2\tan 2x$

Thus proving the identity holds.

Using the identity:

$\sum_{A=0}^{52} \cos^2 A = \frac{1}{2}(52 + 1) = \frac{53}{2}$

Therefore, the final result is:

$= 26.5.$