QUESTION 4 Given that \( \cos 42^{\circ}=k \), determine the following in terms of \( k \). 4.1.1 \( \sin 42^{\circ} \) 4.1.2 \( \cos 21^{\circ} \) (3) \[ \begin{array}{l} \text { Simplify the following expression to a single trigonometric ratio. } \\ \qquad \frac{\cos (-\theta) \cdot \sin \left(90^{\circ}+\theta\right)}{\cos \left(360^{\circ}-\theta\right) \cdot \sin ^{2}\left(540^{\circ}+\theta\right)} \end{array} \] .2 Grade 12 March Control Test \[ \text { Prove the identity : } \frac{\sin 2 x}{1+\cos 2 x}=\tan x \] (3) [15] TOTAL: 50

To find \( \sin 42^{\circ} \), we can use the Pythagorean identity which states that \( \sin^2 \theta + \cos^2 \theta = 1 \). Since \( \cos 42^{\circ} = k \), we have \( \sin^2 42^{\circ} = 1 - k^2 \). Therefore, \( \sin 42^{\circ} = \sqrt{1 - k^2} \). Remember, in a question involving acute angles, the sine function is positive. For \( \cos 21^{\circ} \), we can utilize the angle subtraction identity. We know that \( 21^{\circ} = 42^{\circ} / 2 \), allowing us to apply the half-angle formula: \[ \cos 21^{\circ} = \sqrt{\frac{1 + \cos 42^{\circ}}{2}} = \sqrt{\frac{1 + k}{2}}. \] Now, simplifying the expression \(\frac{\cos (-\theta) \cdot \sin (90^{\circ}+\theta)}{\cos (360^{\circ}-\theta) \cdot \sin^{2}(540^{\circ}+\theta)}\) leads us through some fun exploration of trigonometric identities! We know that \( \cos(-\theta) = \cos(\theta) \) and \( \sin(90^{\circ} + \theta) = \cos(\theta) \). This means the numerator simplifies to \( \cos^2(\theta) \). On the denominator side, \( \cos(360^{\circ} - \theta) = \cos(\theta) \) and \( \sin(540^{\circ} + \theta) = \sin(180^{\circ} + \theta) = -\sin(\theta) \). Hence, \( \sin^2(540^{\circ} + \theta) = \sin^2(\theta) \). Putting this all together gives: \[ \frac{\cos^2(\theta)}{\cos(\theta) \cdot \sin^2(\theta)} = \frac{\cos(\theta)}{\sin^2(\theta)}. \] What a journey through the trigonometric world! To prove the identity \(\frac{\sin 2x}{1 + \cos 2x} = \tan x\), we start by applying the double angle formulas. Remember, \(\sin 2x = 2 \sin x \cos x\) and \(\cos 2x = 1 - 2\sin^2 x\). Thus, \(1 + \cos 2x = 1 + 1 - 2\sin^2 x = 2 - 2\sin^2 x = 2\cos^2 x\). Therefore: \[ \frac{2 \sin x \cos x}{2 \cos^2 x} = \frac{\sin x}{\cos x} = \tan x. \] Voilà! The identity is proven!