QUESTION 4 1.1 Given that \( \cos 42^{\circ}=k \), determine the following in terms of \( k \). 4.1.1 \( \sin 42^{\circ} \) (3) 4.1.2 \( \cos 21^{\circ} \) (3) Simplify the following expression to a single trigonometric ratio. \[ \frac{\cos (-\theta) \cdot \sin \left(90^{\circ}+\theta\right)}{\cos \left(360^{\circ}-\theta\right) \cdot \sin ^{2}\left(540^{\circ}+\theta\right)} \]

To solve for \( \sin 42^{\circ} \) when \( \cos 42^{\circ} = k \), we can use the Pythagorean identity, \( \sin^2 \theta + \cos^2 \theta = 1 \). Thus: \[ \sin 42^{\circ} = \sqrt{1 - \cos^2 42^{\circ}} = \sqrt{1 - k^2} \] For \( \cos 21^{\circ} \), we can make use of the double angle formula: \[ \cos 21^{\circ} = \cos(42/2) = \sqrt{\frac{1 + \cos 42^{\circ}}{2}} = \sqrt{\frac{1 + k}{2}} \] Now, let's simplify the given expression: \[ \frac{\cos (-\theta) \cdot \sin(90^{\circ} + \theta)}{\cos(360^{\circ} - \theta) \cdot \sin^{2}(540^{\circ} + \theta)} \] First, we know that \( \cos(-\theta) = \cos(\theta) \) and \( \sin(90^{\circ} + \theta) = \cos \theta \). Thus, the numerator becomes: \[ \cos(\theta) \cdot \cos(\theta) = \cos^2(\theta) \] For the denominator, we have: \[ \cos(360^{\circ} - \theta) = \cos(\theta) \] And \( \sin(540^{\circ} + \theta) = \sin(180^{\circ} + \theta) = -\sin(\theta) \), so: \[ \sin^2(540^{\circ} + \theta) = \sin^2(\theta) \] Now the denominator becomes: \[ \cos(\theta) \cdot \sin^2(\theta) \] Putting it all together: \[ \frac{\cos^2(\theta)}{\cos(\theta) \cdot \sin^2(\theta)} = \frac{\cos(\theta)}{\sin^2(\theta)} = \cot(\theta) \cdot \csc(\theta) \] The final simplified expression is: \[ \cot(\theta) \cdot \csc(\theta) \]