If \( \sin 12^{\circ}=k \) determine the following in terms of \( K \) \( \begin{array}{ll} \\ \text { a) } \cos 24^{\circ} & \text { (b) } \sin 24^{\circ} \\ \text { (c) } \sin 78^{\circ} & \text { (d) } \sin 42^{\circ} \\ \text { () } \tan 12^{\circ} \\ \text { 2. If } \sin 9^{\circ}=K \text { determine } \\ \text { of value of } \sin 18 \text { in terms } \\ \text { of } K\end{array} \)

If \( \sin 12^{\circ} = k \), we can use double angle and co-function identities! a) \( \cos 24^{\circ} = \sin (90^{\circ} - 24^{\circ}) = \sin 66^{\circ} \). Using the identity \( \sin (a + b) = \sin a \cos b + \cos a \sin b \), we find: \[ = \sin (2 \times 12^{\circ}) = 2 \sin 12^{\circ} \cos 12^{\circ} = 2k \sqrt{1-k^2}. \] b) For \( \sin 24^{\circ} \): \[ \sin 24^{\circ} = \sin(2 \times 12^{\circ}) = 2 \sin 12^{\circ} \cos 12^{\circ} = 2k \sqrt{1-k^2}. \] c) \( \sin 78^{\circ} = \sin (90^{\circ} - 12^{\circ}) = \cos 12^{\circ} = \sqrt{1 - k^2}. \) d) \( \sin 42^{\circ} = \sin (90^{\circ} - 48^{\circ}) = \cos 48^{\circ} \), but also expressed as \( \sin (90^{\circ} - 48^{\circ}) \), which can be calculated but typically involves additional steps, so we express \( \sin 42^{\circ} = \sin(48^{\circ}) \). Finally for \( \tan 12^{\circ} \): \[ \tan 12^{\circ} = \frac{\sin 12^{\circ}}{\cos 12^{\circ}} = \frac{k}{\sqrt{1-k^2}}. \] Now, regarding the second part, if \( \sin 9^{\circ} = K \) and using the double angle formula, we find \( \sin 18^{\circ} = 2 \sin 9^{\circ} \cos 9^{\circ} \). Thus: \[ \sin 18^{\circ} = 2K \sqrt{1 - K^2}. \]