If \( \sin 27^{\circ}=q \), express each of the following in tems of \( q \) \( \begin{array}{ll}\text { a) } \sin 117^{\circ} & \text { b) } \cos \left(-27^{\circ}\right)\end{array} \)

To express \( \sin 117^{\circ} \) in terms of \( q \), we can use the sine angle addition identity. Notably, \( 117^{\circ} = 90^{\circ} + 27^{\circ} \). Hence, \[ \sin 117^{\circ} = \sin(90^{\circ} + 27^{\circ}) = \cos 27^{\circ}. \] Since we know that \( \cos 27^{\circ} = \sqrt{1 - \sin^2 27^{\circ}} = \sqrt{1 - q^2} \), we can express: \[ \sin 117^{\circ} = \sqrt{1 - q^2}. \] For \( \cos(-27^{\circ}) \), we can utilize the property of cosine being an even function: \[ \cos(-27^{\circ}) = \cos(27^{\circ}). \] Again, knowing that \( \cos 27^{\circ} = \sqrt{1 - \sin^2 27^{\circ}} \), we find: \[ \cos(-27^{\circ}) = \sqrt{1 - q^2}. \] In summary, a) \( \sin 117^{\circ} = \sqrt{1 - q^2} \) b) \( \cos(-27^{\circ}) = \sqrt{1 - q^2} \)