4.9 If \( \sin 28^{\circ}=a \) and \( \cos 32^{\circ}=b \), determine the following in terms of \( a \) and/or \( b \) : 4.9 .1 \( \cos 28^{\circ} \quad \begin{array}{ll}4.9 .2 & \cos 64^{\circ} \\ 4.9 .3 & \sin 4^{\circ}\end{array} \)

To express \( \cos 28^{\circ} \) in terms of \( a \), remember the fundamental relationship between sine and cosine: \( \cos(\theta) = \sin(90^{\circ} - \theta) \). Therefore, \( \cos 28^{\circ} = \sin(90^{\circ} - 28^{\circ}) = \sin 62^{\circ} \). Since \( \sin 28^{\circ} = a \), it follows that \( \cos 28^{\circ} = \sin 62^{\circ} = \sqrt{1 - a^2} \). For \( \cos 64^{\circ} \), we can use the identity \( \cos(90^{\circ} - \theta) = \sin(\theta) \). Thus, \( \cos 64^{\circ} = \sin(26^{\circ}) \). Furthermore, since \( 26^{\circ} = 90^{\circ} - 64^{\circ} \), and knowing that \( \sin 28^{\circ} = a \), we can calculate it, but we might need additional conversion to express in terms of \( a \) or \( b \). Now, for \( \sin 4^{\circ} \), we can use a sine subtraction identity: \( \sin 4^{\circ} = \sin(28^{\circ} - 24^{\circ}) \) or other equivalent expressions. Again, this leaves us needing to express in terms of known values without direct calculations. To conclude: 4.9.1 \( \cos 28^{\circ} = \sqrt{1 - a^2} \) 4.9.2 \( \cos 64^{\circ} = \sin 26^{\circ} \), and may need further steps with \( b \). 4.9.3 \( \sin 4^{\circ} \) can be expressed via sine subtraction formulas or approximations as needed.