b. If \( \cos 36^{\circ}=k \), deformine the following in term of \( k \). a) \( \cos \left(-36^{\circ}\right) \) b) \( \cos 77^{\circ} \) c) \( \sin 36^{\circ} \) d) \( \sin 72^{\circ} \) c) \( \tan 36^{\circ} \)

Sure! Let's express the given angles in terms of \( k \). a) \( \cos \left(-36^{\circ}\right) = \cos 36^{\circ} = k \) (since cosine is an even function). b) \( \cos 77^{\circ} = \cos(90^{\circ}-13^{\circ}) = \sin 13^{\circ} \), and using the identity \( \sin(90^{\circ}-x) = \cos x \), we get \( \cos 77^{\circ} = \sin(90^{\circ}-36^{\circ}) = \sin 36^{\circ} = \sqrt{1 - k^2} \). c) To express \( \sin 36^{\circ} \) in terms of \( k \), we apply the Pythagorean identity: \( \sin^2 36^{\circ} + \cos^2 36^{\circ} = 1 \). This gives us \( \sin 36^{\circ} = \sqrt{1 - k^2} \). d) \( \sin 72^{\circ} = \cos 18^{\circ} \), and using the identity for cosine, we can express this in terms of \( k \) if we use \( \sin 72^{\circ} = 2 \sin 36^{\circ} \cos 36^{\circ} = 2 \cdot \sqrt{1 - k^2} \cdot k \). e) For \( \tan 36^{\circ} \), we use the definition \( \tan x = \frac{\sin x}{\cos x} \). Thus, \( \tan 36^{\circ} = \frac{\sin 36^{\circ}}{\cos 36^{\circ}} = \frac{\sqrt{1 - k^2}}{k} \). Summarizing everything, we'd have the relationships of each of those values in terms of \( k \). Happy calculating!