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) \( \tan 36^{\circ} \)

To express the trigonometric functions in terms of \( k \) where \( k = \cos 36^{\circ} \), we can use the properties of trigonometric functions. ### a) \( \cos \left(-36^{\circ}\right) \) Using the even property of cosine: \[ \cos(-x) = \cos(x) \] Thus, \[ \cos(-36^{\circ}) = \cos(36^{\circ}) = k \] ### b) \( \cos 77^{\circ} \) Using the co-function identity: \[ \cos(90^{\circ} - x) = \sin(x) \] We can express \( \cos 77^{\circ} \) as: \[ \cos(77^{\circ}) = \sin(13^{\circ}) \] Now, we can use the sine identity: \[ \sin(36^{\circ}) = \sqrt{1 - \cos^2(36^{\circ})} = \sqrt{1 - k^2} \] Using the sine addition formula, we can find \( \sin(13^{\circ}) \) in terms of \( k \) but it requires additional calculations. For now, we will leave it as \( \sin(13^{\circ}) \). ### c) \( \sin 36^{\circ} \) Using the Pythagorean identity: \[ \sin^2(x) + \cos^2(x) = 1 \] We can express \( \sin(36^{\circ}) \) as: \[ \sin(36^{\circ}) = \sqrt{1 - \cos^2(36^{\circ})} = \sqrt{1 - k^2} \] ### d) \( \sin 72^{\circ} \) Using the co-function identity: \[ \sin(90^{\circ} - x) = \cos(x) \] Thus, \[ \sin(72^{\circ}) = \cos(18^{\circ}) \] To express \( \cos(18^{\circ}) \) in terms of \( k \), we can use the double angle formula: \[ \cos(2x) = 2\cos^2(x) - 1 \] Let \( x = 36^{\circ} \): \[ \cos(72^{\circ}) = 2\cos^2(36^{\circ}) - 1 = 2k^2 - 1 \] Thus, \[ \sin(72^{\circ}) = 2k^2 - 1 \] ### e) \( \tan 36^{\circ} \) Using the definition of tangent: \[ \tan(x) = \frac{\sin(x)}{\cos(x)} \] Thus, \[ \tan(36^{\circ}) = \frac{\sin(36^{\circ})}{\cos(36^{\circ})} = \frac{\sqrt{1 - k^2}}{k} \] ### Summary of Results - a) \( \cos(-36^{\circ} ) = k \) - b) \( \cos(77^{\circ}) = \sin(13^{\circ}) \) (not expressed in terms of \( k \) directly) - c) \( \sin(36^{\circ}) = \sqrt{1 - k^2} \) - d) \( \sin(72^{\circ}) = 2k^2 - 1 \) - e) \( \tan(36^{\circ}) = \frac{\sqrt{1 - k^2}}{k} \)