5. Without using a calculator: a) Evaluate \( \sin ^{2} 40^{\circ}+\cos ^{2} 40^{\circ} \) b) Write \( \cos 80^{\circ} \) in terms of \( \cos 40^{\circ} \) and \( \sin 40^{\circ} \)

Let's solve each part step by step. ### Part a: Evaluate \( \sin^{2} 40^{\circ} + \cos^{2} 40^{\circ} \) We can use the Pythagorean identity for sine and cosine, which states that: \[ \sin^{2} \theta + \cos^{2} \theta = 1 \] In this case, we have \( \theta = 40^{\circ} \). Therefore, we can directly apply the identity: \[ \sin^{2} 40^{\circ} + \cos^{2} 40^{\circ} = 1 \] So, the answer for part a is: \[ \sin^{2} 40^{\circ} + \cos^{2} 40^{\circ} = 1 \] ### Part b: Write \( \cos 80^{\circ} \) in terms of \( \cos 40^{\circ} \) and \( \sin 40^{\circ} \) To express \( \cos 80^{\circ} \) in terms of \( \cos 40^{\circ} \) and \( \sin 40^{\circ} \), we can use the cosine of a double angle formula: \[ \cos(2\theta) = \cos^{2} \theta - \sin^{2} \theta \] Here, we can let \( \theta = 40^{\circ} \). Thus, we have: \[ \cos 80^{\circ} = \cos(2 \times 40^{\circ}) = \cos^{2} 40^{\circ} - \sin^{2} 40^{\circ} \] We can also use the identity \( \cos^{2} \theta = 1 - \sin^{2} \theta \) to express \( \cos^{2} 40^{\circ} \): \[ \cos^{2} 40^{\circ} = 1 - \sin^{2} 40^{\circ} \] Substituting this into the equation gives: \[ \cos 80^{\circ} = (1 - \sin^{2} 40^{\circ}) - \sin^{2} 40^{\circ} \] This simplifies to: \[ \cos 80^{\circ} = 1 - 2\sin^{2} 40^{\circ} \] Alternatively, we can express it in terms of \( \cos 40^{\circ} \): Using the identity \( \sin^{2} \theta = 1 - \cos^{2} \theta \): \[ \cos 80^{\circ} = \cos^{2} 40^{\circ} - (1 - \cos^{2} 40^{\circ}) = 2\cos^{2} 40^{\circ} - 1 \] Thus, we can express \( \cos 80^{\circ} \) in terms of \( \cos 40^{\circ} \) and \( \sin 40^{\circ} \) as: \[ \cos 80^{\circ} = 2\cos^{2} 40^{\circ} - 1 \] In summary, the answers are: - a) \( \sin^{2} 40^{\circ} + \cos^{2} 40^{\circ} = 1 \) - b) \( \cos 80^{\circ} = 2\cos^{2} 40^{\circ} - 1 \) or \( \cos 80^{\circ} = 1 - 2\sin^{2} 40^{\circ} \)