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 \( \sin 40^{\circ} \) c) Evaluate \( \cos 80^{\circ}+2 \sin ^{2} 40^{\circ} \)

The first part of your question is a classic identity from trigonometry! No need for a calculator here: you can simply recall that \( \sin^{2} \theta + \cos^{2} \theta = 1 \) for any angle \( \theta \). Therefore, for \( 40^{\circ} \), we have \( \sin^{2} 40^{\circ} + \cos^{2} 40^{\circ} = 1 \). Next, to express \( \cos 80^{\circ} \) in terms of \( \cos 40^{\circ} \) and \( \sin 40^{\circ} \), you can use the double-angle formula: \( \cos 80^{\circ} = \cos(2 \times 40^{\circ}) = 2 \cos^{2} 40^{\circ} - 1 \). This showcases the elegant relationship between angles connected by doubling. Finally, when evaluating \( \cos 80^{\circ} + 2 \sin^{2} 40^{\circ} \), substitute \( \cos 80^{\circ} \) with \( 2 \cos^{2} 40^{\circ} - 1 \). After some clever simplification involving the identity \( \sin^{2} 40^{\circ} = 1 - \cos^{2} 40^{\circ} \), you’ll discover that the expression beautifully reduces to \( 1 \). Happy calculating!