Răspuns:
Explicație pas cu pas:
La d) și e) sunt greșeli de tipar. Corect este cos 160° și tg 140°
[tex]d) \ \cos 20^{\circ} + \cos 40^{\circ} + \cos 140^{\circ} + \cos 160^{\circ} + \cos 180^{\circ} = \cos 20^{\circ} + \cos 40^{\circ} + \cos(\pi - 40^{\circ}) + \cos(\pi - 20^{\circ}) - 1 = \cos 20^{\circ} + \cos 40^{\circ} - \cos 40^{\circ} - \cos 20^{\circ} - 1 = -1[/tex]
[tex]e) \ tg 20^{\circ} + tg 40^{\circ} + tg 140^{\circ} + tg 160^{\circ} + tg 180^{\circ} = tg 20^{\circ} + tg 40^{\circ} + tg (\pi - 40^{\circ}) + tg (\pi - 20^{\circ}) + 0 = tg 20^{\circ} + tg 40^{\circ} - tg 40^{\circ} - tg 20^{\circ} = 0[/tex]
[tex]a) \ tg\bigg(\dfrac{\pi}{2} + \alpha\bigg) = \dfrac{\sin\bigg(\dfrac{\pi}{2} + \alpha\bigg)}{\cos\bigg(\dfrac{\pi}{2} + \alpha\bigg)} = \dfrac{\cos \alpha}{-\sin \alpha}[/tex]
[tex]ctg \ (2\pi - \alpha) = - ctg \ \alpha = - \dfrac{\cos \alpha}{\sin \alpha}[/tex]
[tex]= \dfrac{1 + \dfrac{\cos \alpha}{\sin \alpha}}{1 - \dfrac{\cos \alpha}{\sin \alpha}} = \dfrac{\sin \alpha + \cos \alpha}{\sin \alpha - \cos \alpha} \ \ \ (1)[/tex]
[tex]tg \ (\pi + \alpha) = tg \ \alpha = \dfrac{\sin \alpha}{\cos \alpha}[/tex]
[tex]ctg \ \bigg(\dfrac{3\pi}{2} - \alpha\bigg) = ctg \ \bigg(\pi + \dfrac{\pi}{2} - \alpha\bigg) = ctg \ \bigg(\dfrac{\pi}{2} - \alpha\bigg) = \dfrac{\cos\bigg(\dfrac{\pi}{2} - \alpha\bigg)}{\sin\bigg(\dfrac{\pi}{2} - \alpha\bigg)} = \dfrac{\sin \alpha}{\cos \alpha}[/tex]
[tex]= \dfrac{\dfrac{\sin \alpha}{\cos \alpha} + 1}{\dfrac{\sin \alpha}{\cos \alpha} - 1} = \dfrac{\sin \alpha + \cos \alpha}{\sin \alpha - \cos \alpha} \ \ \ (2)[/tex]
Din (1) și (2) ⇒ egalitate demonstrată
[tex]b) \ = \dfrac{tg(\pi - \alpha) \cdot \dfrac{1}{tg(\pi - \alpha)} \cdot \cos \alpha}{\cos \alpha \cdot \dfrac{1}{tg\bigg(\dfrac{\pi}{2} + \alpha\bigg)} \cdot tg\bigg(\dfrac{\pi}{2} + \alpha\bigg)} = 1[/tex]
Formule:
[tex]\cos (\pi - \alpha) = - \cos \alpha; \ \ tg (\pi - \alpha) = - tg \ \alpha[/tex]
[tex]\sin\bigg(\dfrac{\pi}{2} \pm \alpha\bigg) = \cos \alpha; \ \ \cos\bigg(\dfrac{\pi}{2} \mp \alpha\bigg) = \pm \sin \alpha[/tex]
[tex]tg \ (\pi + \alpha) = tg \ \alpha; \ \ ctg \ (2\pi - \alpha) = - ctg \ \alpha[/tex]
[tex]tg \ \alpha = \dfrac{\sin \alpha}{\cos \alpha}; \ \ tg \ \alpha = \dfrac{1}{ctg \ \alpha}[/tex]
