Question

Quais os valores das cotangentes dos ângulos de 150,390,570 e 1050 respectivamente:

Answer 1

$\mathrm{cotg\,}\alpha=\dfrac{\cos \alpha}{\mathrm{sen\,}\alpha}$


a) $\mathrm{cotg\,}150^{\circ}$

$=\dfrac{\cos 150^{\circ}}{\mathrm{sen\,}150^{\circ}}\\ \\ =\dfrac{-\cos\left(180^{\circ}-150^{\circ} \right )}{\mathrm{sen}\left(180^{\circ}-150^{\circ} \right )}\\ \\ =-\dfrac{\cos 30^{\circ}}{\mathrm{sen\,}30^{\circ}}\\ \\ =-\mathrm{cotg\,}30^{\circ}\\ \\ =-\sqrt{3}$


b) $\mathrm{cotg\,}390^{\circ}$

$=\dfrac{\cos 390^{\circ}}{\mathrm{sen\,}390^{\circ}}\\ \\ =\dfrac{\cos\left(390^{\circ}-360^{\circ} \right )}{\mathrm{sen}\left(390^{\circ}-360^{\circ} \right )}\\ \\ =\dfrac{\cos 30^{\circ}}{\mathrm{sen\,}30^{\circ}}\\ \\ =\mathrm{cotg\,}30^{\circ}\\ \\ =\sqrt{3}$


c) $\mathrm{cotg\,}570^{\circ}$

$=\dfrac{\cos 570^{\circ}}{\mathrm{sen\,}570^{\circ}}\\ \\ =\dfrac{\cos\left(570^{\circ}-360^{\circ} \right )}{\mathrm{sen}\left(570^{\circ}-360^{\circ} \right )}\\ \\ =\dfrac{\cos 210^{\circ}}{\mathrm{sen\,}210^{\circ}}\\ \\ =\dfrac{-\cos \left(210^{\circ}-180^{\circ}\right)}{-\mathrm{sen\,}\left(210^{\circ}-180^{\circ}\right)}\\ \\ =\dfrac{-\cos 30^{\circ}}{-\mathrm{sen\,}30^{\circ}}\\ \\ =\mathrm{cotg\,}30^{\circ}\\ \\ =\sqrt{3}$


d) $\mathrm{cotg\,}1\,050^{\circ}$

$=\dfrac{\cos 1\,050^{\circ}}{\mathrm{sen\,}1\,050^{\circ}}\\ \\ =\dfrac{\cos\left(-30^{\circ}+3\times360^{\circ} \right )}{\mathrm{sen}\left(-30+3\times360^{\circ} \right )}\\ \\ =\dfrac{\cos \left(-30^{\circ} \right )}{\mathrm{sen\,}\left(-30^{\circ} \right )}\\ \\ =\dfrac{\cos 30^{\circ}}{-\mathrm{sen\,}30^{\circ}}\\ \\ =-\mathrm{cotg\,}30^{\circ}\\ \\ =-\sqrt{3}$