Question

Calcule o valor de cada uma das expressões: 
A) sen 135º + 4 cotg 120º
B) sen 240º - cos 570º
C) sen 330º - cos 2460º

Answer 1

$\text{sen}~135^{\circ}=\text{sen}~(180^{\circ}-135^{\circ})=\text{sen}^45^{\circ}=\dfrac{\sqrt{2}}{2}$.

$\text{tg}~120^{\circ}=-\text{tg}~(180^{\circ}-120^{\circ})=-\text{tg}~60^{\circ}=-\sqrt{3}$

$\text{cotg}~120^{\circ}=-\dfrac{1}{\sqrt{3}}=-\dfrac{\sqrt{3}}{3}$

$4\cdot\text{cotg}~120^{\circ}=4\cdot\left(-\dfrac{\sqrt{3}}{3}\right)=-\dfrac{4\sqrt{3}}{3}$

Assim:

$\text{sen}~135^{\circ}+4\cdot\text{cotg}~120^{\circ}=\dfrac{\sqrt{2}}{2}-\dfrac{4\sqrt{3}}{3}=\dfrac{3\sqrt{2}-8\sqrt{3}}{6}$

b) $\text{sen}~240^{\circ}=-\text{sen}~(240^{\circ}-180^{\circ})=-\text{sen}~60^{\circ}=-\dfrac{\sqrt{3}}{2}$

$\text{cos}~570^{\circ}=\text{cos}~(570^{\circ}-360^{\circ})=\text{cos}~210^{\circ}$

$\text{cos}~570^{\circ}=\text{cos}~(210^{\circ}-180^{\circ})=-\text{sen}~30^{\circ}=-\dfrac{\sqrt{3}}{2}$

Deste modo:

$\text{sen}~240^{\circ}-\text{cos}~570^{\circ}=-\dfrac{\sqrt{3}}{2}-\left(-\dfrac{\sqrt{3}}{2}\right)=-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}=0$


c) $\text{sen}~330^{\circ}=-\text{sen}~(360^{\circ}-330^{\circ})=-\text{sen}~30^{\circ}=-\dfrac{1}{2}$

$360\times6=2~160$

$\text{cos}~2~460^{\circ}=\text{cos}~(2~460^{\circ}-2~160^{\circ})=\text{cos}~300^{\circ}$

$\text{cos}~2~460^{\circ}=\text{cos}~300^{\circ}=\text{cos}~(360^{\circ}-300^{\circ})=\text{cos}~60^{\circ}=\dfrac{1}{2}$

Logo:

$\text{sen}~330^{\circ}-\text{cos}~2~460^{\circ}=-\dfrac{1}{2}-\left(\dfrac{1}{2}\right)=-\dfrac{1}{2}-\dfrac{1}{2}=-1$