Question

(SAS 2014) - Questão 50

Answer 1

Oi Pedro.

Primeiro temos que saber que:

Pi=180°

$[cos(\frac { 7\pi }{ 6 } )-3*sen(\frac { \pi }{ 6 } +\pi )]*tg(\frac { 5\pi }{ 4 } )$

Fazendo a divisão acharemos:

$[cos210^{ o }-3sen210^{ o }]*tg225^{ o }$

Agora temos que saber que:

$cos210^{ o }=30^{ o }\\ sen210^{ o }=30^{ o }\\ tg225^{ o }=45^{ o }$


Lembrando que o cosseno e o seno no 3° quadrante são negativos e a tangente é positiva.

$[-\frac { \sqrt { 3 } }{ 2 } -3(*-\frac { 1 }{ 2 } )]*1\\ \\ -\frac { \sqrt { 3 } }{ 2 } +\frac { 3 }{ 2 } \\ \\ \frac { 3-\sqrt { 3 } }{ 2 }$

Answer 2

$h=\left[\text{cos}~\left(\dfrac{7\pi}{6}\right)-3\cdot\text{sen}~\left(\dfrac{\pi}{6}+\pi\right)\right]\cdot\text{tg}~\left(\dfrac{5\pi}{4}\right)$

Observe que:

$\text{cos}~\left(\dfrac{7\pi}{6}\right)=\text{cos}~210^{\circ}=-\text{cos}~(210^{\circ}-180^{\circ})=-\text{cos}~30^{\circ}=-\dfrac{\sqrt{3}}{2}$.

$\text{sen}~\left(\dfrac{\pi}{6}+\pi\right)=\text{sen}~\left(\dfrac{7\pi}{6}\right)=\text{sen}~210^{\circ}=-\text{sen}~(210^{\circ}-180^{\circ})=-\text{sen}~30^{\circ}=-\dfrac{1}{2}$

$\text{tg}~\left(\dfrac{5\pi}{4}\right)=\text{tg}~225^{\circ}=\text{tg}~(270^{\circ}-225^{\circ})=\text{tg}~45^{\circ}=1$.

Assim:

$h=\left[-\dfrac{\sqrt{3}}{2}-3\cdot\left(-\dfrac{1}{2}\right)\right]\cdot1=-\dfrac{\sqrt{3}}{2}+\dfrac{3}{2}=\dfrac{3-\sqrt{3}}{2}$.

Letra A