Question

* Determine o valor da seguinte expressão :

a) A= sen 2x . cos 3x / tg x /2 . cotg x     ,  sendo x = 60°

Answer 1

Temos que,

$A=\dfrac{\text{sen}~120^{\circ}\cdot\text{cos}~180^{\circ}}{\text{tg}~30^{\circ}\cdot\text{cotg}~60^{\circ}}$

$\text{sen}~60^{\circ}=\dfrac{\sqrt{3}}{2}$,$\text{cos}~60^{\circ}=\dfrac{1}{2}$ e $\text{tg}~60^{\circ}=\sqrt{3}$.

Deste modo, $\text{cotg}~60^{\circ}=\dfrac{1}{\text{tg}~60^{\circ}}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}$.

Além disso, $\text{tg}~\dfrac{60}{2}=\text{tg}~30^{\circ}=\dfrac{\sqrt{3}}{3}$.

Temos que, $\text{sen}~(2x)=2\cdot\text{sen}~x\cdot\text{cos}~x$.

Assim, $\text{sen}~120^{\circ}=2\cdot\text{sen}~60^{\circ}\cdot\text{cos}~60^{\circ}$

$\text{sen}~120^{\circ}=2\cdot\dfrac{\sqrt{3}}{2}\cdot\dfrac{1}{2}$

$\text{sen}~120^{\circ}=\dfrac{2\sqrt{3}}{4}$

$\text{sen}~120^{\circ}=\dfrac{\sqrt{3}}{2}$

Lembrando que, $\text{cos}~180^{\circ}=-1$, temos:

$A=\dfrac{\frac{\sqrt{3}}{2}\cdot(-1)}{\frac{\sqrt{3}}{3}\cdot\frac{\sqrt{3}}{3}}$

$A=\dfrac{-\frac{\sqrt{3}}{2}}{\frac{1}{3}}$

$A=\dfrac{-\sqrt{3}}{2}\cdot\dfrac{3}{1}=\dfrac{-3\sqrt{3}}{2}$