Question

sendo a=5 sobre 8 e 0 <a < $\pi$ sobre 2, calcule sen 2a e cos 2a

Answer 1

Pela relação fundamental, temos $\text{sen}^2~a+\text{cos}^2~a=1$.

Como $a=\dfrac{5}{8}$, segue que, $\left(\dfrac{5}{8}\right)^2+\text{cos}^2~a=1$.

Assim, $\text{cos}^2~a=1-\dfrac{25}{64}$

Temos $1-\dfrac{25}{64}=\dfrac{64-25}{64}=\dfrac{39}{64}$.

Portanto, $\text{cos}~a=\dfrac{\sqrt{39}}{8}$

Temos que:

$\text{sen}(2a)=2\cdot\text{sen}~a\cdot\text{cos}~a$

Portanto, $\text{sen}(2a)=2\cdot\left(\dfrac{5}{8}\right)\cdot\dfrac{\sqrt{39}}{8}$

$\text{sen}~(2a)=\dfrac{10\sqrt{39}}{64}=\dfrac{5\sqrt{39}}{32}$.

Analogamente, temos que:

$\text{cos}(2a)=\text{cos}^2~a-\text{sen}^2~a$

$\text{cos}(2a)=\left(\dfrac{\sqrt{39}}{8}\right)^2-\left(\dfrac{5}{8}\right)^2$

$\text{cos}(2a)=\dfrac{39}{64}-\dfrac{25}{64}=\dfrac{14}{64}$

$\text{cos}(2a)=\dfrac{7}{32}$.