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  • Usuário do Brainly
2014-08-15T20:00:45-03:00
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}.
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