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  • Usuário do Brainly
2014-08-22T14:40:36-03:00
Temos que:

\text{sen}^2~\alpha+\text{cos}^2~\alpha=1

Assim:

\left(\dfrac{4}{5}\right)^2+\text{cos}^2~a=1

\text{cos}~a=\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}.

Do mesmo modo:

\left(\dfrac{12}{13}\right)^2+\text{cos}^2~b=1

\text{cos}~b=\sqrt{1-\dfrac{144}{169}}=\sqrt{\dfrac{25}{169}}=\dfrac{5}{13}.

a) \text{sen}~(a+b)=\text{sen}~a\cdot\text{cos}~b+\text{sen}~b\cdot\text{cos}~a

\text{sen}~(a+b)=\dfrac{4}{5}\cdot\dfrac{5}{13}+\dfrac{12}{13}\cdot\dfrac{3}{5}

\text{sen}~(a+b)=\dfrac{4}{13}+\dfrac{36}{65}

\text{sen}~(a+b)=\dfrac{20+36}{65}=\dfrac{56}{65}

b) \text{cos}~(a-b)=\text{cos}~a\cdot\text{cos}~b+\text{sen}~a\cdot\text{sen}~b

\text{cos}~(a-b)=\dfrac{3}{5}\cdot\dfrac{5}{13}+\dfrac{4}{5}\cdot\dfrac{12}{13}

\text{cos}~(a-b)=\dfrac{3}{13}+\dfrac{48}{65}

\text{cos}~(a-b)=\dfrac{15+48}{65}=\dfrac{63}{65}

c) \text{cos}~(a+b)=\text{cos}~a\cdot\text{cos}~b-\text{sen}~a\cdot\text{sen}~b

\text{cos}~(a+b)=\dfrac{3}{5}\cdot\dfrac{5}{13}-\dfrac{4}{5}\cdot\dfrac{12}{13}

\text{cos}~(a+b)=\dfrac{3}{13}-\dfrac{48}{65}

\text{cos}~(a+b)=\dfrac{15-48}{65}=\dfrac{-33}{65}
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