Question

sabe  que sen a=4/5 (quatro quintos) e sen b= 12/3 (doze terços), com 0<a<(pi/2) e 0<b<(pi/2) calcule:
a) sen (a+b)
b) cos (a-b)
c) cos (a+b)

Answer 1

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

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

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

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