Question

Se o sen (x) = -12/13 e cos (y) = 3/5, calcule
sen(x – y), sabendo que x é um arco do 3º
quadrante e y é um arco do 1º quadrante.
a) -63/65
b) 63/65
c) -16/65
d) 16/65
e) -56/65

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{sen^2\:x + cos^2\:x = 1}$

$\mathsf{cos^2\:x = 1 - sen^2\:x}$

$\mathsf{cos^2\:x = 1 - \left(-\dfrac{12}{13}\right)^2}$

$\mathsf{cos^2\:x = \left(\dfrac{169 - 144}{169}\right)}$

$\mathsf{cos^2\:x = \left(\dfrac{25}{169}\right)}$

$\mathsf{cos\:x = \left(-\dfrac{5}{13}\right)}$

$\mathsf{sen^2\:y + cos^2\:y = 1}$

$\mathsf{sen^2\:y = 1 - cos^2\:y}$

$\mathsf{sen^2\:y = 1 - \left(\dfrac{3}{5}\right)^2}$

$\mathsf{sen^2\:y = \left(\dfrac{25 - 9}{25}\right)}$

$\mathsf{sen^2\:y = \left(\dfrac{16}{25}\right)}$

$\mathsf{sen\:y = \left(\dfrac{4}{5}\right)}$

$\mathsf{sen(x - y) = sen\:x.cos\:y - sen\:y.cos\:x}$

$\mathsf{sen(x - y) = \left(-\dfrac{12}{13}\right).\left(\dfrac{3}{5}\right) - \left(\dfrac{4}{5}\right).\left(-\dfrac{5}{13}\right)}$

$\mathsf{sen(x - y) = \left(-\dfrac{36}{65}\right) - \left(-\dfrac{20}{65}\right)}$

$\mathsf{sen(x - y) = \left(-\dfrac{36}{65}\right) + \left(\dfrac{20}{65}\right)}$

$\mathsf{sen(x - y) = \left(-\dfrac{16}{65}\right)}$