Question

Sabe-se que sen x = 3/5 e sen y = 12/13 , X E 1,° quadrante. Calcule:

A) sen (x + y) =
B) cos (x + y) =​

Answer 1

Resposta:

Explicação passo-a-passo:

Sabemos pelo Identidade fundamental que

sen²x+cos²x=1

cos²x=1-sen²x

$cosx= \sqrt{1-sen^2x}$

cos x= $\sqrt{1-(\frac{3}{5})^2}\\\\cosx= \sqrt{1-\frac{9}{25}}\\\\cosx= \sqrt{\frac{25-9}{25}}\\\\cosx= \sqrt{\frac{16}{25}}\\cosx= \frac{4}{5}$

$cosy= \sqrt{1-(\frac{12}{13}})^2\\\\cosy= \sqrt{1-\frac{144}{169}}\\\\cosy= \sqrt{\frac{169-144}{169}}\\\\cosy= \sqrt{\frac{25}{169}}$$cosy= \frac{5}{13}$

sen(a+b)= sena*cosb+senb*cosa

sen(x+y) = senx*cosy+seny*cosx

=$\frac{3}{5}*\frac{5}{13}+\frac{12}{13}*\frac{4}{5}\\\\\frac{15}{65}+\frac{48}{65}= \frac{63}{65}$

A= $\frac{63}{65}$

cos (a+b)= cosa*cob- sena*senb

$cos(x+y)= \frac{4}{5}*\frac{5}{13}- \frac{3}{5}*\frac{12}{13}\\\\\frac{20}{65}- \frac{36}{65} = \frac{-16}{65}$

B)= $\frac{-16}{65}$