Question

Se x=arc sen( 12/13 ) e y=arc cos( 3/5 ). podemos dizer que sen (x - y) é igual a:

Answer 1

$\boxed{\begin{array}{l}\sf x=arc~sen\bigg(\dfrac{12}{13}\bigg)\iff sen(x)=\dfrac{12}{13}\\\sf y=arc~cos\bigg(\dfrac{3}{5}\bigg)\iff cos(y)=\dfrac{3}{5}\\\underline{\rm C\acute alculo~de~cos(x):}\\\sf sen^2(x)=\dfrac{144}{169}\\\\\sf cos^2(x)=\dfrac{169}{169}-\dfrac{144}{169}=\dfrac{25}{169}\\\\\sf cos(x)=\sqrt{\dfrac{25}{169}}=\dfrac{5}{13}\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm C\acute alculo~do~sen(y):}\\\sf cos^2(y)=\dfrac{9}{25}\\\\\sf sen^2(y)=\dfrac{25}{25}-\dfrac{9}{25}=\dfrac{16}{25}\\\\\sf sen(y)=\sqrt{\dfrac{16}{25}}\\\\\sf sen(y)=\dfrac{4}{5}\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm C\acute alculo~do~sen( x-y):}\\\sf sen(x-y)=sen(x)\cdot cos(y)-sen(y)\cdot cos(x)\\\sf sen(x-y)=\dfrac{12}{13}\cdot\dfrac{3}{5}-\dfrac{4}{5}\cdot\dfrac{5}{13}\\\\\sf sen(x-y)=\dfrac{36}{65}-\dfrac{20}{65}\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf sen(x-y)=\dfrac{16}{65}}}}}\end{array}}$