Question

qual valor de sen (arc cos 3/5 - arc cos 5/13)?

Answer 1

A primeira coisa que devemos fazer é transformar esse seno de diferença de arcos em uma expressão com argumentos únicos "a" e "b":

$sen(a-b)=sena.cosb-senb.cosa\\\\sen(arccos(\frac{3}{5})-arccos(\frac{5}{13}))=sen(arccos(\frac{3}{5})).cos(\arccos(\frac{5}{13}))-sen(\arccos(\frac{5}{13})).cos(arccos(\frac{3}{5})))\\\\=sen(arccos(\frac{3}{5})).\frac{5}{13}-sen(\arccos(\frac{5}{13})).\frac{3}{5}$

Sejam

$arccos(\frac{3}{5})=u,\:\:arccos(\frac{5}{13})=v$

$cos[arccos(\frac{3}{5})]=cos(u)\\\\\frac{3}{5}=cos(u)\\\\cos[arccos(\frac{5}{13})]=cos(v)\\\\\frac{5}{13}=cos(v)$

Desenhando dois triângulos retângulos, podemos observar que:

$5^2=3^2+x^2\\\\x=4\\\\sen(u)=\frac{4}{5}\\\\\\13^2=5^2+x^2\\\\x=12\\\\sen(v)=\frac{12}{13}$

Voltando à expressão original

$sen(arccos(\frac{3}{5})).\frac{5}{13}-sen(\arccos(\frac{5}{13})).\frac{3}{5}\\\\sen(u).\frac{5}{13}-sen(v).\frac{3}{5}\\\\\frac{4}{5}.\frac{5}{13}-\frac{12}{13}.\frac{3}{5}\\\\\frac{4}{13}-\frac{36}{65}$

$\frac{260-468}{845}\\\\\frac{-208}{845}\\\\\frac{-16}{65}$