1.Aflăm cât este cosx întâi de toate.
[tex] cos^{2} x+sin ^{2} x=1 =\ \textgreater \ cos^{2} x=1-sin ^{2} x =\ \textgreater \ cosx= \sqrt{1- sin^{2}x } [/tex] [tex]cosx= \sqrt{1- \frac{9}{25} } = \sqrt{ \frac{16}{25} } = \frac{4}{5} [/tex]
[tex]tgx= \frac{sinx}{cosx} = \frac{ \frac{3}{5} }{ \frac{4}{5} } = \frac{3}{4} [/tex]
[tex]ctgx= \frac{cosx}{sinx} = \frac{ \frac{4}{5} }{ \frac{3}{5} } = \frac{4}{3} [/tex]
2. sin2x=2sinxcosx
aflam sinx
[tex]sinx= \sqrt{1- \frac{144}{169} } = \sqrt{ \frac{169-144}{169} } = \sqrt{ \frac{25}{169} } = \frac{5}{13} [/tex]
[tex]sin2x=2sinxcosx=2* \frac{5}{13} * \frac{-12}{13} = \frac{-120}{169} [/tex]
