Răspuns:
a este în cadranul II, b este în cadranul l
[tex]\sin^2 a + \cos^2 a = 1 \Rightarrow \cos^2 a = 1 - \dfrac{9}{25} = \dfrac{16}{25} \Rightarrow \cos a = \pm \sqrt{\dfrac{4^2}{5^2}} \Rightarrow \bf \cos a = -\dfrac{4}{5}[/tex]
(în cadranul II funcția cosinus ia valori negative)
[tex]\sin^2 b + \cos^2 b = 1 \Rightarrow \sin^2 b = 1 - \dfrac{25}{169} = \dfrac{144}{169} \Rightarrow \bf \sin b = \dfrac{12}{13}[/tex]
Formulele sunt:
[tex]tg \ a = \dfrac{\sin a}{\cos a} = - \dfrac{3}{5} \cdot \dfrac{5}{4} = -\dfrac{3}{4}[/tex]
[tex]ctg \ a = \dfrac{1}{tg \ a} = -\dfrac{4}{3}[/tex]
[tex]\sin (a + b) = \sin a \cos b + \cos a \sin b = \dfrac{3}{5} \cdot \dfrac{5}{13} - \dfrac{4}{5} \cdot \dfrac{12}{13} = -\dfrac{33}{65}[/tex]
[tex]\cos (a - b) = \cos a \cos b + \sin a \sin b = -\dfrac{4}{5} \cdot \dfrac{5}{13} + \dfrac{3}{5} \cdot \dfrac{12}{13} = \dfrac{16}{65}[/tex]
[tex]\sin 2a = 2\sin a \cos a = -2\dfrac{3}{5} \cdot \dfrac{4}{5} = - \dfrac{24}{25}[/tex]
[tex]\cos 2b = 1 - 2\sin^2 b = 1 - 2\cdot\dfrac{144}{169} = -\dfrac{119}{169}[/tex]
[tex]\cos 2a = 1 - 2\sin^2 a = 1 - 2\cdot\dfrac{9}{25} = \dfrac{7}{25}[/tex]
[tex]\sin 2b = 2\sin b \cos b = 2\dfrac{12}{13} \cdot \dfrac{5}{13} = \dfrac{120}{169}[/tex]
