Odpowiedź:
a) cos x = - [tex]\frac{1}{4}[/tex] więc z "jedynki" trygonometrycznej mamy
[tex]sin^2 x = 1 - cos^2 x = 1 - (\frac{-1}{4} )^2 = \frac{15}{16}[/tex]
[tex]sin x = \frac{\sqrt{15} }{\sqrt{16} } = \frac{\sqrt{15} }{4}[/tex] bo x ∈ ( [tex]\frac{\pi }{2}[/tex] , π) II ćwiartka
tg x = sin x : cos x = [tex]\frac{\sqrt{15} }{4} : \frac{-1}{4} = \frac{\sqrt{15} }{4} *( -4) = - \sqrt{15}[/tex]
c)
sin x = [tex]\frac{2}{3}[/tex] i x ∈ ( [tex]\frac{\pi }{2} , \pi )[/tex]
Mamy [tex]cos^2 x = 1 - sin^2 x = 1 - (\frac{2}{3})[/tex]² = [tex]\frac{9}{9} - \frac{4}{9} = \frac{5}{9}[/tex]
więc cos x = - [tex]\sqrt{\frac{5}{9} }[/tex] = - [tex]\frac{\sqrt{5} }{3}[/tex] bo w II ćwiartce cosinus jest ujemny
tg x = sin x : cos x = [tex]\frac{2}{3} : ( - \frac{\sqrt{5} }{3}[/tex] ) = - [tex]\frac{2}{3} *\frac{3}{\sqrt{5} }[/tex] = - [tex]\frac{2}{\sqrt{5} }[/tex] = - [tex]\frac{2\sqrt{5} }{5}[/tex]
Szczegółowe wyjaśnienie:
