sinx = [tex] \frac{2}{3} [/tex]
se cunoastre formula: sin²x+cos²x=1
cos x = √(1-sin²x) = √1-4/9=√(9-4)/9=√(5/9)= + - [tex]\frac{ \sqrt{5} }{3}[/tex]
dar x∈ Cadranului 2 => sin >0
cos<0
Deci cos x = -[tex]\frac{ \sqrt{5} }{3}[/tex]
tgx = [tex] \frac{sinx}{cosx} = \frac{2}{3} * \frac{3}{ \sqrt{5} } = \frac{2}{ \sqrt{5} } [/tex]
ctgx=[tex] \frac{cosx}{sinx} = \frac{1}{tgx} = \frac{ \sqrt{5} }{2} [/tex]
