[tex]\it x\in\left(\dfrac{3\pi}{2},\ 2\pi\right)\Rightarrow sinx<0,\ \ cosx>0.\\ \\ \\ cosx=\sqrt{1-sin^2x}=\sqrt{1-\left(-\dfrac{3}{5}\right)^2}=\sqrt{1-\dfrac{9}{25}}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}\\ \\ \\ tgx=\dfrac{sinx}{cosx}=\dfrac{-\dfrac{3}{5}}{\dfrac{4}{5}}=-\dfrac{3}{4}[/tex]
Cadranul 4, sin e negativ si cos pozitiv, tg este negativa.
cos x = rad(1 - (sin x)^2 ) = rad (1-9/25) = rad((25-9)/25) = rad(16/25) = 4/5
tg x = sin x / cos x = (-3/5) / (4/5) = -3/4
