Question

repede ..................​

Answer 1

 

$\displaystyle\\a)~~~\text{tg}~x=-2,~~x\in\left(\frac{\pi}{2},~\pi\right)~~~\implies~\text{Cadranul 2}\\\\\text{Folosim formulele:}\\\\\sin x = \frac{\text{tg }x}{\pm\sqrt{1+\text{tg}^2x}}\\\\\\\cos x = \frac{1}{\pm\sqrt{1+\text{tg}^2x}}\\\\\\\text{ctg }x=\frac{1}{\text{tg }x}\\\\\\ \text{Rezolvare:}\\\\\sin x = \frac{\text{tg }x}{\pm\sqrt{1+\text{tg}^2x}}=\frac{-2}{-\sqrt{1+(-2)^2}}= \frac{-2}{-\sqrt{1+4}}=\frac{2}{\sqrt{5}}=\boxed{\frac{2\sqrt{5}}{5}}$

$\displaystyle\\\cos x = \frac{1}{\pm\sqrt{1+\text{tg}^2x}}=\frac{1}{-\sqrt{1+(-2)^2}}=\frac{1}{-\sqrt{1+4}}= -\frac{1}{\sqrt{5}}=\boxed{-\frac{\sqrt{5}}{5}}\\\\\\\text{ctg }x=\frac{1}{\text{tg }x}=\frac{1}{-2}=\boxed{-\frac{1}{2}}$

$\displaystyle\\b)~~~\sin x=0,6 ,~~x\in\left(0,~\frac{\pi}{2}\right)~~~\implies~\text{Cadranul 1}\\\\\text{Folosim formulele:}\\\\\cos x=\pm\sqrt{1-\sin^2x}\\\\\text{tg }x=\frac{\sin x}{\cos x}\\\\\text{ctg }x=\frac{\cos x}{\sin x}\\\\\\\text{Rezolvare:}\\\\\\\cos x=\pm\sqrt{1-\sin^2x}=+\sqrt{1-(0,6)^2}=\sqrt{1-0,36}=\boxed{0,64}\\\\\\\text{tg }x=\frac{\sin x}{\cos x}=\frac{0,6}{0,64}=\frac{60}{64}=\boxed{\frac{15}{16}}\\\\\\\text{ctg }x=\frac{\cos x}{\sin x}=\frac{0,64}{0,6}=\boxed{\frac{16}{15}}$