Question

poate cineva sa ma ajute? va rog!
Se considera x e ( 3pi\2 , 2pi ) , astfel incat :
tgx=-1\2 . sa se afle sinx , cosx, ctgx

Answer 1

S-a exprimat sin,cos si ctg in functi de tgx, si cadranul IV unde e x

Answer 2

$x \in (\frac{3 \pi}{2};2 \pi) \\\\ tgx= -\frac{1}{2}\\\\\\ tgx=\frac{sinx}{cosx} \ \ |^2 \\\\ tg^2x= \frac{sin^2 \ x}{cos^2 \ x} \\\\ (-\frac{1}{2})^2= \frac{sin^2 \ x}{1-sin^2 \ x} \\\\ sin ^2 \ x \to t \\\\\\ \frac{1}{4}=\frac{t}{1-t} \\\\ 1-t=4t \\\\ 1=5t \\\\ t=\frac{1}{5} \\\\ sin^2 \ x= \frac{1}{5} \\\\ sinx= \sqrt{\frac{1}{5}}= \frac{1}{\sqrt{5}}= \frac{\sqrt{5}}{5}\\\\\\cosx=\sqrt{1-sin^2 \ x}\\\\cosx=\sqrt{1-(\frac{\sqrt{5}}{5})^2}= \sqrt{1-\frac{\not 5}{\not 25}}= \sqrt{1^{(5}-\frac{1}{5}} \\\\ cosx= \sqrt{\frac{5-1}{5}}=\sqrt{\frac{4}{5}}= \frac{2\sqrt{5}}{5}$


$ctgx= \frac{1}{tgx}= \frac{1}{-\frac{1}{2}}\to -2$