Question

matematica; Xapartime[0 pi/2] sinx=2radical din2/3.sa se derermine;cosx,tgx.ctgx

Answer 1

 

$\displaystyle\bf\\x\in\left[0,~\frac{\pi}{2}\right]~~(Cadranul~1)\\\\sinx=\frac{2\sqrt{2}}{3}\\\\cosx=\sqrt{1-sin^2x}=\sqrt{1-\left(\frac{2\sqrt{2}}{3}\right)^2}=\sqrt{1-\frac{8}{9}}= \sqrt{\frac{9-8}{9}}=\sqrt{\frac{1}{9}}=\boxed{\bf\frac{1}{3}}\\\\\\tgx=\frac{sinx}{cosx}=\frac{~~\dfrac{2\sqrt{2}}{3}~~}{\dfrac{1}{3}}=\frac{2\sqrt{2}}{3}\cdot \frac{3}{1}=\boxed{\bf2\sqrt{2}}\\\\\\ctgx=\frac{1}{tgx}=\frac{1\cdot\sqrt{2}}{2\sqrt{2}\cdot\sqrt{2}}=\frac{\sqrt{2}}{2\cdot2}=\boxed{\bf\frac{\sqrt{2}}{4}}$

 

 

Answer 2

$\it cosx=\sqrt{1-sin^2x}=\sqrt{1-\Big(\dfrac{2\sqrt2}{3}\Big)^2}=\sqrt{1-\dfrac{8}{9}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\\ \\ \\ tgx=\dfrac{sinx}{cosx}=\dfrac{\dfrac{2\sqrt2}{3}}{\dfrac{1}{3}}=2\sqrt2\\ \\ \\ ctgx=\dfrac{1}{tgx}=\dfrac{^{\sqrt2)}1}{\ \ 2\sqrt2}=\dfrac{\sqrt2}{4}$