Question

Un exercitiu destul de usor de trigonometrie, dar am uitat tot. Am atasat poza

Answer 1

 

$\displaystyle\\ \cos x=\frac{5}{13}\\\\{\bf i})~x\in\left(0,~\frac{\pi}{2} \right)~~\text{Cadranul 1}\\\\{\bf ii})~x\in\left(\frac{3\pi}{2},~2\pi\right)~~\text{Cadranul 4}\\\\\text{Si pentru cadranul 1 si pentru cadranul 4 folosim aceleasi formule,}\\\text{dar difera semnul astfel:}\\\\\text{In cadranul 1 sin, cos, tg, ctg sunt pozitive.}\\\text{In cadranul 4 cos este pozitiv, iar sin, tg, ctg sunt negative.}$

$\displaystyle\\\text{Rezolvare in C1}\\\\\cos x=\frac{5}{13}\\\\{\bf i})~x\in\left(0,~\frac{\pi}{2} \right)\\\\\sin x=\pm\sqrt{1-\cos^2x}=+\sqrt{1-\left(\frac{5}{13}\right)^2}=\\\\=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{169-25}{169}}=\sqrt{\frac{144}{169}} =\boxed{\bf\frac{12}{13}}\\\\\\\text{tg }x=\frac{\sin x}{\cos x}=\frac{~\dfrac{12}{13}~}{~\dfrac{5}{13}~}=\frac{12}{13}\cdot\frac{13}{5}=\boxed{\bf\frac{12}{5}}\\\\\\\text{ctg }x=\frac{1}{\text{tg }x}=\frac{1}{\dfrac{12}{5}}=\boxed{\bf\frac{5}{12}}$

$\displaystyle\\\text{Rezolvare in C4}\\\\\cos x=\frac{5}{13}\\\\{\bf ii})~x\in\Big(\frac{3\pi}{2},~2\pi\Big)\\\\\sin x=\pm\sqrt{1-\cos^2x}=-\sqrt{1-\Big(\frac{5}{13}\Big)^2}=\\\\=-\sqrt{1-\frac{25}{169}}=-\sqrt{\frac{169-25}{169}}=-\sqrt{\frac{144}{169}} =\boxed{\bf-\frac{12}{13}}\\\\\text{tg }x=\frac{\sin x}{\cos x}=\frac{~-\dfrac{12}{13}~~~~}{~\dfrac{5}{13}~}=\boxed{\bf-\frac{12}{5}}\\\\\\\text{ctg }x=\frac{1}{\text{tg }x}=\frac{1}{-\dfrac{12}{5}}=\boxed{\bf-\frac{5}{12}}$