How do you determine all trigonometric ratios given sinθ=2/3 ?

Nov 2, 2015

Find the six trig functions ratios, given $\sin x = \frac{2}{3}$

Explanation:

$\sin x = \frac{2}{3}$ --> $x 1 = {41}^{\circ} 81$ and $x 2 = 180 - 41.81 = {138}^{\circ} 19$
${\cos}^{2} x = 1 - {\sin}^{2} x = 1 - \frac{4}{9} = \frac{5}{9}$ --> $\cos x = \pm \frac{\sqrt{5}}{3}$
$\tan x 1 = \tan 41.81 = \left(\frac{2}{3}\right) \left(\frac{3}{\sqrt{5}}\right) = \frac{2}{\sqrt{5}} = \frac{2 \sqrt{5}}{5}$
$\tan x 2 = \tan 138.19 = - \frac{2 \sqrt{5}}{5}$
$\cot 41.81 = \frac{\sqrt{5}}{2}$
$\cot 138.19 = - \frac{\sqrt{5}}{2}$
$\sec 41.81 = \frac{3}{\sqrt{5}} = \frac{3 \sqrt{5}}{5}$
$\sec 138.19 = - \frac{3 \sqrt{5}}{5}$
$\csc 41.81 = \csc 138.19 = \frac{3}{2}$