# Given sin theta = 2/3 and pi/2<theta<pi how do you find the value of other 5 other trigonometric functions?

Oct 20, 2015

$\sin \left(\theta\right) = \frac{2}{3} \textcolor{w h i t e}{\text{XXXXXXX}} \csc \left(\theta\right) = \frac{3}{2}$
$\cos \left(\theta\right) = - \frac{\sqrt{5}}{3} \textcolor{w h i t e}{\text{XXXX}} \sec \left(\theta\right) = - \frac{3}{\sqrt{5}}$
$\tan \left(\theta\right) = - \frac{2}{\sqrt{5}} \textcolor{w h i t e}{\text{XXXX}} \cot \left(\theta\right) = - \frac{\sqrt{5}}{2}$

#### Explanation:

$\frac{\pi}{2} < \theta < \pi$
tells us that $\theta$ is in Quadrant II

$\sin \left(\theta\right) = \frac{2}{3}$
tells us that the side ratios are
$\textcolor{w h i t e}{\text{XXX}}$opposite side: $2$
$\textcolor{w h i t e}{\text{XXX}}$hypotenuse: $3$
and using the Pythagorean Theorem and the fact that we are in Q II
$\textcolor{w h i t e}{\text{XXX}}$adjacent side: $- \sqrt{5}$

By definition:
$\sin = \left(\text{opposite")/("hypotenuse")color(white)("XXXX")csc=("hypotenuse")/("opposite}\right)$

$\cos = \left(\text{adjacent")/("hypotenuse")color(white)("XXXX")sec=("hypotenuse")/("adjacent}\right)$

$\tan = \left(\text{opposite")/("adjacent")color(white)("XXXX")cot=("adjacent")/("opposite}\right)$