given cos(theta)=-2/3 and theta has its terminal side in Quadrant 3, find sin(theta)

Question

jdoe0001 · Answer

$\bf cos(\theta) = -\cfrac{2}{3} \implies \cfrac{\textit{adjacent}}{\textit{hypotenuse}} \cfrac{a}{c}\
\textit{keep in mind that } c^2 = a^2+b^2 \implies c = \sqrt{a^2+b^2}$

jdoe0001 · Answer

so, what do you think is $\bf sin(\theta) \ \ ?$

jdoe0001 · Answer

well, I should have said 
$\bf cos(\theta) = -\cfrac{2}{3} \implies \cfrac{\textit{adjacent}}{\textit{hypotenuse}}\implies \cfrac{a}{c}\
\textit{keep in mind that } c^2 = a^2+b^2 \implies b = \sqrt{c^2+a^2}$

jdoe0001 · Answer

acck

jdoe0001 · Answer

$\bf \textit{keep in mind that} c^2 = a^2-b^2 \implies b = \sqrt{c^2+a^2}$

jdoe0001 · Answer

sorry for the typos

jdoe0001 · Answer

lemme correct myself

jdoe0001 · Answer

$\bf cos(\theta) = -\cfrac{2}{3} \implies \cfrac{\textit{adjacent}}{\textit{hypotenuse}}\implies \cfrac{a}{c}\
\textit{keep in mind that} c^2 = a^2+b^2 \implies b = \sqrt{c^2-a^2}$

jdoe0001 · Answer

geez, much better, anyhow
so what do you think will be $\bf sin(\theta)\ \ \ ?$

anonymous · Answer

i have no clue

jdoe0001 · Answer

well, what do you think is the component "a" and "c"?