[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad \qquad
cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\
-------------------------------\\\\
\textit{the point on the graph is }\left( \stackrel{cosine}{-\cfrac{8}{17}}~,~\stackrel{sine}{\cfrac{15}{17}} \right)[/tex]
now, keep in mind that, the hypotenuse is just the radius unit, so is never negative, so if the cosine is negative, that simply means it has to be the numerator, the 8 is negative, -8. that means
[tex]\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-8}}{\stackrel{hypotenuse}{17}}\qquad \qquad sin(\theta )=\cfrac{\stackrel{opposite}{15}}{\stackrel{hypotenuse}{17}}[/tex]
so, the adjacent is -8, the opposite side is 15, and the hypotenuse is 17, so just plug them in then
[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\quad
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\\\\\\
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\quad
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}[/tex]
