find the value of the 6 trig functions of an angle in standard position whose terminal ray passes through the given point-2,-2)
did you draw the diagram? note that you can get a right triangle with these measurements.
recall that sine = opposite/hypotenuse, cosine = adjacent/hypotenuse, and tangent = opposite/adjacent. that's what SOHCAHTOA is all about.
then remember that secant = 1/cosine, cosecant = 1/sine and cotangent = 1/tangent
thus you can find all the trig ratios. note that you will need Pythagoras' theorem to solve for the missing side of the triangle
You got the coordinates (x,y) = (-2, -2). Apply Pythagorean Theorem and find the value of r.
$\displaystyle x^2 + y^2 = r^2$
Where x being the adjacent side, y being the opposite side, and r is the hypotenuse.
Now that you got the values of x, y, and r, simply replace it in the 6 trigonometric ratios.
$\displaystyle \sin{\theta} = \frac{y}{r}$
$\displaystyle \cos{\theta} = \frac{x}{r}$
$\displaystyle \tan{\theta} = \frac{y}{x}$
$\displaystyle \csc{\theta} = \frac{r}{y} $
$\displaystyle \sec{\theta} = \frac{r}{x}$
$\displaystyle \cot{\theta} = \frac{x}{y}$
EDIT: Jhev, I had the reply menu open and posted before I knew you already answered him.