find the value of the 6 trig functions of an angle in standard position whose terminal ray passes through the given point:(-2,-2)

Printable View

- Jul 4th 2008, 11:46 AMDave1234trigonometry help
find the value of the 6 trig functions of an angle in standard position whose terminal ray passes through the given point:(-2,-2)

- Jul 4th 2008, 12:05 PMJhevon
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 - Jul 4th 2008, 12:16 PMChop Suey
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. :p - Jul 4th 2008, 12:16 PMDave1234i know that but for some reason my answers aren't working out with the diagram.
so can you try and show me the 6 trigs on your answer and then i will see where i went wrong does r=2root2

- Jul 4th 2008, 12:50 PMChop Suey
I've just told you how to do it. And yes, you found r correctly.