Let (3,-2) be a point on the terminal side of . Find the exact values of cosθ , secθ , and cotθ.

Answer 1

cosT= 3/(sqr13)= (3sqr13)/13

secT=1/cosT=(sqr13)/3

cotT=-3/2

plot the point (3,-2) =(x,y) = (a,o) a=adjacent=x, o=opposite=y

it's in quadrant IV

use the pythagorean theorem h^2 =a^2+o^2

draw a diagram with a right triangle, hypotenuse = h = square root of (-2)^2+3^2 =sqr(4+9)= sqr13

adjacent side = a = 3 opposite side = o = -2

cosT = a/h =x/h= 3/sqr13, but rationalize the denominator to get (3sqr13)/13

secT = h/a =h/x= (sqr13)/3

cotT = a/o =x/y= 3/-2 = -3/2