Given a positive angle theta in standard position with the terminal side passing through (1, -3). How do you find the six trigonometric functions of theta?

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Answer 1 · 2016-11-07T16:51:53

Please see the explanation for steps leading to all 6 functions.

Given: #x = 1, and y = -3#

#r = sqrt(x^2 + y^2)#

#r = sqrt(1^2 + (-3)^2)#

#r = sqrt(10)#

#x = rcos(theta) and y = rsin(theta)#

#1 = sqrt(10)cos(theta)#

#cos(theta) = 1/sqrt(10) = sqrt(10)/10#

#-3 = sqrt(10)sin(theta)#

#sin(theta) = -3/sqrt(10) = -3sqrt(10)/10#

#tan(theta) = sin(theta)/cos(theta) = (-3/sqrt(10))/(1/sqrt(10)) = -3#

#cot(theta) = 1/tan(theta) = -1/3#

#sec(theta) = 1/cos(theta) = sqrt(10)#

#csc(theta) = 1/sin(theta) = -sqrt(10)/3#