How do you find the exact value of the five remaining trigonometric function given #costheta=3/5# and the angle is in standard position in quadrant IV?

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Answer 1 · 2016-11-10T18:06:11

Please see the explanation

Given: #cos(theta) = 3/5# and #theta# is in quadrant IV

#sin(theta) = +-sqrt(1 - cos^2(theta))#

#sin(theta) = +-sqrt(1 - (3/5)^2)#

#sin(theta) = +-sqrt(1 - 9/25)#

#sin(theta) = +-sqrt(16/25)#

#sin(theta) = +-4/5#

The sine function is negative in quadrant IV, therefore, we change the #+-# to only #-#:

#sin(theta) = -4/5#

#tan(theta) = sin(theta)/cos(theta)#

#tan(theta) = (-4/5)/(3/5)#

#tan(theta) = -4/(3)#

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

#cot(theta) = -3/4#

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

#sec(theta) = 5/3#

#csc(theta) = 1/sin(theta)#

#csc(theta) = -5/4#