How do you find the exact value of 6 trig function given a point #(-sqrt5, 2)#?

1 Answer
Nov 18, 2016

Answer:

Please see the explanation for steps leading to the six values.

Explanation:

Use the identity: #tan(theta) = y/x#:

#tan(theta) = 2/-sqrt(5)#

Rationalize the denominator:

#tan(theta) = (-2sqrt(5))/5#

Use the identity: #cot(theta) = 1/tan(theta)#

#cot(theta) = -sqrt(5)/2#

Use the identity #tan^2(theta) + 1 = sec^2(theta)#

#4/5 + 1 = sec^2(theta)#

#sec^2(theta) = 9/5#

#sec(theta) = +-3/sqrt(5)#

Rationalize the denominator:

#sec(theta) = +- (3sqrt(5))/5#

Please observe that the point #(-sqrt(5),2)# is in the second quadrant and the secant is negative in the second quadrant, therefore, we drop the + sign:

#sec(theta) = (-3sqrt(5))/5#

Use the identity #cos(theta) = 1/sec(theta)#:

#cos(theta) = -sqrt(5)/3#

Use the identity #tan(theta) = sin(theta)/cos(theta)#:

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

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

#sin(theta) = 2/3#

Use the identity: #csc(theta) = 1/sin(theta)#:

#csc(theta) = 3/2#