How do you find the exact value of the five remaining trigonometric function given #sintheta=1/3# and the angle is in standard position in quadrant II?

1 Answer
NickTheTurtle
Apr 1, 2017

#sin(theta)=1/3#
#cos(theta)=-(2sqrt(2))/3#
#tan(theta)=-sqrt(2)/4#
#csc(theta)=3#
#sec(theta)=-(3sqrt(2))/4#
#cot(theta)=-2sqrt(2)#

Explanation:

First, #csc(theta)=1/sin(theta)=1/(1/3)=3#.

We know that #sin^2(theta)+cos^2(theta)=1#, or #cos(theta)=+-sqrt(1-sin^2(theta))=+-sqrt(1-(1/3)^2)=+-(2sqrt(2))/3#. Now, the angle is in quadrant 2. This means that its cosine value is negative. Thus, #cos(theta)=-(2sqrt(2))/3#.

Then, #sec(theta)=1/cos(theta)=1/(-(2sqrt(2))/3)=-3/(2sqrt(2))=-(3sqrt(2))/4#.

We know that #tan(theta)=sin(theta)/cos(theta)=(1/3)/(-(2sqrt(2))/3)=-1/(2sqrt(2))=-sqrt(2)/4#.

Then, #cot(theta)=1/tan(theta)=1/(-sqrt(2)/4)=-4/sqrt(2)=-2sqrt(2)#.

Related questions
See all questions in Relating Trigonometric Functions
Impact of this question
7813 views around the world
You can reuse this answer
Creative Commons License