How do you determine #costheta# given #tantheta=2/3, pi<theta<(3pi)/2#?

1 Answer
Dec 17, 2016

Answer:

Please see the explanation.

Explanation:

Start with the identity #cos^2(theta) + sin^2(theta) = 1#

Divide both sides by #cos^2(theta)#:

#cos^2(theta)/cos^2(theta) + sin^2(theta)/cos^2(theta) = 1/cos^2(theta)#

Use the identity #sin(theta)/cos(theta) = tan(theta)# and #cos^2(theta)/cos^2(theta) = 1#:

#1 + tan^2(theta) = 1/cos^2(theta)#

Solve for #cos(theta)#:

#cos^2(theta) = 1/(1 + tan^2(theta))#

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

The domain restriction #pi < theta < (3pi)/2# tells us that #theta# is in the third quadrant (where the cosine function is negative), therefore, we change the #+-# to - only:

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

Substitute #(2/3)^2# for #tan^2(theta)#

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

#cos(theta) = -sqrt(1/(1 + 4/9))#

#cos(theta) = -sqrt(1/(13/9))#

#cos(theta) = -sqrt(9/13)#

#cos(theta) = -(3sqrt(13))/13#