Use the information about the angle θ to find the exact value of?

Given: #cot(theta) = 3/4, pi < theta < (3pi)/2#

We need #cos(theta)# for the formulas of the half-angles :

#cos(theta) = cos(cot^-1(x))#

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

Substitute #x = 3/4#:

#cos(theta) = +-(3/4)/sqrt((3/4)^2+1)#

#cos(theta) = +-(3/4)/sqrt((3/4)^2+1)#

#cos(theta) = +-3/sqrt((3^2+4^2)#

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

The cosine function is negative in the third quadrant:

#cos(theta) = -3/5" [1]"#

For the half-angle formulas, please observe that:

#pi < theta < (3pi)/2 to pi/2 < theta/2 < (3pi)/4#

This means that sine and cosine of the half-angles are positive.

#sin(theta/2) = sqrt(1-cos(theta))/2#

Substitute equation [1]:

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

#sin(theta/2) = sqrt(10)/5#

#cos(theta/2) = sqrt(1+cos(theta))/2#

Substitute equation [1]:

#cos(theta/2) = sqrt(1-3/5)/2#

#cos(theta/2) = sqrt(10)/10#

#tan(theta/2) = sin(theta/2)/cos(theta/2)#

#tan(theta/2) = (sqrt(10)/5)/(sqrt(10)/10)#

#tan(theta/2) = 2#

#cot t = 3/4# --> #tan t = 4/3#
First find cos t.
#cos^2 t = 1/(1 +tan^2 t) = 1/(1 + 16/9) = 9/25#
#cos t = +- 3/5#.
Since t is in Quadrant 3, then, cos t is negative
#cos t = - 3/5#.
To find #sin (t/2)# and #cos (t/2)# use these identities:
#2cos^2 (t/2) = 1 + cos t#, and
#2sin^2 (t/2) = 1 - cos t#.
In this case:
a. #2sin^2 (t/2) = 1 - cos t = 1 + 3/5 = 8/5#
#sin^2 (t/2) = 8/10#
#sin (t/2) = +- 2sqrt2/sqrt10#
Because t lies in Q.3, then, #(t/2)# lies in Q.2, and #sin (t/2)# is positive
b. #2cos^2 (t/2) = 1 + cos t = 1 - 3/5 = 2/5#
#cos^2 (t/2) = 2/10#
#cos (t/2) = +- sqrt2/(sqrt10)#
Because #(t/2)# lies in Q. 2, therefor, #cos t/2# is negative.
#cos (t/2) = - sqrt2/(sqrt10)#
c. #tan (t/2) = sin/(cos) = ((2sqrt2)/(sqrt10))(sqrt10/-sqrt2)#.
#tan (t/2) = - 2#