#theta=(3pi)/4#. This angle creates a reference triangle with leg #sqrt2#, leg #sqrt2#, and hypotenuse #2#. Since #(3pi)/4# is in QII #(pi/2<=(3pi)/4<=pi)#, the cosine function is negative. Therefore, #cos(theta)=-(sqrt2)/2#.
To find #cos^2(theta)#:
#cos^2(theta)=(cos(theta))^2#
#cos^2(theta)=(-(sqrt2)/2)^2#
#cos^2(theta)=1/2#
Because #-theta# occurs in quadrant III, #cos(-theta)# is negative, and since it is a 45degrees or #pi/4# angle, it has the same reference triangle, so #cos(-theta)=-(sqrt2)/2# as well.
#2(theta)=2((3pi)/4)#
#2(theta)=(3pi)/2#
Using the unit circle, we know the coordinates of a point on the circle are #(cos,sin)#, and #(3pi)/2# is located on the #y#-axis, #cos(2theta)=0#
