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