First notice that the Theta angles are specified in radians, not degrees. One π (Pi) radians is really 180o and 2π radians is the full circle or 360o.
Imagine the acute angle Θ1 in quadrant I, drop a perpendicular to the x-axis through x=1 and you have a right triangle whose height is sin(Θ1) = k and whose base is 1. Now rotate this triangle about the y-axis into quadrant II. None of the angles or the lengths inside that triangle have changed. Θ2 is now the angle measured from the +x-axis to the hypotenuse of that rotated triangle. So Θ2 = ( π - Θ1) radians. And sin(Θ2) = sin(Θ1) = k by construction.
To prove this analytically, start with the known relationship:
sin (x - y) = sin (x) cos (y) - cos(x) sin(y), substitute in x = π and y = +Θ1
sin (π - Θ1) = sin (π) cos (Θ1) - cos(π) sin(Θ1), and note that sin (π) = 0 and cos(π) = -1 in quadrant II
Thus, sin (π - Θ1) = - (-1) sin(Θ1) = sin(Θ1) = k
For the second part, rotate the quadrant I triangle about the x-axis into Quadrant IV.
Now Θ2 = ( 2π - Θ1) because its the angle that starts at the +x-axis, goes through quadrants I, II, III into IV and ends Θ1 radians shy of the x-axis or 2π. And cos (Θ1) = 1 by construction in quadrants I & IV.
To prove this analytically, start with another known relationship:
cos (x - y) = cos (x) cos (y) - sin (x) sin (y), substitute in x = 2π and y = +Θ1
cos (2π - Θ1) = cos (2π) cos (Θ1) - sin (2π) sin(Θ1), and note that sin (2π) = 0 and cos(2π) = +1 in quadrant IV
Thus, cos (2π - Θ1) = (+1) cos (Θ1) = cos (Θ1) = 1