Russ P. answered • 10/25/14

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Edward, Θ

First notice that the Theta angles are specified in radians, not degrees. One π (Pi) radians is really 180

^{o}and 2π radians is the full circle or 360^{o}.Imagine the acute angle Θ

_{1}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. Θ_{2}is now the angle measured from the +x-axis to the hypotenuse of that rotated triangle. So Θ_{2}= ( π - Θ_{1}) radians. And sin(Θ_{2}) = sin(Θ_{1}) = 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 = +Θ

_{1}sin (π - Θ

_{1}) = sin (π) cos (Θ_{1}) - cos(π) sin(Θ_{1}), and note that sin (π) = 0 and cos(π) = -1 in quadrant IIThus, sin (π - Θ

_{1}) = - (-1) sin(Θ_{1}) = sin(Θ_{1}) = kFor the second part, rotate the quadrant I triangle about the x-axis into Quadrant IV.

Now Θ

_{2}= ( 2π - Θ_{1}) because its the angle that starts at the +x-axis, goes through quadrants I, II, III into IV and ends Θ_{1}radians shy of the x-axis or 2π. And cos (Θ_{1}) = 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 = +Θ

cos (2π - Θ

Thus, cos (2π - Θ1) = (+1) cos (Θ

cos (x - y) = cos (x) cos (y) - sin (x) sin (y), substitute in x = 2π and y = +Θ

_{1}cos (2π - Θ

_{1}) = cos (2π) cos (Θ_{1}) - sin (2π) sin(Θ_{1}), and note that sin (2π) = 0 and cos(2π) = +1 in quadrant IVThus, cos (2π - Θ1) = (+1) cos (Θ

_{1}) = cos (Θ1) = 1