I think that I have almost convinced the readers about the
scalar r = sqrt ( x^2 + y^2 ) >= 0.
Of course, sooner or later thetaθ, used in measuring clockwise or
anticlockwise time-oriented Natural rotations and revolutions, and
serving as a piπ-oriented location-marker, in the marker couple
( r, theta )(r,θ), would be declared as non-negative.
And now, I assume that theta >= 0θ≥0.
For this problem, if 0 <= theta in Q_4, theta/2 in Q-20≤θ∈Q4,θ2∈Q−2.
Accordingly, cos (theta/2) and tan (theta/2) <=0.cos(θ2)andtan(θ2)≤0.
f ( theta ) = +-sqrt(( 1 - cos theta )/( 1 + cos theta )f(θ)=±√1−cosθ1+cosθ
+ sqrt2 sqrt(1/( 1 +- cos (theta/2)))- 1/sqrt2 sqrt( 1 - cos (theta/2) )+√2
⎷11±cos(θ2)−1√2√1−cos(θ2)
= +-sqrt(( 1 - cos theta )/( 1 + cos theta )=±√1−cosθ1+cosθ
+ sqrt2 sqrt(1/( 1 +- 1/sqrt2 sqrt( 1 + cos theta ))+√2
⎷11±1√2√1+cosθ
- 1/sqrt2 sqrt( 1 +- 1/sqrt2 sqrt ( 1 + cos theta ))−1√2√1±1√2√1+cosθ
Sign Disambiguation:
If theta notin Q_4θ∉Q4, choose + for 1st and 2nd terms
and -− for the 3rd. Otherwise. they are opposites.
