I use both r and theta >= 0randθ≥0 only.
So, arctan (y/x) in [ 0, pi ], sans pi/2arctan(yx)∈[0,π],sansπ2
r = theta - 4 sin theta + cos^2theta = thetar=θ−4sinθ+cos2θ=θ + a periodic function
of period 2pi2π. So, r is non-periodic.
Convert using
( x, y ) = r ( cos theta, sin theta ). 0 <= r = sqrt ( x^2 + y^2 ) and(x,y)=r(cosθ,sinθ).0≤r=√x2+y2and
0 <= theta = kpi + arctan ( y/x ), l = 0, 1, 2, 3, ....
sqrt ( x^2 + y^2 ) = kpi + arctan ( y/x )
- 4 x/sqrt ( x^2 + y^2 ) + y^2/(x^2 + y^2 ), giving
0 <= (tan)^(-1) ( y/x ) = kpi + arctan ( y/x )
= 4 x/sqrt ( x^2 + y^2 ) - y^2/(x^2 + y^2 )
+ sqrt ( x^2 + y^2 ), k = 0, 1, 2, 3, ..#, piecewise, for successive
half rotations, about O.
Graph, for k = 0 using conventional arctan ( y/x ) in ( - pi/2, pi/2 ) ).graph{arctan ( y/x )-4 x/sqrt ( x^2 + y^2 ) + y^2/(x^2 + y^2 )=0[-1000 1000 -500 500]}
Graph, for the assumed theta to theta + pi/2# non- negative
arctan ( y/x) in ( 0, pi ), sans pi/2 .
graph{pi/2+arctan ( y/x )+4y/sqrt ( x^2 + y^2 ) + x^2/(x^2 + y^2 )=0[-1000 1000 -500 500]}
Observe that this is confined to Q_3 and Q_4, whereas, the first
that was expected to be in Q_1 and Q_4 is not so.
