I am pretty weak with Mathematica, so please, nothing overly complicated. The ArcTan[x, y] function allows for a full range of $2 \pi$ in the output. However, for the equation $\beta = \tan^{-1} \left( \frac{\tan\gamma}{C} \right)$, $C$ is always positive, so I only get $\pi$ range in $\beta$. But I NEED $\beta$ to be in the same quadrant as $\gamma$. How can I rectify this?


Split the tangent:

ArcTan[c Cos[γ], Sin[γ]]

A quick Manipulate[] for fun:

Manipulate[Graphics[{{Opacity[1/2, ColorData[97, 1]], Disk[{0, 0}, 1, {0, γ}]},
                     {Opacity[1/2, ColorData[97, 2]],
                      Disk[{0, 0}, 1, {0, ArcTan[c Cos[γ], Sin[γ]]}]}}, PlotRange -> 1],
           {{c, 1/2}, 0, 1}, {{γ, π/4}, -π, π}]
  • $\begingroup$ GENIUS! You are a prodigy. $\endgroup$ – Johnver Aug 5 '16 at 2:03
  • $\begingroup$ I believe I'm too old to qualify, but thank you. :) $\endgroup$ – J. M. is in limbo Aug 5 '16 at 2:06

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