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?
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2 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– Johnver2016-08-05 02:03:42 +00:00Commented Aug 5, 2016 at 2:03
- $\begingroup$ I believe I'm too old to qualify, but thank you. :) $\endgroup$J. M.'s missing motivation– J. M.'s missing motivation2016-08-05 02:06:57 +00:00Commented Aug 5, 2016 at 2:06