I have a set of inequalities
Cos[a]Cos[b]>=Cos[t-a]Cos[b]&&Cos[a]Cos[b]>=Cos[t/2]&&Cos[a]Cos[b]>=Sin[t/2]&&a<=t<=Pi
How to solve this to get a range of values for a,b,t
?
RegionPlot3D
shows the region of possible solutions
RegionPlot3D[Cos[a] Cos[b] >= Cos[t - a] Cos[b] && Cos[a] Cos[b] >= Cos[t/2] &&Cos[a] Cos[b] >= Sin[t/2] && a <= t <= Pi, {a, -Pi/2, Pi}, {t, 0, Pi}, {b, -1, 1}, AxesLabel -> {a, t, b}]
{b, -1, 1}
and {a, -Pi/2, Pi}
?
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Commented
Mar 22, 2022 at 11:41
RegionPlot3D[ Cos[a] Cos[b] >= Cos[t - a] Cos[b] && Cos[a] Cos[b] >= Cos[t/2] && Cos[a] Cos[b] >= Sin[t/2] && a <= t <= Pi, {a, -Pi, Pi}, {t, -Pi, Pi}, {b, -Pi, Pi}, AxesLabel -> {a, t, b}]
shows a bigger region.
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Commented
Mar 22, 2022 at 11:48
Reducing trigonometry to algebra by Reduce[ca*cb >= ((ct^2 - st^2)*ca + 2*ct*st*sa)*cb && cb*ca >= ct && cb*ca >= st && ca^2 + sa^2 == 1 && cb^2 + sb^2 == 1 && ct^2 + st^2 == 1, {ct, st}, Reals]
, one obtains a huge and useless output. In order to obtain a concrete result, you have to specify a
and b
, e.g.
a = 1/20; b = Pi/6; Reduce[Cos[a] Cos[b] >= Cos[t - a]*Cos[b] && Cos[a]*Cos[b] >= Cos[t/2] &&
Cos[a]*Cos[b] >= Sin[t/2] && a <= t <= Pi, t, Reals]
2 ArcCos[1/2 Sqrt[3] Cos[1/20]] <= t <= 2 ArcSin[1/2 Sqrt[3] Cos[1/20]]
Try a=0.05;b=Pi/6;
on your own.
ClearAll[b]; a = 1/20; Reduce[ Cos[a] Cos[b] >= Cos[t - a]*Cos[b] && Cos[a]*Cos[b] >= Cos[t/2] && Cos[a]*Cos[b] >= Sin[t/2] && a <= t <= Pi && b > -Pi && b <= Pi, t, Reals]
returns the input.
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Commented
Mar 22, 2022 at 12:20
Reduce[ca*cb >= ((ct^2 - st^2)*ca + 2*ct*st*sa)*cb && cb*ca >= ct && cb*ca >= st && ca^2 + sa^2 == 1 && cb^2 + sb^2 == 1 && ct^2 + st^2 == 1, {ct, st}, Reals]
, one obtains a huge output. $\endgroup$