# Solving a set of trig inequalities

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?

• 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 output. Commented Mar 22, 2022 at 11:37

RegionPlot3Dshows 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}]


• Why {b, -1, 1} and {a, -Pi/2, Pi}? Commented Mar 22, 2022 at 11:41
• -1. 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. Commented Mar 22, 2022 at 11:48
• @user64494 Why this children's egg "-1" again and again? My answer shows part of the solution. Commented Mar 22, 2022 at 12:31
• @UlrichNemann: If you admitted in your answer that the plot in your answer draws only a part of the solution set, I didn't do my downvote. However, you didn't it. Commented Mar 22, 2022 at 15:29
• @user64494 The same old story over and over again... By the way I didn't mention wether my answer shows the complete solution Commented Mar 22, 2022 at 15:36

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.

• It should be noticed that 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. Commented Mar 22, 2022 at 12:20