Solve with inverse trig functions (4-Bar Linkage)

Question

    c = 5.; b = 3.; a = 2.; len = 7.5;
u[t_] := Reduce[
  NSolve[(2 c + a Cos[t] - b Cos[u[t]])^2 + (a Sin[t] - b Sin[u[t]])^2 - 
     len^2 == 0], u, Reals]
Plot[u[t], {t, 0, 2 Pi}]

Please help to obtain a simplified output using Reduce. (4-Bar linkage).

Answer 1

You don't need Reduce , try Weirstrass substitution u[t] -> 2 ArcTan[uu]:

Leaving the parameters undefined try

eq = TrigExpand[(2 c + a Cos[t] - b Cos[u[t]])^2 + (a Sin[t] - 
   b Sin[u[t]])^2 - len^2 == 0]
sol = 2 ArcTan[uu] /.Solve[eq /. u[t] -> 2 ArcTan[uu] // TrigExpand, uu]

which gives the two solution branches u[t] in analytical form. {2 ArcTan[(4 a b Sin[ t] - \[Sqrt](-4 (a^2 + b^2 - 4 b c + 4 c^2 - len^2 - 2 a b Cos[t] + 4 a c Cos[t]) (a^2 + b^2 + 4 b c + 4 c^2 - len^2 + 2 a b Cos[t] + 4 a c Cos[t]) + 16 a^2 b^2 Sin[t]^2))/(2 (a^2 + b^2 + 4 b c + 4 c^2 - len^2 + 2 a b Cos[t] + 4 a c Cos[t]))], 2 ArcTan[(4 a b Sin[ t] + \[Sqrt](-4 (a^2 + b^2 - 4 b c + 4 c^2 - len^2 - 2 a b Cos[t] + 4 a c Cos[t]) (a^2 + b^2 + 4 b c + 4 c^2 - len^2 + 2 a b Cos[t] + 4 a c Cos[t]) + 16 a^2 b^2 Sin[t]^2))/(2 (a^2 + b^2 + 4 b c + 4 c^2 - len^2 + 2 a b Cos[t] + 4 a c Cos[t]))]}

Plot[sol /. {c -> 5, b -> 3, a -> 2, len -> 15/2} //Evaluate, {t, 0, 2 Pi}]

tps://i.sstatic.net/dEkDP.png

Answer 2

Clear["Global`*"]

c = 5; b = 3; a = 2; len = 15/2;

eqn = (2 c + a Cos[t] - b Cos[u])^2 + (a Sin[t] - b Sin[u])^2 - 
    len^2 == 0 // Simplify

(* 227/4 + 40 Cos[t] == 12 (Cos[t - u] + 5 Cos[u]) *)

sol = u /. 
    Assuming[0 <= t <= 2 Pi, 
     Solve[{eqn, 0 <= t < 2 Pi}, u, Reals, Method -> Reduce] // 
      Simplify] /. C[1] -> 0;

u1[t_] = sol[[1]]

u2[t_] = sol[[2]]

FunctionDomain[#[t], t] & /@ {u1, u2}

%[[1]] // N

3.14159 < t < 4.86928 || 1.4139 < t < 3.14159

Plot[{u1[t], u2[t]}, {t, 0, 2 Pi}, PlotPoints -> 100, 
 MaxRecursion -> 5, PlotLegends -> Placed[{u1, u2}, {0.85, 0.7}]]

For comparison,

ContourPlot[Evaluate@eqn,
 {t, 0, 2 Pi}, {u, -1.35, 1.35},
 AspectRatio -> 1/GoldenRatio]