Weakness in Maximize

Question

Solving a minimization problem in version 13 on Windows 10, I obtain

Maximize[{Sin[x + Sin[x]] + Sin[x - Sin[x]] + (Pi/2 - 2)*Sin[Sin[x]], 
x >= 0 && x < 2*Pi}, x]//FullSimplify

{2 Cos[Sin[Root[{Pi Cos[Sin[#]] - 4 Sin[Sin[#]] Sin[#]& , 0.9033391107665128473590217051048716426220.300104378161908}]]]\ Sin[Root[{Pi Cos[Sin[#]] - 4 Sin[Sin[#]] Sin[#]& , 0.9033391107665128473590217051048716426220.300104378161908}]] + 1/2 (-4 + \[Pi]) Sin[Sin[Root[{Pi Cos[Sin[#]] - 4 Sin[Sin[#]] Sin[#]& , 0.9033391107665128473590217051048716426220.300104378161908}]]]\ , {x -> Root[{Pi Cos[Sin[#]] - 4 Sin[Sin[#]] Sin[#]& , 0.9033391107665128473590217051048716426220.300104378161908}]}}

which is not very useful as a symbolic solution.

Along with that, Mathematica is able to find an optimal solution in a closed form and this is its achievement. Indeed,

Maximize[Sin[x+Sin[x]]+Sin[x-Sin[x]]+(Pi/2-2)*Sin[Sin[x]],x]

{1/2 (-2 Sqrt[2] + Sqrt[2] \[Pi]), {x -> -47 \[Pi] - ArcSin[\[Pi]/4]}}

Since the objective function has 2*Pi as its minimal positive period, the same value is taken at

Mod[-47 \[Pi] - ArcSin[\[Pi]/4], 2*Pi]

\[Pi] - ArcSin[\[Pi]/4]

N[%]

2.23825

and

Plot[Sin[x + Sin[x]] + Sin[x - Sin[x]] + (Pi/2 - 2)*Sin[Sin[x]], {x,0, 2*Pi}]

confirms it.

How to obtain {1/2 (-2 Sqrt[2] + Sqrt[2] \[Pi]), {x -> Pi-ArcSin[Pi/4]}} programmatically?

Answer 1

Clear["Global`*"]

expr = Sin[x + Sin[x]] + Sin[x - Sin[x]] + (Pi/2 - 2)*Sin[Sin[x]];

{max, arg} = 
 Maximize[{expr, 0 <= x < 2 Pi}, x] // FullSimplify // Quiet

Maximize will have a much easier time if you simplify the expression first.

expr2 = Simplify[expr, 0 <= x < 2 Pi]

(* 2 Cos[Sin[x]] Sin[x] + 1/2 (-4 + π) Sin[Sin[x]] *)

{max2, arg2} = Maximize[{expr2, 0 <= x < 2 Pi}, x] // FullSimplify

(* {(-2 + π)/Sqrt[2], {x -> ArcSin[π/4]}} *)

max - max2 // N[#, 20] & // Quiet

(* 0.*10^-69 *)

(x /. arg) - (x /. arg2) // N[#, 20] & // Quiet

(* 0.*10^-70 *)

EDIT: Likewise with FunctionPeriod

FunctionPeriod[expr2, x]

(* 2 π *)