Answer: □ We have x⋆y+y⋆x=sinxcosy+cosxsiny=sin(x+y)≤1. Equality is achieved when x=2π and y=0. Indeed, for these values of x and y, we have x⋆y−y⋆x=sinxcosy−cosxsiny=sin(x−y)=sin2π=1
Source: NuminaMath-1.5,
licensed Apache-2.0.
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