Prove, that \(\displaystyle{\sin{{x}}}-{a}^{{3}}{\cos{{x}}}\leq{\frac{{{1}}}{{{3}}}}\sqrt{{{1}+{a}^{{6}}}}\)

afasiask7xg

afasiask7xg

Answered question

2022-03-31

Prove, that
sinxa3cosx131+a6

Answer & Explanation

Jayda Burch

Jayda Burch

Beginner2022-04-01Added 10 answers

Not sure if it is correct.
sinxa3cosx=1+(a3)2(11+a6sin(x)a31+a6cos(x))
=1+a6(cos(ϕ)sin(x)sin(ϕ)cos(x))
=1+a6sin(xϕ)
Where ϕ=arctan(a3). Given, constraint
sin(x)acos(x)xarctan(a)π2
sin(xarctan(a3))sin(arctan(a)arctan(a3))
=sin(arctanaa31+a4))
=aa3(aa3)2+(1+a4)
From calculus, that thing seems to have maximum value of 13

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