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单词 ConcavityOfSineFunction
释义

concavity of sine function


Theorem 1.

The sine function is concave on the interval [0,π].

Proof.

Suppose that x and y lie in the interval [0,π/2].Then sinx, siny, cosx, and cosy are allnon-negative. Subtracting the identities

sin2x+cos2x=1

and

sin2y+cos2y=1

from each other, we conclude that

sin2x-sin2y=cos2y-cos2x.

This implies that sin2x-sin2y0 if and only ifcos2y-cos2x0, which is equivalentPlanetmathPlanetmath to statingthat sin2xsin2y if and only if cos2xcos2y. Taking square roots, we conclude thatsinxsiny if and only ifcosxcosy.

Hence, we have

(sinx-siny)(cosx-cosy)0.

Multiply out both sides and move terms to conclude

sinxcosx+sinycosysinxcosy+sinycosx.

Applying the angle addition and double-angle identities for thesine function, this becomes

12(sin(2x)+sin(2y))sin(x+y).

This is equivalent to stating that, for allu,v[0,π],

12(sinu+sinv)sin(u+v2),

which implies that sin is concave in the interval [0,π].∎

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更新时间:2025/5/4 11:24:10