Advanced book on Mathematics Olympiad

238 4 Geometry and Trigonometry

673.For positive integersndefineF (n)=xnsin(nA)+ynsin(nB)+znsin(nC), where

x, y, z, A, B, Care real numbers andA+B+C=kπfor some integerk. Prove

that ifF( 1 )=F( 2 )=0, thenF (n)=0 for all positive integersn.

674.The continuous real-valued functionφ(t)is defined fort ≥0 and is absolutely

integrable on every bounded interval. Define

P=

∫∞

0

e−(t+iφ(t))dt and Q=

∫∞

0

e−^2 (t+iφ(t))dt.

Prove that

| 4 P^2 − 2 Q|≤ 3 ,

with equality if and only ifφ(t)is constant.

4.2.3 Trigonometric Substitutions................................

The fact that the circlex^2 +y^2 =1 can be parametrized by trigonometric functions

asx =costandy =sintgives rise to the standard substitutionx=acost(orx=

asint) in expressions of the form

√

a^2 −x^2. Our purpose is to emphasize less standard

substitutions, usually suggested by the similarity between an algebraic expression and

a trigonometric formula. Such is the case with the following problem from the 61st

W.L. Putnam Mathematical Competition, 2000.

Example.Letf:[− 1 , 1 ]→Rbe a continuous function such thatf( 2 x^2 − 1 )= 2 xf (x)

for allx∈[− 1 , 1 ]. Show thatfis identically equal to zero.

Solution.Here the expression 2x^2 −1 should remind us of the trigonometric formula

2 cos^2 t− 1 =cos 2t, suggesting the substitutionx=cost,t∈[ 0 ,π]. The functional

equation from the statement becomesf(cos 2t)=2 costf (cost).

First, note that settingx=0 andx=1, we obtainf( 1 )=f(− 1 )=0. Now let us

defineg:R→R,g(t)=f(sincostt). Then for anytnot a multiple ofπ,

g( 2 t)=

f(2 cos^2 t− 1 )

sin( 2 t)

=

2 costf (cost)

2 sintcost

=

f(cost)

sint

=g(t).

Also,g(t+ 2 π)=g(t). In particular, for any integersnandk,

g

(

1 +

nπ

2 k

)

=g( 2 k+^1 + 2 nπ )=g( 2 k+^1 )=g( 1 ).

Becausefis continuous,gis continuous everywhere except at multiples ofπ. The set

{ 1 +nπ 2 k |n, k∈Z}is dense on the real axis, and sogmust be constant on its domain.