Advanced book on Mathematics Olympiad

656 Geometry and Trigonometry

Herex=sina,y=cosa,z=−^15. It follows that eitherx+y+z=0orx=y=z. The
latter would imply sina=cosa=−^15 , which violates the identity sin^2 a+cos^2 a=1.
Hencex+y+z=0, implying sina+cosa=^15. Then 5(sina+cosa)=1, and so

sin^2 a+2 sinacosa+cos^2 a=

1

25

.

It follows that 1+2 sinacosa= 0 .04; hence

5 (sina+cosa)+2 sinacosa= 0. 04 ,

as desired.
Conversely,

5 (sina+cosa)+2 sinacosa= 0. 04

implies

125 (sina+cosa)= 1 −50 sinacosa.

Squaring both sides and setting 2 sinacosa=byields

1252 + 1252 b= 1 − 50 b+ 252 b^2 ,

which simplifies to

( 25 b+ 24 )( 25 b− 651 )= 0.

We obtain 2 sinacosa=−^2425 , or 2 sinacosa=^65125. The latter is impossible because
sin 2a≤1. Hence 2 sinacosa=− 0 .96, and we obtain sina+cosa= 0 .2. Then

5 (sin^3 a+cos^3 a)+3 sinacosa= 5 (sina+cosa)(sin^2 a−sinacosa+cos^2 a)

+3 sinacosa

=sin^2 a−sinacosa+cos^2 a+3 sinacosa

=(sina+cosa)^2 =( 0. 2 )^2 = 0. 04 ,

as desired.
(Mathematical Reflections, proposed by T. Andreescu)

665.If we setbk=tan(ak−π 4 ),k= 0 , 1 ,...,n, then

tan

(

ak−

π

4

+

π

4

)

=

1 +tan(ak−π 4 )

1 −tan(ak−π 4 )

=

1 +bk

1 −bk

.