Advanced book on Mathematics Olympiad

Geometry and Trigonometry 653

658.The relation from the statement can be transformed into

tan^2 b=

tan^2 a+ 1

tan^2 a− 1

=−

1

cos 2a

.

This is further equivalent to

sin^2 b

1 −sin^2 b

=

1

2 sin^2 a− 1

.

Eliminating the denominators, we obtain

2 sin^2 asin^2 b= 1 ,

which gives the desired sinasinb=±

√

2

2 =±sin 45

◦.

(Romanian Mathematical Olympiad, 1959)

659.We have

f(x)=sinxcosx+sinx+cosx+ 1 =

1

2

(sinx+cosx)^2 −

1

2

+sinx+cosx+ 1

=

1

2

[(sinx+cosx)^2 + 2 (sinx+cosx)+ 1 ]=

1

2

[(sinx+cosx)+ 1 ]^2.

This is a function ofy=sinx+cosx, namelyf(y)=^12 (y+ 1 )^2. Note that

y=cos

(π

2

−x

)

+cosx=2 cos

π

4

cos

(

x−

π

4

)

=

√

2 cos

(

x−

π

4

)

.

Soyranges between−

√

2 and

√

2. Hencef(y)ranges between 0 and^12 (

√

2 + 1 )^2.

660.Relate the secant and the cosecant to the tangent and cotangent:

sec^2 x=tan^2 x+ 1 ≥2 tanx and csc^2 x=cot^2 x+ 1 ≥2 cotx,

where the inequalities come from the most particular case of AM–GM. It follows that

sec^2 nx+csc^2 nx≥ 2 n(tannx+cotnx).

Now observe that

tannx+cotnx=tannx+

1

tannx

≥ 2 ,

again by the AM–GM inequality. We obtain

sec^2 nx+csc^2 nx≥ 2 n+^1 ,