Geometry and Trigonometry 6652 tanun+ 1 =bn+ 1 =
2 tanun
2 +√
4 +4 tanun=
4 tanun
2 +cos^2 un= 2 ·sinun
1 +cosun=2 tanun
2.
Therefore,un+ 1 =u 2 n, which together withu 0 =π 4 impliesun= 2 nπ+ 2 ,n≥0. Hence
bn=2 tan 2 nπ+ 2 forn≥0.
Returning to the problem, we recall that sine and tangent are decreasing on( 0 ,π 2 )
and their limit at 0 is 0. This takes care of (a).
For (b), note that the functions sinx/xand tanx/xare increasing, respectively,
decreasing, on( 0 , 2 π). Hence 2nan = π 2 sin 2 nπ+ 2 / 2 nπ+ 2 is increasing, and 2nbn =
π
2 tan
π
2 n+^2 /π
2 n+^2 is decreasing. Also, sincelim
x→ 0sinx
x=lim
x→ 0tanx
x= 1 ,
it follows that
nlim→∞^2 nan=π
2
nlim→∞sin 2 nπ+ 2
π
2 n+^2=
π
2,
and similarly limn→∞ 2 nbn=π 2. This answers (b).
The first inequality in (c) follows from the fact that tanx>sinxforx∈( 0 ,π 2 ). For
the second inequality we use Taylor series expansions. We have
tanx−sinx=x−x^3
12
+o(x^4 )−x+x^3
6
+o(x^4 )=x^3
12
+o(x^4 ).Hence
bn−an= 2(
tanπ
2 n+^2
−sinπ
2 n+^2)
=
π^3
6 · 26·
1
8 n
+o(
1
24 n)
.
It follows that forC>π
3
6 · 26 we can findn^0 such thatbn−an<
C
8 nforn≥n^0. ChooseC
such that the inequality also holds for (the finitely many)n<n 0. This concludes (c).
(8th International Competition in Mathematics for University Students, 2001)
684.Writingxn=tananfor 0◦<an< 90 ◦, we have
xn+ 1 =tanan+√
1 +tan^2 an=tanan+secan=
1 +sinan
cosan=tan(
90 ◦+an
2)
.
Becausea 1 = 60 ◦, we havea 2 = 75 ◦,a 3 = 82. 5 ◦, and in generalan= 90 ◦−^30
◦
2 n−^1 ,
whence