Sec. 12.2 Circular functions 243
The sequences un,vnhave the following properties:-
(i)(un)is non-decreasing.
(ii)(vn)is non-increasing.
(iii)For 0 <x≤ 360 and n≥ 2 , we have 0 <un(x)≤vn(x).
(iv) As n→∞,un(x)and vn(x)tend to a common limitμ(x)and
un(x)≤μ(x)≤vn(x).
Proof.
(i) For
un(x)= 2 n+^1 s
( x
2 n+^2
)
c
( x
2 n+^2
)
≤un+ 1 (x).
(ii) For
vn(x)=
2 n+^1 s
( x
2 n+^2
)
c
( x
2 n+^2
)
2 c
( x
2 n+^2
) 2
− 1
=vn+ 1 (x)
c
( x
2 n+^2
) 2
2 c
( x
2 n+^2
) 2
− 1
>vn+ 1 (x)
asc( 2 nx+ 2 )^2 <1andso2c( 2 nx+ 2 )^2 − 1 <c( 2 nx+ 2 )^2 , whilec( 2 nx+ 1 )>0 and thus
2 c( 2 nx+ 2 )^2 − 1 >0.
(iii) For
un(x)=vn(x)c
( x
2 n+^1
)
≤vn(x).
(iv) By (i), (ii) and (iii),un(x)≤v 3 (x)so we have a non-decreasing sequence
which is bounded above. Hence there is someμ(x)such that limn→∞un(x)=μ(x).It
follows thats( 2 nx+ 1 )→ 0 (n→∞)and hence
c
( x
2 n+^1
)
= 1 − 2 s
( x
2 n+^2
) 2
→ 1 (n→∞).
But then
vn(x)
un(x)
=c
( x
2 n+^1
)− 1
→ 1 (n→∞),
and so as well limn→∞vn(x)=μ(x).Asun(x)is non-decreasing andvn(x)is non-
increasing we must haveun(x)≤μ(x)≤vn(x).
For x≥ 0 ,y≥ 0 ,x+y≤ 360 ,
μ(x+y)=μ(x)+μ(y).
Proof.For
un(x+y)= 2 ns
(
x+y
2 n+^1
)
= 2 ns
( x
2 n+^1
+
y
2 n+^1
)
= 2 ns
( x
2 n+^1
)
c
( y
2 n+^1
)
+c
( x
2 n+^1
)
2 ns
( y
2 n+^1
)
→μ(x). 1 + 1 .μ(y)(n→∞).