Geometry with Trigonometry

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→∞).