Geometry with Trigonometry

Sec. 12.3 Derivatives of cosine and sine functions 245

must exist. But the formulaμ(^1 n)=μ(n^1 )shows thatμ(^1 n)→ 0 (n→∞)and so we
must havel=0. Astn−y→0 it follows thatμ(tn−y)→ 0 (n→∞)and then

nlim→∞μ(y)=nlim→∞[tnμ(^1 )]+nlim→∞μ(tn−y)

yieldsμ(y)=yμ( 1 )+0.
To conclude we takep=μ( 1 )and note that this is positive as shown above.

12.2.2 Definition ofπ.............................

Definition. We denoteμ( 360 )byπ. Then 360p=πso thatp= 360 π and so

μ(x)=

π

360

x.

12.3 Derivativesofcosineandsinefunctions................

12.3.1 .....................................

With the notation of12.1,

tlim→ 0 +

s(t)

t

=

π

180

.

Proof. We have seen above that for 0<x≤360, takingn=2 in 12.2.1(iv),

0 <s

(x

8

)

≤

π

180

x

8

≤

s(x 8 )

c(x 8 )

,

and so for 0<t<45,

0 <s(t)≤

π

180

t≤

s(t)

c(t)

.

From the first two inequalities here we infer thats(t)→ 0 (t→ 0 +)and hence

c(t)= 1 − 2 s

(t

2

) 2

→ 1 (t→ 0 +).

As
π
180

c(t)≤

s(t)

t

≤

π

180

,

the result follows.

For 0 <x< 360 the derivatives c′(x)and s′(x)exist and are given by

c′(x)=−

π

180

s(x),s′(x)=

π

180

c(x).