8.3. THE SPECIAL FUNCTIONS OF ELEMENTARY CALCULUS 161

for all x. Also sin(−x) = −sin(x) while cos(−x) = cos(x) and the usual trig. identitieshold,

sin(x+ y) = sin(x)cos(y)+ sin(y)cos(x) (8.8)cos(x+ y) = cos(x)cos(y)− sin(x)sin(y) (8.9)

Proof: That sin′ (x) = cos(x) and cos′ (x) =−sin(x) follows right away from differen-tiating the power series term by term using Theorem 8.2.1. It follows from the series thatcos(0) = 1 and sin(0) = 0 and sin(−x) = −sin(x) while cos(−x) = cos(x) because theseries for sin(x) only involves odd powers of x while the series for cos(x) only involveseven powers.

For x ∈ R, let f (x) = cos2 (x)+ sin2 (x) , it follows from what was just discussed thatf (0) = 1. Also from the chain rule, f ′ (x) = 2cos(x)(−sin(x))+2sin(x)cos(x) = 0 andso by Corollary 7.8.5, f (x) is constant for all x ∈R. But f (0) = 1 so the constant can onlybe 1. Thus

cos2 (x)+ sin2 (x) = 1 (8.10)

as claimed.It only remains to verify the identities. Consider 8.8 first. Fixing y and considering both

sides as a function of x, it follows from the above that both sides of the identity satisfy theinitial value problem

y′′+ y = 0, y(0) = sin(y) ,y′ (0) = cos(y)

Therefore, the difference satisfies the initial value problem of Lemma 8.3.3. Therefore, bythis lemma, the difference equals 0. The next identity is handled similarly.

Note that 8.10 shows that the ordered pair (cosx,sinx) lies on the unit circle with centerat (0,0).

Proposition 8.3.5 The following important limits hold for a,b ̸= 0.

limx→0

sin(ax)bx

=ab, lim

x→0

1− cos(x)x

= 0.

Proof: From the definition of sin(x) given above, sin(ax)bx =

∑∞k=0 (−1)k (ax)2k+1

(2k+1)!

bx=

ax+∑∞k=1 (−1)k (ax)2k+1

(2k+1)!

bx=

a+∑∞k=1 (−1)k (ax)2k

(2k+1)!

b

Now∣∣∣∑∞

k=1 (−1)k (ax)2k

(2k+1)!

∣∣∣ ≤ ∑∞k=1 |ax|2k = ∑

∞k=1

(|ax|2

)k=(|ax|2

1−|ax|

)whenever |ax| < 1.

Thus limx→0 ∑∞k=1 (−1)k (ax)2k

(2k+1)! = 0 and so limx→0sin(ax)

bx = ab .The other limit can be han-

dled similarly.It is possible to verify the functions are periodic.

Lemma 8.3.6 There exists a positive number a, such that cos(a) = 0.

Proof: To prove this, note that cos(0) = 1 and so if it is false, it must be the case thatcos(x) > 0 for all positive x since otherwise, it would follow from the intermediate valuetheorem there would exist a point, x where cosx = 0. Assume cos(x) > 0 for all x. Then