Sines and Cosines
The following problems show that the two functions, S and C, defined below have the same
properties as our two favorite trig functions sin x and cos x.
Define S(x) and C(x) as follows:
∞
n
S(x) = ∑ (−1)
n=0
x2n+1
,
(2n + 1)!
∞
C(x) = ∑ (−1)
n=0
n
x2n
.
(2n)!
1. Show that the series we use to define S and C converge for all x.
Easiest way is to use the ratio test. Checking S first and then C we have
RRRR x2n+3 RRRR
x2
(2n+3)!
∣=0
lim RRRR x2n+1 RRRR = lim ∣
n→∞ RR
n→∞ (2n + 3)(2n + 2)
RR (2n+1)! RRRR
RRR x2n+2 RRR
x2
R (2n+2)! R
lim RRRR x2n RRRR = lim ∣
∣=0
n→∞ RR
n→∞ (2n + 2)(2n + 1)
RR (2n)! RRRR
Thus, both series converge absolutely for all x
2. Show that the following differentiation formulas are valid:
d
[S(x)] = C(x),
dx
d
[C(x)] = −S(x) .
dx
∞
dS
d ∞
x2n+1
x2n+1
n
n d
=
( ∑ (−1)
) = ∑ (−1)
(
)
dx
dx n=0
(2n + 1)!
dx (2n + 1)!
n=0
∞
n
= ∑ (−1)
n=0
dC
dx
=
x2n
= C(x)
(2n)!
∞
2n
d ∞
x2n
n x
n d
( ∑ (−1)
) = ∑ (−1)
(
)
dx n=0
(2n)!
dx (2n)!
n=0
∞
n
= ∑ (−1) (2n)
n=0
∞
n+1
= ∑ (−1)
n=0
x2n−1 ∞
x2n−1
n
= ∑ (−1)
(2n)! n=1
(2n − 1)!
x2n+1
= −S(x)
(2n + 1)!
3. Show that S 2 (x) + C 2 (x) = 1 for all x. Hint: what is the derivative of S 2 + C 2 ? Use
this equality to deduce that for any x we have ∣S(x)∣ ≤ 1 and ∣C(x)∣ ≤ 1.
We first show that the derivative of S 2 (x) + C 2 (x) equals zero, which means this
expression is constant for all x. Since S(0) = 0 and C(0) = 1, this constant must be 1.
d
(S 2 (x) + C 2 (x)) = 2SS ′ + 2CC ′ = 2SC − 2CS = 0 .
dx
The two inequalities are easy consequences of the above.
√
√
√
∣S(x)∣ = S 2 (x) ≤ S 2 (x) + C 2 (x) = 1 = 1 ,
with a similar argument for C(x).
4. By expanding S(x + y) about x, show that S(x + y) = S(x)C(y) + S(y)C(x). Use this
to show that C(x) satisfies C(x + y) = C(x)C(y) − S(y)S(x). Hint: you may find the
following useful:
∞
∞
∞
∞
∞
∑ f (xk ) = ∑ f (x4j ) + ∑ f (x4j+1 ) + ∑ f (x4j+2 ) + ∑ f (x4j+3 )
j=0
k=0
j=0
j=0
j=0
Note: this assumes the original series converges absolutely.
The Taylor series of S(x + y) about the point x equals
S(x + y) = S(x) + S ′ (x)y + S (2) (x)
∞
= ∑ S 4j
j=0
∞
yk
y2
+ ⋯ = ∑ S k (x)
2
k!
k=0
∞
∞
∞
y 4j+1
y 4j+2
y 4j+3
y 4j
+ ∑ S 4j+1
+ ∑ S 4j+2
+ ∑ S 4j+3
(4j)! j=0
(4j + 1)! j=0
(4j + 2)! j=0
(4j + 3)!
∞
= ∑ S(x)
j=0
∞
∞
∞
y 4j
y 4j+1
y 4j+2
y 4j+3
+ ∑ C(x)
+ ∑ (−S(x))
+ ∑ (−C(x))
(4j)! j=0
(4j + 1)! j=0
(4j + 2)! j=0
(4j + 3)!
∞
= S(x) ∑ (
j=0
∞
y 4j
y 4j+2
y 4j+1
y 4j+3
−
) + C(x) ∑ (
−
)
(4j)! (4j + 2)!
(4j + 3)!
j=0 (4j + 1)!
= S(x)C(y) + C(x)S(y)
We used the fact that the Taylor series expansion of S(x) actually equals S. One can
show that this is the case by using the integral formula for the error in approximating
S with its nth order Taylor polynomial.
To derive the second identity differentiate the sin x identity.
C(x + y) =
d
d
[S(x + y)] =
[S(x)C(y) + C(x)S(y)]
dx
dx
= C(x)C(y) − S(x)S(y)
As an aside the cosine identity can be used to prove the trigonometric version of the
Pythagorean theorem.
1 = C(0) = C(x − x) = C(x)C(−x) − S(x)S(−x) = C 2 (x) + S 2 (x) .
2
π
.
2
We know that C(0) = 1, S(0) = 0, and the derivative derivative of C(x), C ′ (x) = −S(x).
Thus, for positive values of x in a small interval about 0, C(x) has a negative derivative
and is a decreasing function. If C(x) never takes on the value zero, then it must
always be positive, which means S(x) is always an increasing function. Thus, there is
a κ > 0 and a δ > 0 such that S(x) ≥ κ for all x ≥ δ. This implies that for all x ≥ δ,
C ′ (x) = −S(x) ≤ −κ, and this in turn implies that
5. Show there is a smallest positive x0 for which C(x0 ) = 0. Denote this value by
x
C(x) − C(δ) = ∫
(−S ′ (t)) dt ≤ −κ(x − δ) ,
δ
but the limit of this as x → ∞ is −∞ ,which forces C(x) to become negative. Thus,
there is a minimum value of x, for which C(x) = 0. Note too, that at this x we must
have S(x) = 1).
The above argument demonstrates that C(x) does have a smallest positive zero. Unfortunately, it does not help us determine how big π/2 is. The following argument shows
that C(2) < 0, from which we can conclude that π/2 < 2.
∞
C(2) = ∑ (−1)
n=0
n
∞
∞
2n
2n
22n
n 2
n 2
= 1 − 2 + ∑ (−1)
= −1 + ∑ (−1)
(2n)!
(2n)!
(2n)!
n=2
n=2
The infinite sum in the right hand side of the above equation is an alternating series,
and its absolute value is less than or equal to the fist term in the series. That is,
∞
n
∣ ∑ (−1)
n=2
24 2
22n
∣ ≤
= <1
(2n)!
4! 3
Thus,
∞
C(2) = −1 + ∑ (−1)
n
n=2
≤ −1 +
∞
2n
22n
n 2
≤ −1 + ∣ ∑ (−1)
∣
(2n)!
(2n)!
n=2
2
1
=− .
3
3
6. Show that for all x we have S(x + 2π) = S(x) and C(x + 2π) = C(x).
Remember that π/2 is the smallest positive value for which C(x) = 0. Thus,
S(x + 2π) = S ((x + 3π/2) + π/2) = S(x + 3π/2)C(π/2) + C(x + 3π/2)S(π/2)
= C(x + 3π/2) = C(x + 2π/2)C(π/2) − S(x + 2π/2)S(π/2)
= −S(x + π) = − [S(x + π/2)C(π/2) + C(x + π/2)S(π/2)]
= −C(x + π/2) = − [C(x)C(π/2 − S(x)S(π/2]
= S(x)
Easiest way to obtain the periodicity of C(X) is to differentiate the periodicity equation of
sine.
d
d
C(x + 2π) =
[S(x + 2π)] =
[S(x)] = C(x) .
dx
dx
3