8.8 *Elementary 'Iranscendental Functions (Project) 539(d) 'Vx E JR., C(x + y) = C(x)C(y) - S(x)S(y);
(e) 'Vx E JR., S(x - y) = S(x)C(y) - C(x)S(y);
(f) 'Vx E JR., C(x - y) = C(x)C(y) + S(x)S(y).
Proof. To prove (a), let f(x) = S^2 (x) + C^2 (x), and prove that f'(x) = 0
for all x. This means that f must be a constant function. Find that constant
and you will be done.
To prove (c) and (d) we can use a similar approach, but with more com-
plicated functions. Let y be a fixed real number, and define F and G by
F(x) = S(x + y) - S(x)C(y) - C(x)S(y),
G(x) = C(x + y) - C(x)C(y) + S(x)S(y).Then prove that 'Vx E JR., F^2 (x) + G^2 (x) = 0 by the method used to prove (a).
Parts (e) and (f) follow easily from previously proved identities.
For a more challenging alternative approach, one could try to prove ( c) and
( d) by using Cauchy's product series formula on both terms of the right side of
the equation and adding the results. •
Lemma 8.8.14 There exists a positive real number t such that C(t) < 0.
Proof. Suppose that Vt> 0, C(t) ~ 0. Then Sis monotone increasing on
(O,oo) since S'(x) = C(x). Now,
S(l) = 1 - fi + tr -fi + · · · > 0.
Let x > l. By the mean value theorem, :Ju E (1,x) 3C(x) - C(l) = -S(u)
x-1
:S -S(l) since S is monotone increasing.Thus, C(x) - C(l) :S -(x - l)S(l), so
C(x) :S C(l) - (x - l)S(l).Now, when x is sufficiently large, C(l) - (x - l)S(l) < 0. Thus, :Jx > 0 3
C(x) < 0. Contradiction. Therefore, :Jt > 0 3 C(t) < 0. •
Lemma 8.8.15 There is a smallest positive real number t such that C(t) = 0.
Proof. Let A= {t ~ 0: C(t) = O}. To see that A is nonempty, use Lemma
8.8.14 and the intermediate value theorem. Show that :J u = inf A , and then
use Exercise 5.1.21, Theorem 3.2.8, and Exercise 3.2.7 to show that u EA. •