376
IV - Powers, Exponentials, Logarithms, Trigonometric Functions
(14.14)
cos (x + a) = - sinx,
sin(x + a) = cosx
for any x real or complex.
These relations show that, on C, all multiples of 4a, i.e. of 271", are periods
of the trigonometric functions. In fact, they have no other in lR (nor, as we
shall see later, in C, leaving tan aside). For let 4b E lR be a period. By adding
a number of the form 4ka, with k E Z, we can assume 0 ~ b < a, and so we
need to show that b = o.
Now, since cos4b = cosO = 1, the relation cosx = 2 cos 2 (x/2) - 1 shows
that either cos 2b = 1, or cos 2b = -1. The second hypothesis implies cos b = 0
by virtue of the same relation; since b < a, least root ~ 0 of the cosine, this
is absurd. If on the other hand cos 2b = 1, then cos b = 1 or -1; but the
inequalities (9) show that -1 < cosx < 1 for 0 < x < 2 as we saw above,
with strict inequalities; thus b = 0, qed.
This result also shows that a is the only root of cos x between 0 and 2.
Indeed, let a' be another root between 0 and 2. Since sin a' > 0 by (11), we
have sin a' = 1, and since cos a' = 0 we see (addition formulae) that 4a' is a
period like 4a. So a' = ka for an integer k > O. If k = 1, we have finished.
If k > 1, then a ~ a' /2 < 1 then, by (9), cosx is obviously> 1/2 between 0
and 1, qed.
We can even, better than nothing, verify that 2a (= 71") > 3 i.e. that
the number a introduced above is > 3/2. Since cosa = 0, (8) shows that
1 - a 2 /2 < 0, i.e. a > V2; foiled. But the argument leading to (8) shows as
well that
(0 < x < 2),
whence, by a short calculation, cos 3/2 > 679/2560 > O. Since we have seen
that cos 2 < 0, the function cosx must have a root between 3/2 and 2, and
since its only root between 0 and 2 is a, we obtain the required inequality.
To conclude this section we ought to rewrite the formulae (12), (13) and
(14), replacing a by 71"/2, but one may assume that the reader would then say,
like a Parisian humorist d propos a best seller: "I haven't read it, I haven't
seen it, though I've heard it mentioned".
(viii) Arguments of a complex number. Every non zero complex number
can be put in the form
(14.15)
z=r(u+iv)
in a unique way, with r = Izl real> 0, and u and v real satisfying
(14.16)
whence lui ::; 1. Since the function cosine takes the values 1 and -1 and
is continuous, the intermediate value theorem, already invoked, ensures the
existence of atE IR such that u = cos t. Then sin 2 t = v 2 , so either v = sin t,
IV - Powers, Exponentials, Logarithms, Trigonometric Functions
(14.14)
cos (x + a) = - sinx,
sin(x + a) = cosx
for any x real or complex.
These relations show that, on C, all multiples of 4a, i.e. of 271", are periods
of the trigonometric functions. In fact, they have no other in lR (nor, as we
shall see later, in C, leaving tan aside). For let 4b E lR be a period. By adding
a number of the form 4ka, with k E Z, we can assume 0 ~ b < a, and so we
need to show that b = o.
Now, since cos4b = cosO = 1, the relation cosx = 2 cos 2 (x/2) - 1 shows
that either cos 2b = 1, or cos 2b = -1. The second hypothesis implies cos b = 0
by virtue of the same relation; since b < a, least root ~ 0 of the cosine, this
is absurd. If on the other hand cos 2b = 1, then cos b = 1 or -1; but the
inequalities (9) show that -1 < cosx < 1 for 0 < x < 2 as we saw above,
with strict inequalities; thus b = 0, qed.
This result also shows that a is the only root of cos x between 0 and 2.
Indeed, let a' be another root between 0 and 2. Since sin a' > 0 by (11), we
have sin a' = 1, and since cos a' = 0 we see (addition formulae) that 4a' is a
period like 4a. So a' = ka for an integer k > O. If k = 1, we have finished.
If k > 1, then a ~ a' /2 < 1 then, by (9), cosx is obviously> 1/2 between 0
and 1, qed.
We can even, better than nothing, verify that 2a (= 71") > 3 i.e. that
the number a introduced above is > 3/2. Since cosa = 0, (8) shows that
1 - a 2 /2 < 0, i.e. a > V2; foiled. But the argument leading to (8) shows as
well that
(0 < x < 2),
whence, by a short calculation, cos 3/2 > 679/2560 > O. Since we have seen
that cos 2 < 0, the function cosx must have a root between 3/2 and 2, and
since its only root between 0 and 2 is a, we obtain the required inequality.
To conclude this section we ought to rewrite the formulae (12), (13) and
(14), replacing a by 71"/2, but one may assume that the reader would then say,
like a Parisian humorist d propos a best seller: "I haven't read it, I haven't
seen it, though I've heard it mentioned".
(viii) Arguments of a complex number. Every non zero complex number
can be put in the form
(14.15)
z=r(u+iv)
in a unique way, with r = Izl real> 0, and u and v real satisfying
(14.16)
whence lui ::; 1. Since the function cosine takes the values 1 and -1 and
is continuous, the intermediate value theorem, already invoked, ensures the
existence of atE IR such that u = cos t. Then sin 2 t = v 2 , so either v = sin t,