1. Transcendency Of The Trigonometric Functions
In this section were going to show that the trigonometric
functions and their inverses are transcendental. For the definition of
transcendental functions, see Section 6.3.1 Definition 3.1.
We see in Section
6.3.1 Part 3 that sin x is a transcendental function.
Intuitively it can't be expressed as a finite combination of
normal algebraic operations applied explicitly to polynomials in x, neither can cos x and tan x. Now we're going to confirm the
transcendency of these basic trigonometric functions sin x, cos x, and tan x, and of their respective reciprocals csc x, sec x,
and cot x. But first we deal with the reciprocals of functions.
Theorem 1.1 Transcendency Of Reciprocals Of Transcendental Functions
A function is transcendental iff its reciprocal is transcendental.
Suppose f is transcendental. If its reciprocal 1/ f is algebraic, then there exist polynomials pi(x)'s, i = n, n 1, n 2, , 1, 0,
such that the equation:
is satisfied for all x in dom(1/ f ). Multiplying both sides of the above equation by f n(x) we get:
pn(x) + pn1(x) f (x) + pn2(x) f 2 (x) + ... + p1(x) f n1(x) + p0(x) f n(x) = 0,
p0(x) f n(x) + p1(x) f n1(x) + ... + pn2(x) f 2(x) + pn1(x) f (x) + pn(x) = 0,
which shows that f (x) is algebraic. This is a contradiction.
So 1/ f is transcendental. Conversely, if 1/ f
its reciprocal 1/(1/ f ) = f is transcendental.
Transcendency Of The Trigonometric Functions
Let's consider our now-familiar equation:
pn(x) yn + pn1(x) yn1 + pn2(x) yn2 + + p1(x) y + p0(x) = 0.
Its left-hand side is a polynomial in function y with coefficients pi(x)s that are
polynomials in real variable x
coefficients. Of course it depends on both x and y. So lets denote it by P(x, y), so that:
P(x, y) = pn(x) yn + pn1(x) yn1 + pn2(x) yn2 + + p1(x) y + p0(x).
The degree of a polynomial pi(x) doesnt have to be i or anything else. It
can be any integer greater than or equal to 0. The
largest exponent n of y is called the y-degree of P(x, y). This P corresponds to the particular set of the pis. For a different
set of the pjs, theres a different P. In the particular case where y = sin x:
P(x, sin x) = pn(x) sinn x + pn1(x) sinn1 x + pn2(x) sinn2 x + + p1(x) sin x + p0(x),
is the (sin x)-degree
of P(x, sin x). Recall that when
there are conditions on n,
is the y-degree,
not the subscript in
pn(x), which is simply a naming subscript.
The trigonometric functions sin x,
and csc x
are transcendental. They belong to a class of
We'll prove that sin x is transcendental. The proofs for cos x and tan x are similar to that for sin x, and are left as problem & solution 4 and 5 respectively. Because, by Theorem 1.1, a reciprocal of a transcendental function is also transcendental, cot x,
sec x, and csc x are transcendental.
2. Transcendency Of The Inverse Trigonometric Functions
We're now going to show and state as a theorem that the
inverse function of a transcendental function is also transcendental.
As a consequence, since the principal-value trigonometric functions sin x, cos x, tan x, cot x, sec x, and csc x are
transcendental, their respective inverses arcsin x, arccos x, arctan x, arccot x, arcsec x, and arccsc x are also
transcendental. This property will be stated as a corollary of the theorem.
Theorem 2.1 Transcendency Of The Inverse Functions Of Transcendental Functions
A function is transcendental iff its inverse function is transcendental.
The inverse relation of a function may or may not be a function. Of course we require that the inverse be a function. That's
why in the theorem we specify inverse function, we don't just say "inverse".
Adding these n + 1
identities we get an identity where the left-hand side is 0. As for the
right-hand side, factor all the like
powers of f (x), and re-arrange the terms in decreasing order of power of f (x). Then, exchanging the sides, we obtain an
equation of the form:
qm(x)( f (x))m + qm1(x)( f (x))m1 + qm2(x)( f (x))m2 + q1(x) f (x) + q0(x) = 0,
where the qi(x)'s are polynomials, m is the maximum of
the mk's, and q0(x)
= na0xn + n1a0xn1 + n2a0xn2 +
The above equation is satisfied by for all x in dom( f ). This shows that f (x) is algebraic, a contradiction. Thus f 1 must be
transcendental. Conversely, if f 1 is transcendental, then its inverse function ( f 1) 1 = f is transcendental.
Corollary 2.1 Transcendency Of The Inverse Trigonometric Functions
The inverse trigonometric functions arcsin x, arccos x, arctan x, arccot x, arcsec x, and arccsc x are transcendental.
Problems & Solutions
1. Is it possible that:
Justify your answer.
No, because cos x is transcendental and thus can't be
constructed by a finite combination of algebraic operations.
It can be shown that:
as done in Section
15.3 Part 3. The function cos x can be expressed as an infinite combination of additions and
alternatingly. For any non-0 x, any finite portion of that infinite combination can only be an approximate value and can never
be the exact value of cos x. For x = 0, as cos 0 = 1, any finite portion containing the first term 1 is the exact value of cos 0.
2. Give an example where the sum of two
transcendental functions is transcendental. Hint: In Section
6.3.1 Theorem 3.1 it
was shown that for any non-0 constant c, a function f (x) is transcendental iff c f (x) is transcendental.
Since sin x is transcendental, 2 sin
x is transcendental.
x + 2 sin
x = 3 sin
x and 3 sin
x is transcendental.
sin x + 2 sin x is transcendental.
3. Give an example where the sum of two transcendental functions is algebraic. Hint: See Hint of Problem & Solution 2.
Since sin x is transcendental, sin
x = (1) sin
x is transcendental.
x + ( sin
x) = 0 for all x and the function f (x) =
0 for all x is algebraic. So sin x + ( sin x) is algebraic.
4. Prove that the function cos x is transcendental.
5. Prove that the function tan x is transcendental.