construction of tangent function from addition formula

construction of tangent function from addition formula

It is possible to define

trigonometric functions

rigorously using a
procedure based upon the

addition formula

for the

tangent function

.
The idea is to first note a few purely

algebraic

facts and then use
these to show that a certain limiting process

converges

to a

function

which

satisfies

the

properties

of the tangent function, from which
the remaining trigonometric functions may be defined by purely
algebraic operations.

Theorem 1

.

If

x

is a positive real number, then

0

<

1

+

1

x

2

-

1

x

<

1

(Here and henceforth, the

square root

sign denotes
the

positive

square root.)

Proof.

Let

y

=

1

/

x

. Then

y

is also a positive real number.
We have the following

inequalities

:

y

2

<

1

+

y

2

<

1

+

2

⁢

y

+

y

2

Taking square roots:

y

<

1

+

y

2

<

1

+

y

Subtracting

y

:

0

≤

1

+

y

2

-

y

<

1

Remembering the

definition

of

y

, this is the
inequality which we set out to demonstrate.
∎

Definition 1

.

Define the

algebraic functions

s

:

{

(

x

,

y

)

∈

R

2

∣

x

⁢

y

≠

1

}

→

R

and

h

:

(

0

,

∞

)

→

(

0

,

1

)

and

g

:

(

0

,

1

)

→

(

0

,

1

)

as follows:

s

⁢

(

x

,

y

)

=

x

+

y

1

-

x

⁢

y

(1)

h

⁢

(

x

)

=

1

+

1

x

2

-

1

x

(2)

g

⁢

(

x

)

=

h

⁢

(

1

+

x

1

-

x

)

=

x

2

-

2

⁢

x

+

2

+

x

-

1

x

+

1

(3)

Theorem 2

.

s

⁢

(

s

⁢

(

x

,

y

)

,

z

)

=

s

⁢

(

x

,

s

⁢

(

y

,

z

)

)

Proof.

Calculemus! On the one hand,

s

⁢

(

s

⁢

(

x

,

y

)

,

z

)

=

x

+

y

1

-

x

⁢

y

+

z

1

-

x

+

y

1

-

x

⁢

y

⁢

z

=

x

+

y

+

z

-

x

⁢

y

⁢

z

1

-

x

⁢

y

-

y

⁢

z

-

z

⁢

x

On the other hand,

s

⁢

(

x

,

s

⁢

(

y

,

z

)

)

=

x

+

y

+

z

1

-

y

⁢

z

1

-

x

⁢

y

+

z

1

-

y

⁢

z

=

x

+

y

+

z

-

x

⁢

y

⁢

z

1

-

x

⁢

y

-

y

⁢

z

-

z

⁢

x

These quantities are equal.
∎

Theorem 3

.

s

⁢

(

h

⁢

(

x

)

,

h

⁢

(

x

)

)

=

x

Proof.

Calculemus rursum!

s

⁢

(

h

⁢

(

x

)

,

h

⁢

(

x

)

)

=

2

⁢

1

+

1

x

2

-

2

x

1

-

(

1

+

1

x

2

-

1

x

)

2

=

2

⁢

1

+

1

x

2

-

2

x

1

-

(

1

+

2

x

2

-

2

x

⁢

1

+

1

x

2

)

=

2

⁢

1

+

1

x

2

-

2

x

-

1

x

⁢

(

2

x

-

2

⁢

1

+

1

x

2

)

=

x

∎

Theorem 4

.

s

⁢

(

h

⁢

(

x

)

,

h

⁢

(

y

)

)

=

h

⁢

(

s

⁢

(

x

,

y

)

)

Theorem 5

.

For all

x

>

0

, we have

h

⁢

(

x

)

<

x

.

Proof.

Since

x

>

0

, we have

x

2

+

1

<

x

4

+

2

⁢

x

2

+

1

.

By the binomial identity, the right-hand side equals

(

x

+

1

)

2

.
Taking square roots of both sides,

x

2

+

1

<

x

2

+

1

.

Subtracting

1

from both sides,

x

2

+

1

-

1

<

x

2

.

Dividing by

x

on both sides,

1

+

1

x

2

-

1

x

<

x

,

or

h

⁢

(

x

)

<

x

.
∎

Theorem 6

.

Let

a

be a positive real number. Then the

sequence

a

,

h

⁢

(

a

)

,

h

⁢

(

h

⁢

(

a

)

)

,

h

⁢

(

h

⁢

(

h

⁢

(

a

)

)

)

,

h

⁢

(

h

⁢

(

h

⁢

(

h

⁢

(

a

)

)

)

)

,

h

⁢

(

h

⁢

(

h

⁢

(

h

⁢

(

h

⁢

(

a

)

)

)

)

)

,

…

converges to

0

.

Proof.

By the foregoing

theorem

, this sequence is

decreasing

. Hence, it
must converge to its

infimum

. Call this infimum

b

. Suppose that

b

>

0

. Then, since

h

is

continuous

, we must have

h

⁢

(

b

)

=

b

,
which is not possible by the foregoing theorem. Hence, we must
have

b

=

0

, so the sequence converges to

0

.
∎

Having made these preliminary observations, we may now begin making
the construction of the trigonometric function. We begin by defining
the tangent function for successive bisections of a right angle.

Definition 2

.

Define the sequence

{

t

n

}

n

=

0

∞

as follows:

t

0

=

1

t

n

+

1

=

h

⁢

(

t

n

)

By the forgoing theorem, this is a

decreasing sequence

which tends
to zero. These will be the values of the tangent function at
successive bisections of the right angle. We now use our function

s

to construct other values of the tangent function.

Definition 3

.

Define the sequence

{

r

m

⁢

n

}

by the following

recursions

:

r

m

⁢

0

=

0

r

m

⁢

n

+

1

=

s

⁢

(

r

m

⁢

n

,

t

m

)

There is a subtlety involved in this definition (which is why we
did not specify the range of

m

and

n

). Since

s

⁢

(

x

,

y

)

is
only

well-defined

when

x

⁢

y

≠

1

, we do not know that

r

m

⁢

n

is well defined for all

m

and

n

. In particular, if it should
happen that

r

m

⁢

n

is well defined for some

m

and

n

but that

r

m

⁢

n

⁢

t

m

=

1

, then

r

m

⁢

k

will be undefined for all

k

>

m

.

Theorem 7

.

Suppose that

r

m

⁢

n

,

r

m

⁢

n

′

, and

r

m

⁢

n

+

n

′

are all well-defined.
Then

r

m

⁢

n

+

n

′

=

s

⁢

(

r

m

⁢

n

,

r

m

⁢

n

′

)

.

Proof.

We proceed by

induction

on

n

′

. If

n

′

=

0

, then

r

m

⁢

0

is defined to
be

0

, and it is easy to see that

s

⁢

(

r

m

⁢

n

,

0

)

=

r

m

⁢

n

.

Suppose, then, that we know that

r

m

⁢

n

+

n

′

-

1

=

s

⁢

(

r

m

⁢

n

,

r

m

⁢

n

′

-

1

)

. By definition,

r

m

⁢

n

′

=

s

⁢

(

r

m

⁢

n

′

-

1

,

t

m

)

and,
by theorem 2, we have

s

⁢

(

r

m

⁢

n

,

s

⁢

(

r

m

⁢

n

′

-

1

,

t

m

)

)

=

s

⁢

(

s

⁢

(

r

m

⁢

n

,

r

m

⁢

n

′

-

1

)

,

t

m

)

=

s

⁢

(

r

m

⁢

n

+

n

′

-

1

,

t

m

)

=

r

m

⁢

n

+

n

′

∎

Theorem 8

.

If

n

≤

2

m

, then

r

m

⁢

n

is well-defined,

r

m

⁢

n

≤

1

, and

r

m

-

1

⁢

n

=

r

m

⁢

2

⁢

n

.

Proof.

We shall proceed by induction on

m

. To begin, we note that

r

00

≤

1

because

r

00

=

0

. Also note that, if

m

=

0

, then

n

=

0

is the only value for
which the condition

n

≤

2

m

happens to be satisfied. The condition

r

m

-

1

⁢

n

=

r

m

⁢

2

⁢

n

is not relevant when

n

=

0

.

Suppose that we know that, for a certain

m

, when

n

≤

2

m

, then

r

m

⁢

n

is
well-defined and

r

m

⁢

n

≤

1

. We will now make an induction on

n

to show that
if

n

≤

2

m

+

1

, then

r

m

+

1

⁢

n

is well-defined,

r

m

⁢

n

≤

1

and

r

m

⁢

n

=

r

m

+

1 2

⁢

n

. When

n

=

0

, we have, by definition,

r

m

+

1 0

=

0

so the quantity is defined and it is

obvious

that

r

m

⁢

n

≤

1

and

r

m

⁢

n

=

r

m

+

1 2

⁢

n

.

Suppose we know that, for some number

n

<

2

m

, we find that

r

m

+

1 2

⁢

n

is well-defined,

strictly

less than

1

and equals

r

m

+

1 2

⁢

n

. By theorem 4,
since

r

m

⁢

n

≤

1

and

r

m

⁢

n

+

1

≤

1

, we may conclude that

h

⁢

(

r

m

⁢

n

)

<

1

and

h

⁢

(

r

m

⁢

n

+

1

)

<

1

, which implies that

h

⁢

(

r

m

⁢

n

)

⁢

h

⁢

(

r

m

⁢

n

+

1

)

≠

1

,
so

s

⁢

(

h

⁢

(

r

m

⁢

n

)

,

h

⁢

(

r

m

⁢

n

+

1

)

)

is well-defined. By definition,

r

m

⁢

n

+

1

=

s

⁢

(

r

m

⁢

n

,

t

m

)

, so

h

⁢

(

r

m

⁢

n

+

1

)

=

s

⁢

(

h

⁢

(

r

m

⁢

n

)

,

h

⁢

(

t

m

)

)

. Recall that

h

⁢

(

t

m

)

=

t

m

+

1

. By theorem 1, we have

s

(

h

(

r

m

⁢

n

)

,

s

(

h

(

r

m

⁢

n

)

,

t

m

+

1

)

)

=

s

(

s

(

h

(

r

m

⁢

n

)

,

h

(

r

m

⁢

n

)

)

,

t

m

+

1

)

)

.

By theorem 2,

s

⁢

(

h

⁢

(

r

m

⁢

n

)

,

h

⁢

(

r

m

⁢

n

)

)

equals

r

m

⁢

n

which, in turn, by our

induction hypothesis

, equals

r

m

+

1

⁢

n

. Combining the results of this
paragraph, we may conclude that:

s

⁢

(

h

⁢

(

r

m

⁢

n

)

,

h

⁢

(

r

m

⁢

n

+

1

)

)

=

s

⁢

(

r

m

+

1 2

⁢

n

,

t

m

+

1

)

,

which means that

r

m

+

1 2

⁢

n

+

1

is defined and equals

s

⁢

(

h

⁢

(

r

m

⁢

n

)

,

h

⁢

(

r

m

⁢

n

+

1

)

)

.

Moreover, by definition,

s

⁢

(

h

⁢

(

r

m

⁢

n

)

,

h

⁢

(

r

m

⁢

n

+

1

)

)

=

h

⁢

(

r

m

⁢

n

)

+

h

⁢

(

r

m

⁢

n

+

1

)

1

-

h

⁢

(

r

m

⁢

n

)

⁢

h

⁢

(

r

m

⁢

n

+

1

)

Since

r

m

⁢

n

+

1

>

r

m

⁢

n

, we have

h

⁢

(

r

m

⁢

n

+

1

)

>

h

⁢

(

r

m

⁢

n

)

as well. This
implies that the

numerator

is less than

2

⁢

h

⁢

(

r

m

⁢

n

+

1

)

and that the denominator
is

greater

than

1

-

h

(

r

m

⁢

n

+

1

2

. Hence, we have

r

m

+

1 2

⁢

n

+

1

<

s

(

h

(

r

m

⁢

n

+

1

,

h

(

r

m

⁢

n

+

1

)

=

h

(

r

m

⁢

n

+

1

<

1

.

Since, as we have just shown,

r

m

+

1 2

⁢

n

+

1

<

1

and, as we already know,

t

m

+

1

<

1

, we have

r

m

+

1 2

⁢

n

+

1

⁢

t

m

+

1

<

1

, so

r

m

+

1 2

⁢

n

+

2

is well-defined. Furthermore, we may evaluate this quantity using theorem 1:

s

⁢

(

r

m

+

1 2

⁢

n

+

1

,

t

m

+

1

)

=

s

⁢

(

s

⁢

(

r

m

⁢

n

,

t

m

+

1

)

,

t

m

+

1

)

=

s

⁢

(

r

m

⁢

n

,

s

⁢

(

t

m

+

1

,

t

m

+

1

)

)

=

s

⁢

(

r

m

⁢

n

,

t

m

)

=

r

m

⁢

n

+

1

Hence, we have

r

m

+

12

⁢

m

+

2

=

r

m

⁢

n

+

1

.

∎

Title

construction of tangent function from addition formula

Canonical

name

ConstructionOfTangentFunctionFromAdditionFormula

Date of creation

2013-03-22 16:58:39

Last modified on

2013-03-22 16:58:39

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

24

Author

rspuzio (6075)

Entry type

Derivation

Classification

msc 26A09

Generated on Fri Feb 9 19:46:42 2018 by

LaTeXML