construction of tangent function from addition formula

It is possible to define trigonometric functionsDlmfMathworldPlanetmath rigorously using a procedure based upon the addition formulaPlanetmathPlanetmath 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 convergesPlanetmathPlanetmath to a functionMathworldPlanetmath 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


(Here and henceforth, the square rootMathworldPlanetmath sign denotes the positive square root.)


Let y=1/x. Then y is also a positive real number. We have the following inequalitiesMathworldPlanetmath:


Taking square roots:


Subtracting y:


Remembering the definition of y, this is the inequality which we set out to demonstrate. ∎

Definition 1.

Define the algebraic functionsMathworldPlanetmath s:{(x,y)R2xy1}R and h:(0,)(0,1) and g:(0,1)(0,1) as follows:

s(x,y) =x+y1-xy (1)
h(x) =1+1x2-1x (2)
g(x) =h(1+x1-x)=x2-2x+2+x-1x+1 (3)
Theorem 2.



Calculemus! On the one hand,


On the other hand,


These quantities are equal. ∎

Theorem 3.



Calculemus rursum!

s(h(x),h(x)) =21+1x2-2x1-(1+1x2-1x)2

Theorem 4.


Theorem 5.

For all x>0, we have h(x)<x.


Since x>0, we have


By the binomial identity, the right-hand side equals (x+1)2. Taking square roots of both sides,


Subtracting 1 from both sides,


Dividing by x on both sides,


or h(x)<x. ∎

Theorem 6.

Let a be a positive real number. Then the sequenceMathworldPlanetmath


converges to 0.


By the foregoing theoremMathworldPlanetmath, this sequence is decreasing. Hence, it must converge to its infimumMathworldPlanetmath. Call this infimum b. Suppose that b>0. Then, since h is continuousMathworldPlanetmathPlanetmath, 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 {tn}n=0 as follows:

t0 =1
tn+1 =h(tn)

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 {rmn} by the following recursions:

rm0 =0
rmn+1 =s(rmn,tm)

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 xy1, we do not know that rmn is well defined for all m and n. In particular, if it should happen that rmn is well defined for some m and n but that rmntm=1, then rmk will be undefined for all k>m.

Theorem 7.

Suppose that rmn, rmn, and rmn+n are all well-defined. Then rmn+n=s(rmn,rmn).


We proceed by inductionMathworldPlanetmath on n. If n=0, then rm0 is defined to be 0, and it is easy to see that s(rmn,0)=rmn.

Suppose, then, that we know that rmn+n-1=s(rmn,rmn-1). By definition, rmn=s(rmn-1,tm) and, by theorem 2, we have

s(rmn,s(rmn-1,tm)) =s(s(rmn,rmn-1),tm)

Theorem 8.

If n2m, then rmn is well-defined, rmn1, and rm-1n=rm 2n.


We shall proceed by induction on m. To begin, we note that r001 because r00=0. Also note that, if m=0, then n=0 is the only value for which the condition n2m happens to be satisfied. The condition rm-1n=rm 2n is not relevant when n=0.

Suppose that we know that, for a certain m, when n2m, then rmn is well-defined and rmn1. We will now make an induction on n to show that if n2m+1, then rm+1n is well-defined, rmn1 and rmn=rm+1 2n. When n=0, we have, by definition, rm+1 0=0 so the quantity is defined and it is obvious that rmn1 and rmn=rm+1 2n.

Suppose we know that, for some number n<2m, we find that rm+1 2n is well-defined, strictly less than 1 and equals rm+1 2n. By theorem 4, since rmn1 and rmn+11, we may conclude that h(rmn)<1 and h(rmn+1)<1, which implies that h(rmn)h(rmn+1)1, so s(h(rmn),h(rmn+1)) is well-defined. By definition, rmn+1=s(rmn,tm), so h(rmn+1)=s(h(rmn),h(tm)). Recall that h(tm)=tm+1. By theorem 1, we have


By theorem 2, s(h(rmn),h(rmn)) equals rmn which, in turn, by our induction hypothesis, equals rm+1n. Combining the results of this paragraph, we may conclude that:

s(h(rmn),h(rmn+1))=s(rm+1 2n,tm+1),

which means that rm+1 2n+1 is defined and equals s(h(rmn),h(rmn+1)).

Moreover, by definition,


Since rmn+1>rmn, we have h(rmn+1)>h(rmn) as well. This implies that the numerator is less than 2h(rmn+1) and that the denominator is greater than 1-h(rmn+12. Hence, we have rm+1 2n+1<s(h(rmn+1,h(rmn+1)=h(rmn+1<1.

Since, as we have just shown, rm+1 2n+1<1 and, as we already know, tm+1<1, we have rm+1 2n+1tm+1<1, so rm+1 2n+2 is well-defined. Furthermore, we may evaluate this quantity using theorem 1:

s(rm+1 2n+1,tm+1) =s(s(rmn,tm+1),tm+1)

Hence, we have rm+12m+2=rmn+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