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

(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:

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 functions $s\mathrm{:}\mathrm{\{}\mathrm{(}x\mathrm{,}y\mathrm{)}\mathrm{\in }{\mathrm{R}}^{\mathrm{2}}\mathrm{\mid }x\mathit{}y\mathrm{\ne }\mathrm{1}\mathrm{\}}\mathrm{\to }\mathrm{R}$ and $h\mathrm{:}\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{\infty }\mathrm{)}\mathrm{\to }\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{)}$ and $g\mathrm{:}\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{)}\mathrm{\to }\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{)}$ as follows:

$s(x,y)$ $={\displaystyle \frac{x+y}{1-xy}}$ (1)

$h(x)$ $=\sqrt{1+{\displaystyle \frac{1}{x^2}}}-{\displaystyle \frac{1}{x}}$ (2)

$g(x)$ $=h\left({\displaystyle \frac{1+x}{1-x}}\right)={\displaystyle \frac{\sqrt{x^2-2x+2}+x-1}{x+1}}$ (3)

Theorem 2.

$s(s(x,y),z)=s(x,s(y,z))$

Proof.

Calculemus! On the one hand,

$$s(s(x,y),z)=\frac{\frac{x+y}{1-xy}+z}{1-\frac{x+y}{1-xy}z}=\frac{x+y+z-xyz}{1-xy-yz-zx}$$

On the other hand,

$$s(x,s(y,z))=\frac{x+\frac{y+z}{1-yz}}{1-x\frac{y+z}{1-yz}}=\frac{x+y+z-xyz}{1-xy-yz-zx}$$

These quantities are equal. ∎

Theorem 3.

$s(h(x),h(x))=x$

Proof.

Calculemus rursum!

$s(h(x),h(x))$ $={\displaystyle \frac{2\sqrt{1+\frac{1}{x^2}}-\frac{2}{x}}{1-{\left(\sqrt{1+\frac{1}{x^2}}-\frac{1}{x}\right)}^2}}$

$={\displaystyle \frac{2\sqrt{1+\frac{1}{x^2}}-\frac{2}{x}}{1-\left(1+\frac{2}{x^2}-\frac{2}{x}\sqrt{1+\frac{1}{x^2}}\right)}}$

$={\displaystyle \frac{2\sqrt{1+\frac{1}{x^2}}-\frac{2}{x}}{-\frac{1}{x}\left(\frac{2}{x}-2\sqrt{1+\frac{1}{x^2}}\right)}}=x$

∎

Theorem 4.

$s(h(x),h(y))=h(s(x,y))$

Theorem 5.

For all $x\mathrm{>}\mathrm{0}$, we have .

Proof.

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 . ∎

Theorem 6.

Let $a$ be a positive real number. Then the sequence

$$a,h(a),h(h(a)),h(h(h(a))),h(h(h(h(a)))),h(h(h(h(h(a))))),\mathrm{\dots }$$

converges to $\mathrm{0}$.

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 ${\mathrm{\{}{t}_n\mathrm{\}}}_{n\mathrm{=}\mathrm{0}}^{\mathrm{\infty }}$ 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 $\mathrm{\{}{r}_{m\mathit{}n}\mathrm{\}}$ by the following recursions:

$r_{m0}$ $=0$

$r_{mn+1}$ $=s(r_{mn},t_m)$

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 $xy\ne 1$, we do not know that $r_{mn}$ is well defined for all $m$ and $n$. In particular, if it should happen that $r_{mn}$ is well defined for some $m$ and $n$ but that $r_{mn}{t}_m=1$, then $r_{mk}$ will be undefined for all $k>m$.

Theorem 7.

Suppose that $r_{m\mathit{}n}$, $r_{m\mathit{}{n}^{\mathrm{\prime }}}$, and $r_{m\mathit{}n\mathrm{+}{n}^{\mathrm{\prime }}}$ are all well-defined. Then $r_{m\mathit{}n\mathrm{+}{n}^{\mathrm{\prime }}}\mathrm{=}s\mathit{}\mathrm{(}{r}_{m\mathit{}n}\mathrm{,}{r}_{m\mathit{}{n}^{\mathrm{\prime }}}\mathrm{)}$.

Proof.

We proceed by induction on $n^{\prime }$. If $n^{\prime }=0$, then $r_{m0}$ is defined to be $0$, and it is easy to see that $s(r_{mn},0)=r_{mn}$.

Suppose, then, that we know that $r_{mn+n^{\prime }-1}=s(r_{mn},r_{m{n}^{\prime }-1})$. By definition, $r_{m{n}^{\prime }}=s(r_{m{n}^{\prime }-1},t_m)$ and, by theorem 2, we have

$s(r_{mn},s(r_{m{n}^{\prime }-1},t_m))$ $=s(s(r_{mn},r_{m{n}^{\prime }-1}),t_m)$

$=s(r_{mn+n^{\prime }-1},t_m)$

$=r_{mn+n^{\prime }}$

∎

Theorem 8.

If $n\mathrm{\le }{\mathrm{2}}^m$, then $r_{m\mathit{}n}$ is well-defined, $r_{m\mathit{}n}\mathrm{\le }\mathrm{1}$, and $r_{m\mathrm{-}\mathrm{1}\mathit{}n}\mathrm{=}{r}_{m\mathit{}\mathrm{\hspace{0.17em}2}\mathit{}n}$.

Proof.

We shall proceed by induction on $m$. To begin, we note that $r_{00}\le 1$ because $r_{00}=0$. Also note that, if $m=0$, then $n=0$ is the only value for which the condition $n\le 2^m$ happens to be satisfied. The condition $r_{m-1n}=r_{m\mathrm{\hspace{0.17em}2}n}$ is not relevant when $n=0$.

Suppose that we know that, for a certain $m$, when $n\le 2^m$, then $r_{mn}$ is well-defined and $r_{mn}\le 1$. We will now make an induction on $n$ to show that if $n\le 2^{m+1}$, then $r_{m+1n}$ is well-defined, $r_{mn}\le 1$ and $r_{mn}=r_{m+\mathrm{1\hspace{0.17em}2}n}$. When $n=0$, we have, by definition, $r_{m+\mathrm{1\hspace{0.17em}0}}=0$ so the quantity is defined and it is obvious that $r_{mn}\le 1$ and $r_{mn}=r_{m+\mathrm{1\hspace{0.17em}2}n}$.

Suppose we know that, for some number , we find that $r_{m+\mathrm{1\hspace{0.17em}2}n}$ is well-defined, strictly less than $1$ and equals $r_{m+\mathrm{1\hspace{0.17em}2}n}$. By theorem 4, since $r_{mn}\le 1$ and $r_{mn+1}\le 1$, we may conclude that and , which implies that $h(r_{mn})h(r_{mn+1})\ne 1$, so $s(h(r_{mn}),h(r_{mn+1}))$ is well-defined. By definition, $r_{mn+1}=s(r_{mn},t_m)$, so $h(r_{mn+1})=s(h(r_{mn}),h(t_m))$. Recall that $h(t_m)=t_{m+1}$. By theorem 1, we have

$$s(h(r_{mn}),s(h(r_{mn}),t_{m+1}))=s(s(h(r_{mn}),h(r_{mn})),t_{m+1})).$$

By theorem 2, $s(h(r_{mn}),h(r_{mn}))$ equals $r_{mn}$ which, in turn, by our induction hypothesis, equals $r_{m+1n}$. Combining the results of this paragraph, we may conclude that:

$$s(h(r_{mn}),h(r_{mn+1}))=s(r_{m+\mathrm{1\hspace{0.17em}2}n},t_{m+1}),$$

which means that $r_{m+\mathrm{1\hspace{0.17em}2}n+1}$ is defined and equals $s(h(r_{mn}),h(r_{mn+1}))$.

Moreover, by definition,

$$s(h(r_{mn}),h(r_{mn+1}))=\frac{h(r_{mn})+h(r_{mn+1})}{1-h(r_{mn})h(r_{mn+1})}$$

Since $r_{mn+1}>r_{mn}$, we have $h(r_{mn+1})>h(r_{mn})$ as well. This implies that the numerator is less than $2h(r_{mn+1})$ and that the denominator is greater than $1-h(r_{mn+1}^2$. Hence, we have .

Since, as we have just shown, and, as we already know, , we have , so $r_{m+\mathrm{1\hspace{0.17em}2}n+2}$ is well-defined. Furthermore, we may evaluate this quantity using theorem 1:

$s(r_{m+\mathrm{1\hspace{0.17em}2}n+1},t_{m+1})$ $=s(s(r_{mn},t_{m+1}),t_{m+1})$

$=s(r_{mn},s(t_{m+1},t_{m+1}))$

$=s(r_{mn},t_m)$

$=r_{mn+1}$

Hence, we have $r_{m+12m+2}=r_{mn+1}$.

∎

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