construction of tangent function from addition formula

It is possible to define trigonometric functions rigorously using aprocedure based upon the addition formula for the tangent function.The idea is to first note a few purely algebraic facts and then usethese to show that a certain limiting process converges to a functionwhich satisfies the properties of the tangent function, from whichthe remaining trigonometric functions may be defined by purelyalgebraic operations.

Theorem 1.

If $x$ is a positive real number, then

(Here and henceforth, the square root sign denotesthe 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 theinequality 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, itmust 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 musthave $b=0$, so the sequence converges to $0$.∎

Having made these preliminary observations, we may now begin makingthe construction of the trigonometric function. We begin by definingthe 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 tendsto zero. These will be the values of the tangent function atsuccessive 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 wedid not specify the range of $m$ and $n$). Since $s(x,y)$ isonly 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 shouldhappen 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 tobe $0$, 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 forwhich 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 $r_{mn}$ iswell-defined and $r_{mn}\le 1$. We will now make an induction on $n$ to show thatif $n\le 2^{m+1}$, 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 ourinduction hypothesis, equals $r_{m+1n}$. Combining the results of thisparagraph, 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. Thisimplies that the numerator is less than $2h(r_{mn+1})$ and that the denominatoris 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 $r_{m+12m+2}=r_{mn+1}$.

∎