# construction of tangent function from addition formula

###### Theorem 1.

If $x$ is a positive real number, then

 $0<\sqrt{1+{1\over x^{2}}}-{1\over x}<1$

(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  :

 $y^{2}<1+y^{2}<1+2y+y^{2}$

Taking square roots:

 $y<\sqrt{1+y^{2}}<1+y$

Subtracting $y$:

 $0\leq\sqrt{1+y^{2}}-y<1$

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

###### Definition 1.

Define the algebraic functions  $s\colon\{(x,y)\in\mathbb{R}^{2}\mid xy\neq 1\}\to\mathbb{R}$ and $h\colon(0,\infty)\to(0,1)$ and $g\colon(0,1)\to(0,1)$ as follows:

 $\displaystyle s(x,y)$ $\displaystyle={x+y\over 1-xy}$ (1) $\displaystyle h(x)$ $\displaystyle=\sqrt{1+{1\over x^{2}}}-{1\over x}$ (2) $\displaystyle g(x)$ $\displaystyle=h\left({1+x\over 1-x}\right)={\sqrt{x^{2}-2x+2}+x-1\over x+1}$ (3)
###### Theorem 2.

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

###### Proof.

Calculemus! On the one hand,

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

On the other hand,

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

These quantities are equal. ∎

###### Theorem 3.

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

###### Proof.

Calculemus rursum!

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

###### Theorem 4.

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

###### Theorem 5.

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

###### Proof.

Since $x>0$, we have

 $x^{2}+1

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

 $\sqrt{x^{2}+1}

Subtracting $1$ from both sides,

 $\sqrt{x^{2}+1}-1

Dividing by $x$ on both sides,

 $\sqrt{1+{1\over x^{2}}}-{1\over x}

or $h(x). ∎

###### Theorem 6.

Let $a$ be a positive real number. Then the sequence  $a,h(a),h(h(a)),h(h(h(a))),h(h(h(h(a)))),h(h(h(h(h(a))))),\ldots$

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 $\{t_{n}\}_{n=0}^{\infty}$ as follows:

 $\displaystyle t_{0}$ $\displaystyle=1$ $\displaystyle t_{n+1}$ $\displaystyle=h(t_{n})$

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 $\{r_{mn}\}$ by the following recursions:

 $\displaystyle r_{m0}$ $\displaystyle=0$ $\displaystyle r_{m\,n+1}$ $\displaystyle=s(r_{mn},t_{m})$

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 $xy\neq 1$, 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 $r_{mn}$, $r_{mn^{\prime}}$, and $r_{m\,n+n^{\prime}}$ are all well-defined. Then $r_{m\,n+n^{\prime}}=s(r_{mn},r_{mn^{\prime}})$.

###### 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 $r_{m\,n+n^{\prime}-1}=s(r_{mn},r_{m\,n^{\prime}-1})$. By definition, $r_{mn^{\prime}}=s(r_{m\,n^{\prime}-1},t_{m})$ and, by theorem 2, we have

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

###### Theorem 8.

If $n\leq 2^{m}$, then $r_{mn}$ is well-defined, $r_{mn}\leq 1$, and $r_{m-1\,n}=r_{m\,2n}$.

###### Proof.

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

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

Suppose we know that, for some number $n<2^{m}$, we find that $r_{m+1\,2n}$ is well-defined, strictly less than $1$ and equals $r_{m+1\,2n}$. By theorem 4, since $r_{mn}\leq 1$ and $r_{m\,n+1}\leq 1$, we may conclude that $h(r_{mn})<1$ and $h(r_{m\,n+1})<1$, which implies that $h(r_{mn})h(r_{m\,n+1})\neq 1$, so $s(h(r_{mn}),h(r_{m\,n+1}))$ is well-defined. By definition, $r_{m\,n+1}=s(r_{mn},t_{m})$, so $h(r_{m\,n+1})=s(h(r_{mn}),h(t_{m}))$. 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 $r_{m+1\,n}$. Combining the results of this paragraph, we may conclude that:

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

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

Moreover, by definition,

 $s(h(r_{mn}),h(r_{m\,n+1}))={h(r_{mn})+h(r_{m\,n+1})\over 1-h(r_{mn})h(r_{m\,n+% 1})}$

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

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

 $\displaystyle s(r_{m+1\,2n+1},t_{m+1})$ $\displaystyle=s(s(r_{m\,n},t_{m+1}),t_{m+1})$ $\displaystyle=s(r_{m\,n},s(t_{m+1},t_{m+1}))$ $\displaystyle=s(r_{m\,n},t_{m})$ $\displaystyle=r_{m\,n+1}$

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

Title construction of tangent function from addition formula ConstructionOfTangentFunctionFromAdditionFormula 2013-03-22 16:58:39 2013-03-22 16:58:39 rspuzio (6075) rspuzio (6075) 24 rspuzio (6075) Derivation msc 26A09