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# 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

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

(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 $h(x)<x$.

###### Proof.

Since $x>0$, we have

$x^{2}+1<x^{4}+2x^{2}+1.$ |

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

$\sqrt{x^{2}+1}<x^{2}+1.$ |

Subtracting $1$ from both sides,

$\sqrt{x^{2}+1}-1<x^{2}.$ |

Dividing by $x$ on both sides,

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

or $h(x)<x$. ∎

###### Theorem 6.

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

$a,h(a),h(h(a)),h(h(h(a))),h(h(h(h(a)))),h(h(h(h(h(a))))),\ldots$ |

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

Since, as we have just shown, $r_{{m+1\,2n+1}}<1$ 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 $r_{{m+12m+2}}=r_{{m\,n+1}}$.

∎

## Mathematics Subject Classification

26A09*no label found*

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