# 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 |