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# 13.6: Pythagorean theorem

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Recall that $$\cosh$$ denotes hyperbolic cosine; that is, the function defined by

$$\cosh x := \dfrac{e^x+e^{-x}}2.$$

Theorem $$\PageIndex{1}$$ Hyperbolic Pythagorean theorem

Assume that $$ACB$$ is an h-triangle with right angle at $$C$$. Set

$$a = BC_h,$$     $$b = CA_h$$     and     $$c = AB_h$$.

Then

$\cosh c = \cosh a \cdot \cosh b.$

The formula 13.6.1 will be proved by means of direct calculations. Before giving the proof, let us discuss the limit cases of this formula.

Note that $$\cosh x$$ can be written using the Taylor expansion

$$\cosh x=1+\dfrac{1}{2}\cdot x^2+\dfrac{1}{24}\cdot x^4+\dots.$$

It follows that if $$a$$, $$b$$, and $$c$$ are small, then

$$\begin{array} {rcl} {1 + \dfrac{1}{2} \cdot c^2} & \approx & {\cosh c = \cosh a \cdot \cosh b \approx} \\ {} & \approx & {(1 + \dfrac{1}{2} \cdot a^2)(1 + \dfrac{1}{2} \cdot b^2) \approx} \\ {} & \approx & {1 + \dfrac{1}{2} \cdot (a^2 + b^2).} \end{array}$$

In other words, the original Pythagorean theorem (Theorem 6.2.1) is a limit case of the hyperbolic Pythagorean theorem for small triangles.

For large $$a$$ and $$b$$ the terms $$e^{-a}$$, $$e^{-b}$$, and $$e^{-a-b+\ln 2}$$ are neglectable. In this case we have the following approximations:

$$\begin{array} {\cosh a \cdot \cos h b} & \approx & {\dfrac{e^a}{2} \cdot \dfrac{e^b}{2} =} \\ {} & = & {\dfrac{e^{a+b-\ln 2}}{2} \approx} \\ {} & \approx & {\cosh (a+b-\ln 2)} \end{array}$$

Therefore $$c \approx a+b-\ln 2$$.

Exercise $$\PageIndex{1}$$

Assume that $$ACB$$ is an h-triangle with right angle at $$C$$. Set $$a=BC_h$$, $$b=CA_h$$, and $$c=AB_h$$. Show that

$$c+\ln 2>a+b.$$

Hint

Apply the hyperbolic Pythagorean theorem and the definition of hyperbolic cosine. The following observations should help:

• $$x \mapsto e^x$$ is an increasing positive function.
• By the triangle inequality, we have $$-c \le a - b$$ and $$-c \le b - a$$ In the proof of the hyperbolic Pythagorean theorem we use the following formula from Exercise 12.9.2:

$$\cosh AB_h=\dfrac{AB\cdot A'B'+AB'\cdot A'B}{AA'\cdot BB'},$$

here $$A$$, $$B$$ are h-points distinct from the center of absolute and $$A'$$, $$B'$$ are their inversions in the absolute. This formula is derived in the hints.

Proof of Theorem $$\PageIndex{1}$$. We assume that absolute is a unit circle. By the main observation (Theorem 12.3.1) we can assume that $$C$$ is the center of absolute. Let $$A'$$ and $$B'$$ denote the inverses of $$A$$ and $$B$$ in the absolute.

Set $$x=BC$$, $$y=AC$$. By Lemma Theorem 12.3.2

$$a = \ln \dfrac{1 + x}{1 - x}$$,                             $$b = \ln \dfrac{1 + y}{1 - y}.$$

Therefore

$\begin{array} {rclcrcl} {\cosh a} & = & {\dfrac{1}{2} \cdot (\dfrac{1 + x}{1 - x} + \dfrac{1 - x}{1 + x}) =} & \ \ \ \ & {\cosh b} & = & {\dfrac{1}{2} \cdot (\dfrac{1 + y}{1 - y} + \dfrac{1 - y}{1 + y}) =} \\ {} & = & {\dfrac{1 + x^2}{1-x^2},} & \ \ \ \ & {} & = & {\dfrac{1 + y^2}{1 - y^2}.} \end{array}$

Note that

$$B'C = \dfrac{1}{x},$$                             $$A'C = \dfrac{1}{y}.$$

Therefore

$$BB' = \dfrac{1}{x} - x$$,                                 $$AA' = \dfrac{1}{y} - y.$$

Since the triangles $$ABC$$, $$A'BC$$, $$AB'C$$, $$A'B'C$$ are right, the original Pythagoran theorem (Theorem 6.2.1) implies

$$\begin{array} {rcl} {AB} & = & {\sqrt{x^2 + y^2}} \\ {A'B} & = & {\sqrt{x^2 + \dfrac{1}{y^2}},} \end{array}$$               $$\begin{array} {rcl} {AB'} & = & {\sqrt{\dfrac{1}{x^2} + y^2,}} \\ {A'B'} & = & {\sqrt{\dfrac{1}{x^2} + \dfrac{1}{y^2}}.} \end{array}$$

According to Exercise 12.9.2,

$\begin{array} {rcl} {\cosh c} & = & {\dfrac{AB \cdot A'B' + AB' \cdot A'B}{AA' \cdot BB'} = } \\ {} & = & {\dfrac{\sqrt{x^2 + y^2} \cdot \sqrt{\dfrac{1}{x^2} + \dfrac{1}{y^2}} + \sqrt{\dfrac{1}{x^2} + y^2} \cdot \sqrt{x^2 + \dfrac{1}{y^2}}}{(\dfrac{1}{y} - y) \cdot (\dfrac{1}{x} - x)} =} \\ {} & = & {\dfrac{x^2 + y^2 + 1 + x^2 \cdot y^2}{(1 - y^2) \cdot (1 - x^2)} =} \\ {} & = & {\dfrac{1 + x^2}{1 - x^2} \cdot \dfrac{1 + y^2}{1 - y^2}.} \end{array}$

Finally note that 13.6.2 and 13.6.3 imply 13.6.1.