
# 12.9: Hyperbolic trigonometry


In this section we give formulas for h-distance using hyperbolic functions. One of these formulas will be used in the proof of the hyperbolic Pythagorean theorem (Theorem 13.6.1).

Recall that $$\cosh$$, $$\sinh$$, and $$\tanh$$ denote hyperbolic cosine, hyperbolic sine, and hyperbolic tangent; that is, the functions defined by

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

$$\tanh x := \dfrac{\sinh x}{\cosh x}.$$

These hyperbolic functions are analogous to sine and cosine and tangent.

Exercise $$\PageIndex{1}$$

Prove the following identities:

$$\cosh' x=\sinh x$$;     $$\sinh'x=\cosh x$$;     $$(\cosh x)^2-(\sinh x)^2=1.$$

Theorem $$\PageIndex{1}$$ Double-argument identities

The identities

$$\cosh (2 \cdot x) = (\cosh x)^2 + (\sinh x)^2$$     and     $$\sinh (2 \cdot x) = 2 \cdot \sinh x \cdot \cosh x$$

hold for any real value $$x$$.

Proof

$$\begin{array} {rcl} {(\sinh x)^2 + (\cosh x)^2} & = & {(\dfrac{e^x - e^{-x}}{2})^2 + (\dfrac{e^x + e^{-x}}{2})^2 =} \\ {} & = & {\dfrac{e^{2 \cdot x} + e^{-2 \cdot x}}{2} =} \\ {} & = & {\cosh (2 \cdot x);} \end{array}$$

$$\begin{array} {rcl} {2 \cdot \sinh x \cdot cosh x} & = & {2 \cdot (\dfrac{e^x - e^{-x}}{2}) \cdot (\dfrac{e^x + e^{-x}}{2})} \\ {} & = & {\dfrac{e^{2 \cdot x} - e^{-2 \cdot x}}{2}} \\ {} & = & {\cosh (2 \cdot x).} \end{array}$$

Advanced Exercise $$\PageIndex{2}$$

Let $$P$$ and $$Q$$ be two h-poins distinct from the center of absolute. Denote by $$P'$$ and $$Q'$$ the inverses of $$P$$ and $$Q$$ in the absolute.

(a) $$\cosh [\dfrac{1}{2} \cdot PQ_h] = \sqrt{\dfrac{PQ' \cdot P'Q}{PP' \cdot QQ'}};$$

(b) $$\sinh [\dfrac{1}{2} \cdot PQ_h] = \sqrt{\dfrac{PQ \cdot P'Q'}{PP' \cdot QQ'}};$$

(c) $$\tanh [\dfrac{1}{2} \cdot PQ_h] = \sqrt{\dfrac{PQ \cdot P'Q'}{PQ' \cdot P'Q}};$$

(d) $$\cosh PQ_h = \dfrac{PQ \cdot P'Q' + PQ' \cdot P'Q}{PP' \cdot QQ'}$$.

Hint

By Corollary 10.6.1 and Theorem 10.2.1, the right hand sides in the identities survive under an inversion in a circle perpendicular to the absolute.

As usual we assume that the absolute is a unit circle. Suppose that $$O$$ denotes the h-midpoint of $$[PQ]_h$$. By the main observation (Theorem 12.3.1) we can assume that $$O$$ is the center of the absolute. In this case $$O$$ is also the Euclidean midpoint of $$[P Q]$$. (Instead, we may move $$Q$$ to the center of absolute. In this case the derivations are simpler. But since $$Q'Q = Q'P = QP = \infty$$, one has to justify that $$\dfrac{\infty}{\infty} = 1$$ every time.)

Set $$a = OP = OQ$$; in this case we have

$$\begin{array} {rcl} {PQ} & = & {2 \cdot a,} \\ {P'Q'} & = & {2 \cdot \dfrac{1}{a},} \end{array}$$             $$\begin{array} {l} {PP' = QQ' = \dfrac{1}{a} - a,} \\ {PQ' = QP' = \dfrac{1}{a} + a.} \end{array}$$

and

$$PQ_h = \ln \dfrac{(1 + a)^2}{(1 - a)^2} = 2 \cdot \ln \dfrac{1 + a}{1 - a}.$$

Therefore

$$\begin{array} {rcl} {\cosh [\dfrac{1}{2} \cdot PQ_h]} & = & {\dfrac{1}{2} \cdot (\dfrac{1 + a}{1 - a} + \dfrac{1 - a}{1 + a})=} \\ {} & = & {\dfrac{1 + a^2}{1 - a^2};} \end{array}$$                    $$\begin{array} {rcl} {\sqrt{\dfrac{PQ' \cdot P'Q}{PP' \cdot QQ'}}} & = & {\dfrac{\dfrac{1}{a} + a}{\dfrac{1}{a} - a} =} \\ {} & = & {\dfrac{1 + a^2}{1 - a^2}.} \end{array}$$

Hence the part (a) follows. Similarly,

$$\begin{array} {rcl} {\sinh [\dfrac{1}{2} \cdot PQ_h]} & = & {\dfrac{1}{2} \cdot (\dfrac{1 + a}{1 - a} - \dfrac{1 - a}{1 + a}) =} \\ {} & = & {\dfrac{2 \cdot a}{1 - a^2};} \end{array}$$                    $$\begin{array} {rcl} {\sqrt{\dfrac{PQ \cdot P'Q'}{PP' \cdot QQ'}}} & = & {\dfrac{2}{\dfrac{1}{a} - a} =} \\ {} & = & {\dfrac{2 \cdot a}{1 - a^2}.} \end{array}$$

Hence the part (b) follows.

The parts (c) and (d) follow from (a), (b), the definition of hyperbolic tangent, and the double-argument identity for hyperbolic cosine, see Theorem $$\PageIndex{1}$$.