The hyperbolic functions
cosh
x
,
sinh
x
satisfy similar (but not exactly equivalent) identities to those satisfied by
cos
x
,
sin
x
. We note first some basic notation similar to that employed with trigonometric functions:
cosh
n
x
means
(
cosh
x
)
n
sinh
n
x
means
(
sinh
x
)
n
n
≠
−
1
In the special case that
n
=
−
1
we
do not
use
cosh
−
1
x
and
sinh
−
1
x
to mean
1
cosh
x
and
1
sinh
x
respectively. The notation
cosh
−
1
x
and
sinh
−
1
x
is reserved for the
inverse functions
of
cosh
x
and
sinh
x
respectively.
Show that
cosh
2
x
−
sinh
2
x
≡
1
for all
x
.
First, express
cosh
2
x
in terms of the exponential functions
e
x
,
e
−
x
:
Answer
1
4
(
e
x
+
e
−
x
)
2
≡
1
4
[
(
e
x
)
2
+
2
e
x
e
−
x
+
(
e
−
x
)
2
]
≡
1
4
[
e
2
x
+
2
e
x
−
x
+
e
−
2
x
]
≡
1
4
[
e
2
x
+
2
+
e
−
2
x
]
Similarly, express
sinh
2
x
in terms of
e
x
and
e
−
x
:
Answer
1
4
(
e
x
−
e
−
x
)
2
≡
1
4
[
(
e
x
)
2
−
2
e
x
e
−
x
+
(
e
−
x
)
2
]
≡
1
4
[
e
2
x
−
2
e
x
−
x
+
e
−
2
x
]
≡
1
4
[
e
2
x
−
2
+
e
−
2
x
]
Finally determine
cosh
2
x
−
sinh
2
x
using the results from (1) and (2):
Answer
cosh
2
x
−
sinh
2
x
≡
1
4
[
e
2
x
+
2
+
e
−
2
x
]
−
1
4
[
e
2
x
−
2
+
e
−
2
x
]
≡
1
As an alternative to the calculation in this Task we could, instead, use the relations
e
x
≡
cosh
x
+
sinh
x
e
−
x
≡
cosh
x
−
sinh
x
and remembering the algebraic identity
(
a
+
b
)
(
a
−
b
)
≡
a
2
−
b
2
, we see that
(
cosh
x
+
sinh
x
)
(
cosh
x
−
sinh
x
)
≡
e
x
e
−
x
≡
1
that is
cosh
2
x
−
sinh
2
x
≡
1
The fundamental identity relating hyperbolic functions is:
cosh
2
x
−
sinh
2
x
≡
1
This is the hyperbolic function equivalent of the trigonometric identity:
cos
2
x
+
sin
2
x
≡
1
Show that
cosh
(
x
+
y
)
≡
cosh
x
cosh
y
+
sinh
x
sinh
y
.
First, express
cosh
x
cosh
y
in terms of exponentials:
Answer
e
x
+
e
−
x
2
e
y
+
e
−
y
2
≡
1
4
[
e
x
e
y
+
e
−
x
e
y
+
e
x
e
−
y
+
e
−
x
e
−
y
]
≡
1
4
(
e
x
+
y
+
e
−
x
+
y
+
e
x
−
y
+
e
−
x
−
y
)
Now express
sinh
x
sinh
y
in terms of exponentials:
Answer
e
x
−
e
−
x
2
e
y
−
e
−
y
2
≡
1
4
(
e
x
+
y
−
e
−
x
+
y
−
e
x
−
y
+
e
−
x
−
y
)
Now express
cosh
x
cosh
y
+
sinh
x
sinh
y
in terms of a hyperbolic function:
Answer
cosh
x
cosh
y
+
sinh
x
sinh
y
≡
1
2
(
e
x
+
y
+
e
−
(
x
+
y
)
)
which we recognise as
cosh
(
x
+
y
)
Other hyperbolic function identities can be found in a similar way. The most commonly used are listed in the following Key Point.
Hyperbolic Identities
∙
cosh
2
−
sinh
2
≡
1
∙
cosh
(
x
+
y
)
≡
cosh
x
cosh
y
+
sinh
x
sinh
y
∙
sinh
(
x
+
y
)
≡
sinh
x
cosh
y
+
cosh
x
sinh
y
∙
sinh
2
x
≡
2
sinh
x
cosh
y
∙
cosh
2
x
≡
cosh
2
x
+
sinh
2
x
or
cosh
2
x
≡
2
cosh
2
−
1
or
cosh
2
x
≡
1
+
2
sinh
2
x