For any xeR cosh⁴x -sinh⁴-x = cosh(2x)
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cosh
2
x
=
2
cosh
2
x
−
1
By the definition of hyperbolic functions
cosh
(
2
x
)
=
e
2
x
+
e
−
2
x
2
cosh
x
=
e
x
+
e
−
x
2
cosh
2
x
=
(
e
x
+
e
−
x
2
)
2
=
1
4
(
e
x
+
e
−
x
)
2
=
1
4
(
e
2
x
+
e
−
2
x
+
2
)
Therefore,
2
cosh
2
x
−
1
=
2
⋅
1
4
(
e
2
x
+
e
−
2
x
+
2
)
−
1
=
e
2
x
+
e
−
2
x
2
+
1
−
1
=
cosh
(
2
x
)
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