Chapter 1 | Functions and Graphs
111
5. cosh 2 x −sinh 2 x =1 6. 1−tanh 2 x = sech 2 x 7. coth 2 x −1=csch 2 x
8. sinh( x ± y ) = sinh x cosh y ±cosh x sinh y 9. cosh( x ± y ) =cosh x cosh y ±sinh x sinh y
Example 1.40 Evaluating Hyperbolic Functions a. Simplify sinh(5ln x ). b. If sinh x =3/4, find the values of the remaining five hyperbolic functions.
Solution a. Using the definition of the sinh function, we write sinh(5ln x ) = e 5ln x − e −5ln x 2 = e ln ⎛ ⎝ x 5
⎞ ⎠
⎛ ⎝ x −5
⎞ ⎠
ln
x 5 − x −5
− e
=
2
2 .
b. Using the identity cosh 2 x −sinh 2 x =1, we see that cosh 2 x =1+ ⎛ ⎝ 3 4
⎞ ⎠
2
= 25 16
.
Since cosh x ≥1 for all x , we must have cosh x =5/4. Then, using the definitions for the other hyperbolic functions, we conclude that tanh x =3/5, csch x =4/3, sech x =4/5, and coth x =5/3.
Simplify cosh(2ln x ).
1.34
Inverse Hyperbolic Functions From the graphs of the hyperbolic functions, we see that all of them are one-to-one except cosh x and sech x . If we restrict the domains of these two functions to the interval [0, ∞), then all the hyperbolic functions are one-to-one, and we can define the inverse hyperbolic functions . Since the hyperbolic functions themselves involve exponential functions, the inverse hyperbolic functions involve logarithmic functions.
Definition Inverse Hyperbolic Functions
sinh −1 x =arcsinh x = ln ⎛ tanh −1 x =arctanh x = 1 2 ln ⎛ sech −1 x =arcsech x = ln ⎛ ⎝
⎝ x + x 2 +1 ⎞ ⎠
cosh −1 x =arccosh x = ln ⎛
⎝ x + x 2 −1 ⎞ ⎠
⎞ ⎠
⎛ ⎝ x +1 x −1
⎞ ⎠
⎝ 1+ x 1− x
coth −1 x =arccot x = 1 ⎞ ⎠ ⎟ csch −1 x =arccsch x = ln 2 ln ⎛ ⎝ ⎜ 1
⎞ ⎠ ⎟
⎜ 1+ 1− x 2 x
x 2
x + 1+ | x |
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