Exponential and Logarithmic Functions: Learn It 5

Hyperbolic Functions Cont.

Identities Involving Hyperbolic Functions

Just as trigonometric functions have identities that allow for the simplification and transformation of expressions, hyperbolic functions also possess their own set of identities.

hyperbolic function identities

Hyperbolic Reflection Identities:

  • cosh(x)=coshx
  • sinh(x)=sinhx

Hyperbolic Pythagorean Identities:

  • cosh2xsinh2x=1

Hyperbolic Squared Identities:

  • 1tanh2x=sech2x
  • coth2x1=csch2x

Hyperbolic Addition Formulas:

  • sinh(x±y)=sinhxcoshy±coshxsinhy
  • cosh(x±y)=coshxcoshy±sinhxsinhy

Exponential Definitions of Hyperbolic Functions

  • coshx+sinhx=ex
  • coshxsinhx=ex
  1. Simplify sinh(5lnx).
  2. If sinhx=34, find the values of the remaining five hyperbolic functions.

Inverse Hyperbolic Functions

From the graphs of the hyperbolic functions, we see that all of them are one-to-one except coshx and sechx. 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.

inverse hyperbolic functions

sinh1x=arcsinh x=ln(x+x2+1)cosh1x=arccosh x=ln(x+x21)tanh1x=arctanh x=12ln(1+x1x)coth1x=arccot x=12ln(x+1x1)sech1x=arcsech x=ln(1+1x2x)csch1x=arccsch x=ln(1x+1+x2|x|)

Let’s look at how to derive the first equation, sinh1x=arcsinh x=ln(x+x2+1). The others follow similarly.

Suppose y=sinh1x. Then, x=sinhy and, by the definition of the hyperbolic sine function, x=eyey2

Multiplying both sides of this equation by 2 to get rid of the fraction and setting that equal to zero, we get

ey2xey=0

Multiplying this equation by ey, we obtain

e2y2xey1=0

This can be solved like a quadratic equation, with the solution

ey=2x±4x2+42=x±x2+1

Since ey>0, the only solution is the one with the positive sign.

Applying the natural logarithm to both sides of the equation, we conclude that

y=ln(x+x2+1)

Evaluate each of the following expressions

sinh1(2)
tanh1(14)