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Hyperbolic Functions MA 341

Fall 2010 | 4 P a g e

Now from above we know that this area is 12

. Thus,

0

0

0

1 1ln 2

2 2

ln 2

2

x

x

e x

Thus, the value of ox can be expressed in terms of the area ,

0 .2

ex

Now, we want to find the coordinates of C . Since B C OC , then the slope of

B C has slope 1 , with the equation

1

1( ).2 o

o

y x xx

At the point C , y x so substituting we get

1

.2 4

o

o

xx y

x

So, the length

1/22 2

0

0 0

1 1 12 4 2 2 4

1 12 , 0 whenever 2 / 2.

2 4 2 4

o o

o o o

oo o

x xB C x

x x x

x xx

x x

Now, / 2ox e and sinhB C . Therefore,

2sinh 2 .

4 22 2

e e ee

Similarly,

1/22 21 1

2 4 2 4

1cosh 2

2 4

22 .

4 22 2

o o

o o

o

o

x xOC

x x

xx

e e ee

What do the graphs of these functions look like? We could plot them on the calculator, but because of their pattern, we want to try some algebra first.

Hyperbolic Functions MA 341

Fall 2010 | 5 P a g e

Note that

2 2 22 2

sinh ( )2 4

x x x xe e e ex

and

2 2 22 2

cosh ( )2 4

x x x xe e e ex ,

so that

2 2 2 2

2 2 2 2cosh ( ) sinh (

2 21

4 4 4 4)

x x x xe e e ex x

That is, we get that these functions satisfy the equation 2 2 1u v .

We also have seen the basic hyperbolic trigonometric identity: 2 2cosh ( ) sinh ( ) 1x x

Once we have the hyperbolic sine and hyperbolic cosine defined, then we define the other four functions as:

sinh( )tanh( )

cosh( )

x x

x x

x e ex

x e e

cosh( )coth( )

sinh( )

x x

x x

x e ex

x e e

, 0x

1 2sech( )

cosh( ) x xxx e e

1 2csch( )

sinh( ) x xxx e e

, 0x

Based on these definitions and the basic hyperbolic trigonometric identity, we find a large number of hyperbolic trigonometric identities that are analogous to the usual trigonometric identities

Trigonometric Identity Hyperbolic Trigonometric Identity 2 2cos sin 1x x 2 2cosh sinh 1x x

2 21 tan secx x 2 21 tanh sechx x sin( ) sin cos cos sinx y x y x y sinh( ) sinh cosh cosh sinhx y x y x y cos( ) cos cos sin sinx y x y x y cosh( ) cosh cosh sinh sinhx y x y x y

tan tantan( )

1 tan tanx y

x yx y

tanh tanh

tanh( )1 tanh tanh

x yx y

x y

2 1 cossin

2 2x x 2 cosh 1

sinh2 2x x

2 1 coscos

2 2x x 2 cosh 1

cosh2 2x x

2 1 costan

2 1 cosx x

x

2 cosh 1tanh

2 cosh 1x x

x

Hyperbolic Functions MA 341

Fall 2010 | 6 P a g e

sintan

2 1 cosx x

x

sinhtanh

2 cosh 1x x

x

1 costan

2 sinx x

x

cosh 1

tanh2 sinhx x

x

sin 2sin cos2 2x x

x sinh 2sinh cosh2 2x x

x

1 12 2sin sin 2sin ( )cos ( )x y x y x y 1 1

2 2sinh sinh 2sinh ( )cosh ( )x y x y x y 1 12 2cos cos 2cos ( )cos ( )x y x y x y 1 1

2 2cosh cosh 2cosh ( )cosh ( )x y x y x y 1 12 2cos cos 2sin ( )sin ( )x y x y x y 1 1

2 2cosh cosh 2sinh ( )sinh ( )x y x y x y

Certain Values Hyperbolic Trigonometric Functions Just as with the circular functions and certain basic angles for which we want to know the values, there are certain base values of the hyperbolic functions that we might want to know:

sinh 0 0 coth 0

cosh 0 1 sech 0 1

tanh 0 0 csch 0

undefined

undefined

This is a good start. Are there other values of which we should be aware? Well, we might expect that it will involve logarithms. For example,

ln2 - ln2

ln2 - ln2

- 2 - 1 2 3 5sinh(ln2) coth(ln2)2 2 4 3

2 1 2 5 4cosh(ln2) sech(ln2)

2 2 4 5

sinh(ln2) 3 4tanh(ln2) csch(ln2)

cosh(ln2) 5 3

e e

e e

What other properties do they seem to have in common? We should consider at least three more things: inverse functions and derivatives and graphs. Inverse Hyperbolic Trigonometric Functions Since the hyperbolic trigonometric functions are defined in terms of exponentials, we might expect that the inverse hyperbolic functions might involve logarithms. Let us first consider the inverse function to the hyperbolic sine: arcsinh(x). By the definition of an inverse function, arcsinh( )y x means that sinh( )x y . Thus,

2

y yex

e

2y y xe e

2y y y ye e e xe

Hyperbolic Functions MA 341

Fall 2010 | 7 P a g e

2 12 0y yxee Let yu e , then this equation becomes

2 12 0xuu 2

22 4 41

2u

x xx x

2 1y xe x 2ln( 1)xy x , and it must be the positive square root because 0ye .

2arcsinh( ) ln( 1)x xx We should not find this too surprising. We would expect the others to be similar. Doing similar work, we find that:

2arccosh( ) ln( 1)x xx

1 1 1ln ln(1 ) ln(1 )

1 2 2arctanh( )

xx

xx x

1 1 1ln ln( 1) ln( 1)

1arccoth( )

2 2x

x xx

x

2

arcsech( )1 1

lnx

xx

2

arccsch( )1 1

lnx

xx

Wow! No, you don’t have to memorize all of this!! Derivatives of Hyperbolic Trigonometric Functions Just like everything else, we would expect the derivatives of the hyperbolic trigonometric functions to take an analogous route to those of the regular trigonometric functions. We can easily find the derivatives since they are defined in terms of the exponential function.

)1 1 1

sinh( ) ( ( )) ( ) cosh( )2 2

(2 2

x xx x x x x xd d e e d

x e e e e xdx dx dx

e e

)1 1 1

cosh( ) ( ( )) ( ) sinh( )2 2

(2 2

x xx x x x x xd d e e d

x e e e e xdx dx dx

e e

Thus we have found that these functions have a nice derivative periodicity – and we do not have to take negative signs into account:

sinh( ) cosh( )d

x xdx

Hyperbolic Functions MA 341

Fall 2010 | 8 P a g e

cosh( ) sinh( )d

x xdx

From these, we can compute the other derivatives – expecting analogous results.

22 2

sinh( ) cosh cosh sinh sinh 1tanh( ) ( )

cosh( ) (cosh ) (ce

s )c

hs h

od d x x x x x

x xdx dx x x x

22 2

cosh( ) sinh sinh cosh cosh 1coth( ) ( )

sinh( ) (sinh ) (csch

sinh )d d x x x x x

x xdx dx x x x

2

1 sinhsech( ) ( )tanh( )

cosh( ) (sec

cosh )h

d d xx x x

dx dx x x

2

1 coshcsch( ) ( )coth( )

sinh( ) (csc

sinh )h

d d xx x x

dx dx x x

These are pretty close to what we expected. Graphs of Hyperbolic Trigonometric Functions We wait until now to look at the graphs so that we can use the derivative to help us. First, let us look at cosh(x).

cosh( ) 02

x xe ex

since the numerator is always

positive.

Since sinh( ) cosh( )d

x xdx

, and cosh( ) 0x for all x, we

see that the hyperbolic sine function is always increasing. We also note that it has no critical points, since its derivative is always defined and is never 0. Now, looking at the graph it is not too surprising to find that it looks like the figure to the right.

Now that we know what the hyperbolic sine looks like, we can analyze the hyperbolic cosine. Since its derivative is 0 at 0x , we know that it has a critical point there. Since the second derivative is always positive this critical point must be a local minimum. It is not hard to show that it is a global minimum. The graph looks like the figure on the left. This may look somewhat like a parabola to you, but it actually grows faster than any parabola. The interesting thing is that it does describe a physical setting. If you put two pegs at the same height some distance apart and let a rope hang between the two pegs so that the ends just hang

over the pegs (not tied to them), then this rope takes on a shape called a catenary. The catenary is a hyperbolic cosine shape. Examples are electrical wires hanging between power poles.

Figure 1:

x

y

x

y

Figure 2:

Hyperbolic Functions MA 341

Fall 2010 | 9 P a g e

tanh( )y x coth( )y x sec ( )hy x csch( )y x

Figure 3: A Web of Hyperbolic Trigonometric Functions

Power Series Expansions for Hyperbolic Trigonometric Functions

Since the exponential function has a power series expansion

0 !

kx

k

xe

k and since the

hyperbolic functions are defined in terms of the exponential function, we find that the power series expansions for the hyperbolic functions are:

0 0

2 2 1

0 0 0

2

0

1 12 2

1 1 12 2 2

cosh2

( )! !

! (2 )! (2 1)!

cosh(2 )!

n n

n n

n k k

n k k

k

x x

k

ex

x xn n

x x xn k k

xx

k

e

x

y

x

y

x

y

x

y

x

y

Hyperbolic Functions MA 341

Fall 2010 | 10 P a g e

Likewise the hyperbolic sine function has a power series expansion

2 1

0

sinh(2 1)!

k

k

xx

k

Now, these look vaguely familiar – the more vague the longer it has been since you studied power series. As a reminder let me point out that the Maclaurin series for the circular functions are:

2 1

0

2

0

(-1)sin

(2 1)!

(-1)cos

(2 )!

k k

k

k k

k

xx

k

xx

k

So, the Maclaurin series for the hyperbolic functions and those for the circular functions are very similar. Now, there is even more of a relationship between these two – using complex numbers. Recall that we let i2 = –1. In the power series expansion for cos x, let’s replace x by ix.

2 2 2 2

0 0 0

2 1 2 1 2 1 2 1

0 0 0

(-1) ( ) (-1)cos( ) cosh( )

(2 )! (2 )! (2 )!

(-1) ( ) (-1)sin( ) sinh( )

(2 1)! (2 1)! (2 1)!

n n n n n n

n n n

n n n n n n

n n n

ix i x xix x

n n n

ix i x ixix i x

n n n

Therefore, we see that we now have a relationship between the circular and the hyperbolic functions as follows:

cosh( ) cos( ) cos

sinh( ) - sin( ) sin

xx ix

i

xx i ix i

i

Uses for Hyperbolic Functions 1. Antiderivatives These inverse hyperbolic trigonometric functions often appear in antiderivative formulas instead of logarithms. As an example, we get

2

2ln - 9

- 9

dxx x C

x

using the substitution x = 3 sec u. On the other hand we get

-1

2cosh

3- 9

dx xC

x

using the substitution x = 3 cosh u. This second integral is actually easier to compute symbolically – if you remember to use hyperbolic functions. 2. Applications What shape does a chain take when hanging freely between two pegs? This does not mean that it is fastened at the two pegs, but is, in fact, free to move at the pegs. For the most part we would think that this shape is the shape of a parabola.

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Fall 201

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