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FP3 Notes (Edexcel)



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Hyperbolic Functions Definition of the six hyperbolic functions in terms of exponentials. Graphs and properties of the

hyperbolic functions. For example, ( )1cosh e e2

x xx −= + , 1 2sechcosh e ex xx

x −= =+

.

Students should be able to derive and use simple identities such as 2 2cosh sinh 1x x− ≡ and

2 2cosh sinh cosh 2x x x+ ≡ and to solve equations such as cosh sinha x b x c+ = . Inverse hyperbolic functions, their graphs, properties and logarithmic equivalents. For example,

2arsinh ln 1x x x = + + and students may be required to prove this and similar results.

There are six hyperbolic functions which are all linked to the exponential function.

The two basic hyperbolic functions are as follows:

( )1sinh e e2

x xx −= − ( )1cosh e e2

x xx −= + .

From these it follows that

( )( )

( )( )

1 22

212

e e e e esinh e 1tanhcosh e 1e e e e e

x x x x x x

xx x x x x

xxx

− −

− −

− − −= = = =

++ +

( )12

1 1 2sechcosh e ee e x xx x

xx −−

= = =++

( )12

1 1 2cosechsinh e ee e x xx x

xx −−

= = =−−

( )( )

1 22

212

e ecosh e 1cothsinh e 1e e

x x x

xx x

xxx

−

−

+ += = =

−−



Consider their graphs

If ( )1f ( ) sinh e e2

x xx x −= = − then:

f (0) 0f ( ) f ( )

1f ( ) e as 2

x

x x

x x

=− = −

→ →∞

So it is an odd function (i.e. has rotational symmetry order 2 about origin).

( )1sinh e e2

x xy x −= = −

If ( )1f ( ) cosh e e2

x xx x −= = + then:

f (0) 1f ( ) f ( )

1f ( ) e as 2

x

x x

x x

=− =

→ →∞

So it is an even function (i.e. has symmetry about y axis).

( )1cosh e e2

x xy x −= = +



Put together on the same graph gives:

Consider 2 2

2 2

sinh e e e 1 1 ef ( ) tanhcosh e e e 1 1 e

x x x x

x x x x

xx xx

− −

− −

− − −= = = = =

+ + +

As , tanh 1x x→∞ → And as , tanh 1x x→−∞ →

Also, ( ) ( )

1 2sech1 e ee e2

x xx x

x−

−= =

++



and ( ) ( )

1 2cosech1 e ee e2

x xx x

x−

−= =

−−

From the definitions of ( )1sinh e e2

x xx −= − and ( )1cosh e e2

x xx −= + :

( ) ( )

( )

( )

( )

2 22 2

2 2 2 2

2 2

2 2

1 1cosh sinh e e e e4 41 e e 2 e e 241 2e 2e41 e e2cosh 2

x x x x

x x x x

x x

x x

x x

x

− −

− −

−

−

+ = + + −

= + + + + −

= +

= +

=

. This, and other formulae are given in the formula book. Note how similar they look to the trigonometric identities. Osborn’s law states that trigonometric identities can be turned into hyperbolic identities by replacing each trigonometric function with is equivalent hyperbolic function and changing the sign of every product of two sines. Examples 2 2cos sin cos 2x x x− = becomes 2 2cosh sinh cosh 2x x x+ = . Other identities are

(a) 1sinhcosh 22 =− xx (b) xxx coshsinh22sinh = (c) xxx 22 sinhcosh2cosh +=



Solving equations involving hyperbolics Example Solve 2cosh sinh 2x x− = . Replace the hyperbolic functions with exponentials.

So ( ) ( )1 12 22 2

x x x xe e e e− − + − − =

, and so ( ) ( )2 4x x x xe e e e− −+ − − =

This gives 3 4x xe e−+ = .

Replace xe with z gives 3 4zz

+ = and so 2 4 3 0z z− + = .

Factorising this gives ( )( )1 3 0z z− − = and so 1z = or 3z = . Hence 0x = or ln 3x = . Inverse Hyperbolics Example If zx sinh= then arsinh z x=

Using the definition of sinh gives ( )1 e e2

z zx −= − and so, rearranging, gives 2e 2 e 1 0z zx− − = .

Solving this gives 2

22 4 4e 12

z x x x x± += = ± + .

Since e 0z > it follows that 2e 1z x x= + + and so ( )1ln 2 ++= xxz . Since arsinh z x= , as in formula book,

( )2arsinh ln 1x x x= + + . There is no restriction on

the domain. Note that 2 1x x+ + is always positive, whether x is positive or negative, so therefore,

( )2arsinh ln 1x x x= + + exists for all values of x.

When solving problems which involve solving sinh x a= , simply use the formula book, which

states that ( )2arsinh ln 1x x x= + + .

sinhy x=

arsinh y x=

y x=



Example If coshx z= then arcosh z x=

Using the definition of cosh gives ( )1 e +e2

z zx −= and so, rearranging, gives 2e 2 e 1 0z zx− + = .

Solving this gives 2

22 4 4e 12

z x x x x± −= = ± − .

So ( )2ln 1z x x= ± − but which sign should be taken?

( )( ) ( )2 2 2 21 1 1 1x x x x x x+ − − − = − − =

( ) ( )2

2

111

x xx x

− − =+ −

( ) ( ) ( )2 2

2

1ln 1 ln ln 11

x x x xx x

− − = = − + − + −

.

If the domain of coshx z= has been limited to 0z ≥ then, since

( ) ( )2 2ln 1 ln 1 0x x x x− − = − + − < , it follows that ( )2ln 1z x x= + − .

So ( )2arcosh ln 1x x x= + − . The domain of this, is

the same as the range of cosh z , so the domain of

( )2arcosh ln 1x x x= + − is 1x ≥ .

When solving problems which involve solving sinh x a= , simply use the formula book, which

states that ( )2arcosh ln 1x x x= + − .

coshy x=

arcosh y x=

y x=



Example If tanhx z= then artanh z x=

2

2

sinh e e e 1cosh e e e 1

z z z

z z z

zxz

−

−

− −= = =

+ +

( )

( )

2 2

2

2

e 1 e 1

e 1 11e

1

z z

z

z

x

x xx

x

− = +

− = +

+=

−

So 1 1ln2 1

xzx

+ = − , that is, as in the

formula book, 1 1artanh ln2 1

xxx

+ = − for 1<x

When solving problems which involve solving tanh x a= , simply use the formula book, which

states that 1 1artanh ln2 1

xxx

+ = − for 1<x .

Questions involving inverse hyperbolics Example Solve 2cosh 8x = .

2cosh 8x = and so cosh 8x = ± . cosh 0x > , so cosh 8x = .

This could be solve by writing ( )1cosh e e2

x xx −= + but is very long-winded…

Thinking about the graph shows that there will be two solutions….

Formula book states that ( )2arcosh ln 1x x x= + −

So ( ) ( )ln 8 8 1 ln 8 7x = + − = + is one solution.

Symmetry of graph means that the other solution is ( )ln 8 7x = − + .

tanhy x=

artanh y x=

y x=



Example Solve 2sinh 5x = .

2sinh 5x = and so sinh 5x = ± .

This could be solve by writing ( )1sinh e e2

x xx −= − but is very long-winded…

Thinking about the graph shows that there will be two solutions….

Formula book states that ( )2arsinh ln 1x x x= + +

Solving sinh 5x = gives ( ) ( )ln 5 5 1 ln 5 6x = + + = + as one solution.

Solving sinh 5x = − gives ( ) ( ) ( )ln 5 5 1 ln 5 6 ln 6 5x = − − + = − + = − as the other

solution.

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