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Hyperbolic Functions Summary

1. Uses of the hyperbolic functions:

a. Engineering applications (catenary curves)

b. Finding certain antiderivatives.

c. Solving differential equations, such as

y(x) = y(x), which has solution y(x) = Asinh(x) + Bcosh(x)

vs. y(x) = -y(x), which has solution y(x) = Asin(x) + Bcos(x)

2. Definition of the hyperbolic functions:

2sinh

xx eex

2cosh

xx eex

1

1

cosh

sinhtanh

2

2

x

x

xx

xx

e

e

ee

ee

x

xx

1

1

sinh

coshcoth

2

2

x

x

xx

xx

e

e

ee

ee

x

xx

xx eexx

2

cosh

1sech cosech x

1

sinh x

2

ex e x

3. Some identities involving hyperbolic functions:

1sinhcosh 22 uu

Note: if x = cosh u and y = sinh u, then

122 yx is the equation of a hyperbola

sinh2

cos sin cos sin

2

sin

i ie ei

i i

i

uuu coshsinh22sinh 1cosh22cosh 2 uu

cosh (s ± t) = cosh s cosh t ± sinh s sinh t sinh (s ± t) = sinh s cosh t ± cosh s sinh t

4. Osbornes’s Rule: a trigonometry identity can be converted to an equivalent identity for hyperbolic

functions by expanding, exchanging trigonometric functions with their hyperbolic counterparts, and

then flipping the sign of each term involving the product of two hyperbolic sines.

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5. Graphs of the hyperbolic functions:

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6. Inverses of the hyperbolic functions and their formulae:

arsinh x ln x x2 1 arcosh x ln x x2 1 x 1

artanh x 1

2ln

1 x

1 x

x 1

7. Proof of the formula for the inverse hyperbolic sine:

y arsinh x

x sinh y ey e y

2

2x ey e y

2xey e2 y 1

e2 y 2xey 1 0

ey 2

2x ey 1 0

ey 2x 4x2 4(1)(1)

2(1) x x2 1

y arsinh x ln x x2 1

8. Derivatives of the hyperbolic functions:

xeeee

dx

dx

dx

d xxxx

cosh22

sinh

xxdx

dsinhcosh

xxdx

d 2sech tanh d

dxcoth x -cosech2x

xxxdx

dtanhsech sech

d

dxcosech x cosech xcoth x

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9. Derivatives of the inverse hyperbolic functions:

d

dxarsinh x

1

1 x2

d

dxarcosh x

1

x2 1

d

dxartanh x

1

1 x2

10. Proof of the derivative of the inverse hyperbolic sine:

y arsinh x

x sinh y

d

dxsinh y

d

dxx

cosh ydy

dx 1

dy

dx

1

cosh y

1

1 sinh2 y

dy

dx

1

1 x2

d

dxarsinh x

d

dxln x x2 1

1

x x2 1

d

dxx x2 1

1x

x2 1

x x2 1

x x2 1

x x2 1

x x2 1 x2

x2 1 x

x2 x2 1

x2 1 x2

x2 1

1

x2 1 x2

x2 1

1

1 x2

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11. Recall that the derivative formulas can be used to obtain integral formulas, like

sinh xdx cosh x c

sinh xdx cosh x C cosh xdx sinh x C

sech2xdx tanh x C cosech2xdx coth x C

sech x tanh xdx sech x C cosech x coth xdx cosech x C

1

(1 x2 )dx arsinh x C

1

(x2 1)dx arcosh x C, x 1

1

(a x2 )dx arsinh

x

a

C 1

(x2 a2 )dx arcosh

x

a

C, x a

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Solving Equations involving Hyperbolic Functions

Type 1

To solve Type 1 equations you need to use the definitions of the hyperbolic functions and you usually end up

with a quadratic equation in ex. It’s easy to recognise Type 1 because they have no double argument nor are

the hyperbolic functions squared.

Example of Type 1

Solve the equation, giving your answer in a logarithmic form

8coshx 4sinhx 13

Type 2

For Type 2, you need to use a hyperbolic identity in order to solve the equation. Unlike Type 1, you don’t

change everything to ex. It’s easy to recognise Type 2 because they either have a hyperbolic function

squared or a double argument.

Example of Type 2

Solve the equation, giving your answer in a logarithmic form

sinh2 x 3cosh x 9

Another example of Type 2

Solve the equation, giving your answer in a logarithmic form

cosh2x coshx 2

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HYPERBOLICS EXAM QUESTIONS

FP3 2009

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FP3 2010

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FP3 2011

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FP3 2012