FMSP FP3 Revision Day 1
FMSP FP3 Revision Day 2
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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FMSP FP3 Revision Day 8
FMSP FP3 Revision Day 9
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
FMSP FP3 Revision Day 13
FP3 2012