NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
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Proofs
Lecture 9
• Hyperbolic functions are the complex analogues of
the trigonometric functions.
(8.7)
(8.8)
Relation to trigonometric functions
cos x = cosh(ix)
i sinh x = sin(ix)
1
2
(e−x + ex )
Hence,
cos(ix) = cosh x
⇒ cos(i2 x) = cosh(ix)
• Other hyperbolic functions can be obtained from
these:
sinh x
1
tanh x =
=
cosh x
coth x
1
sech x =
cosh x
1
cosech x =
sinh x
cosh x = cos(ix)
=
• cos(ix) = cosh(x)
• These are the two fundamental hyperbolic functions.
i sin x = sinh(ix)
1
2
eix + e−ix
2
2
⇒ cos(ix) = 12 ei x + e−i x
• cos x =
Hyperbolic Functions
• By definition
cosh x = 12 (ex + e−x )
sinh x = 12 (ex − e−x )
NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
(8.9)
cos(−x) = cosh(ix)
But
cos(−x) = cos(x) = cosh(ix)
2
1 i2 x
e − e−i x
2i
1 −x
sin(ix) =
e − ex
2i
• sin(ix) =
Multiply RHS top and bottom by i
i x
−i −x
sin(ix) =
e − ex =
e − e−x
2
2
Therefore,
sin(ix) = i sinh x
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
• ⇒ sin(i2 x) = i sinh(ix)
sin(−x) = i sinh(ix)
−sin(x) = i sinh(ix)
Multiply both sides by −i
i sin x = sinh(ix)
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
119
Identities of Hyperbolic Functions
• Trig identities for cos, sin, etc. can be transformed to
their hyperbolic analogue by changing the trig
functions to their corresponding hyperbolic ones and
remembering to change the sign of any term
involving sinh2 .
E.g. cos2 x + sin2 x = 1
Therefore cos2 (ix) + sin2 (ix) must = 1.
• Using eqns. 8.9 this becomes
cosh2 (x) + i2 sinh2 (x) = 1
⇒ cosh2 x − sinh2 x = 1
(8.10)
• Similarly we can transform
1 + tan2 x = sec2 x
into
1 − tanh2 x = sech 2 x
and
becomes
cot2 x + 1 = cosec 2 x
− coth2 x + 1 = −cosech 2 x
⇒ coth2 x − 1 = cosech 2 x
NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
120
Inverse Hyperbolic Functions
• Again analogous to trig functions.
• If y = cosh x then
x = cosh−1 y
also have sinh−1 , tanh−1 , cosech
Graphs of Hyperbolic Functions
−1
, sech
−1
, coth−1 .
NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
Writing Inverse Hyperbolic
Functions in Closed Form
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
Example 2:
• y = tanh x, x = tanh−1 y
y=
Example 1:
125
sinh x
ex − e−x
= x
cosh x
e + e−x
y(ex + e−x ) = ex − e−x
• y = cosh x, x = cosh−1 y
cosh x = 12 (ex + e−x )
sinh x = 12 (ex − e−x )
yex + ye−x = ex − e−x
from (eqn 8.1)
from (eqn 8.2)
Add these equations together
(y + 1)e−x = (1 − y)ex
1+y
1−y
1+y
2x = ln
1−y
e2x =
cosh x + sinh x = 12 ex + 12 e−x + 12 ex − 12 e−x
⇒ cosh x + sinh x = ex
x = ln[cosh x + sinh x]
p
x = ln[cosh x + cosh2 x − 1]
cosh−1 y = ln [y +
ye−x + e−x = ex − yex
x = tanh
−1
p
y 2 − 1]
1
y = ln
2
1+y
1−y