The relationship between trigonometric and hyperbolic functions 161
But from equation (5), cos jA = cosh A
and from equation (6), sin jA = j sinh A.
Hence
cosh
2 A
j 2 sinh 2 A
+ 1 =
1
j 2 sinh 2 A
and since j 2 =−1, −
cosh 2 A
sinh
2 A
+ 1 =−
1
sinh
2 A
Multiplying throughout by −1, gives:
cosh 2 A
sinh
2 A
− 1 =
1
sinh
2 A
i.e. coth
2 A − 1 = cosech
2 A
Problem 4. By substituting jA and jB for θ and
φ respectively in the trigonometric identity for
cos θ − cos φ, show that
cosh A − cosh B
= 2 sinh
A + B
2
sinh
A − B
2
cos θ − cos φ = −2 sin
θ + φ
2
sin
θ − φ
2
(see Chapter 17, page 172)
thus cos jA − cos jB
= −2 sin j
A + B
2
sin j
A − B
2
But from equation (5), cos jA = cosh A
and from equation (6), sin jA = j sinh A
Hence, cosh A − cosh B
= −2 j sinh
A + B
2
j sinh
A − B
2
= −2 j
2 sinh
A + B
2
sinh
A − B
2
But j 2 =−1, hence
cosh A − cosh B = 2 sinh
A + B
2
sinh
A− B
2
Problem 5. Develop the hyperbolic identity
corresponding to sin 3θ = 3 sinθ − 4 sin
3
θ by
writing jA for θ.
Substituting jA for θ gives:
sin 3 jA = 3 sin jA − 4 sin
3 jA
and since from equation (6),
sin jA = j sinh A,
j sinh 3A = 3 j sinh A − 4 j
3 sinh
3 A
Dividing throughout by j gives:
sinh 3A = 3 sinh A − j
2 4 sinh
3 A
But j 2 =−1, hence
sinh 3A = 3 sinh A + 4 sinh
3 A
[An examination of Problems 3 to 5 shows that whenever the trigonometric identity contains a term which
is the product of two sines, or the implied product
of two sine (e.g. tan 2 θ = sin 2 θ/cos 2 θ, thus tan 2 θ is
the implied product of two sines), the sign of the corresponding term in the hyperbolic function changes.
This relationship between trigonometric and hyperbolic
functions is known as Osborne’s rule, as discussed in
Chapter 5, page 45].
Now try the following exercise
Exercise 71 Further problems on
hyperbolic identities
In Problems 1 to 9, use the substitution A = j θ (and
B = j φ) to obtain the hyperbolic identities corresponding to the trigonometric identities given.
1. 1 + tan 2 A = sec 2 A
[1 − tanh 2 θ = sech 2 θ]
2. cos(A + B) = cos A cos B − sin A sin B
cosh(θ + φ)
= cosh θ cosh φ + sinh θ sinh φ
3. sin(A − B) = sin A cos B − cos A sin B
sinh(θ + φ) = sinh θ cosh φ
− cosh θ sinh φ
4. tan 2 A =
2 tan A
1 − tan 2 A
tanh 2θ =
2 tanhθ
1 + tanh
2
θ
But from equation (5), cos jA = cosh A
and from equation (6), sin jA = j sinh A.
Hence
cosh
2 A
j 2 sinh 2 A
+ 1 =
1
j 2 sinh 2 A
and since j 2 =−1, −
cosh 2 A
sinh
2 A
+ 1 =−
1
sinh
2 A
Multiplying throughout by −1, gives:
cosh 2 A
sinh
2 A
− 1 =
1
sinh
2 A
i.e. coth
2 A − 1 = cosech
2 A
Problem 4. By substituting jA and jB for θ and
φ respectively in the trigonometric identity for
cos θ − cos φ, show that
cosh A − cosh B
= 2 sinh
A + B
2
sinh
A − B
2
cos θ − cos φ = −2 sin
θ + φ
2
sin
θ − φ
2
(see Chapter 17, page 172)
thus cos jA − cos jB
= −2 sin j
A + B
2
sin j
A − B
2
But from equation (5), cos jA = cosh A
and from equation (6), sin jA = j sinh A
Hence, cosh A − cosh B
= −2 j sinh
A + B
2
j sinh
A − B
2
= −2 j
2 sinh
A + B
2
sinh
A − B
2
But j 2 =−1, hence
cosh A − cosh B = 2 sinh
A + B
2
sinh
A− B
2
Problem 5. Develop the hyperbolic identity
corresponding to sin 3θ = 3 sinθ − 4 sin
3
θ by
writing jA for θ.
Substituting jA for θ gives:
sin 3 jA = 3 sin jA − 4 sin
3 jA
and since from equation (6),
sin jA = j sinh A,
j sinh 3A = 3 j sinh A − 4 j
3 sinh
3 A
Dividing throughout by j gives:
sinh 3A = 3 sinh A − j
2 4 sinh
3 A
But j 2 =−1, hence
sinh 3A = 3 sinh A + 4 sinh
3 A
[An examination of Problems 3 to 5 shows that whenever the trigonometric identity contains a term which
is the product of two sines, or the implied product
of two sine (e.g. tan 2 θ = sin 2 θ/cos 2 θ, thus tan 2 θ is
the implied product of two sines), the sign of the corresponding term in the hyperbolic function changes.
This relationship between trigonometric and hyperbolic
functions is known as Osborne’s rule, as discussed in
Chapter 5, page 45].
Now try the following exercise
Exercise 71 Further problems on
hyperbolic identities
In Problems 1 to 9, use the substitution A = j θ (and
B = j φ) to obtain the hyperbolic identities corresponding to the trigonometric identities given.
1. 1 + tan 2 A = sec 2 A
[1 − tanh 2 θ = sech 2 θ]
2. cos(A + B) = cos A cos B − sin A sin B
cosh(θ + φ)
= cosh θ cosh φ + sinh θ sinh φ
3. sin(A − B) = sin A cos B − cos A sin B
sinh(θ + φ) = sinh θ cosh φ
− cosh θ sinh φ
4. tan 2 A =
2 tan A
1 − tan 2 A
tanh 2θ =
2 tanhθ
1 + tanh
2
θ