The relationship between trigonometric and hyperbolic functions 161
But from equation (5), cosjA=coshA
and from equation (6), sinjA=jsinhA.
Hence
cosh^2 A
j^2 sinh^2 A+ 1 =1
j^2 sinh^2 Aand since j^2 =− 1 ,−
cosh^2 A
sinh^2 A+ 1 =−1
sinh^2 AMultiplying throughout by−1, gives:
cosh^2 A
sinh^2 A− 1 =1
sinh^2 Ai.e. coth^2 A− 1 =cosech^2 A
Problem 4. By substituting jAand jBforθand
φrespectively in the trigonometric identity for
cosθ−cosφ, show thatcoshA−coshB=2sinh(
A+B
2)
sinh(
A−B
2)cosθ−cosφ=−2sin(
θ+φ
2)
sin(
θ−φ
2)(see Chapter 17, page 172)thus cosjA−cosjB
=−2sinj(
A+B
2)
sinj(
A−B
2)But from equation (5), cosjA=coshA
and from equation (6), sinjA=jsinhA
Hence, coshA−coshB
=− 2 jsinh(
A+B
2)
jsinh(
A−B
2)=− 2 j^2 sinh(
A+B
2)
sinh(
A−B
2)But j^2 =−1, hence
coshA−coshB=2sinh(
A+B
2)
sinh(
A−B
2)Problem 5. Develop the hyperbolic identity
corresponding to sin3θ=3sinθ−4sin^3 θby
writingjAforθ.SubstitutingjAforθgives:sin3jA=3sinjA−4sin^3 jAand since from equation (6),sinjA=jsinhA,jsinh3A= 3 jsinhA− 4 j^3 sinh^3 ADividing throughout byjgives:sinh3A=3sinhA−j^2 4sinh^3 ABut j^2 =−1, hencesinh3A=3sinhA+4sinh^3 A[An examination of Problems 3 to 5 shows that when-
ever 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 θis
the implied product of two sines), the sign of the cor-
responding term in the hyperbolic function changes.
This relationshipbetween trigonometricand hyperbolic
functions is known as Osborne’s rule, as discussed in
Chapter 5, page 45].Now try the following exerciseExercise 71 Further problemson
hyperbolic identitiesIn Problems 1 to 9, use the substitutionA=jθ(and
B=jφ) to obtain the hyperbolic identities corre-
sponding to the trigonometric identities given.- 1+tan^2 A=sec^2 A
[1−tanh^2 θ=sech^2 θ] - cos(A+B)=cosAcosB−sinAsinB
[
cosh(θ+φ)
=coshθcoshφ+sinhθsinhφ
]- sin(A−B)=sinAcosB−cosAsinB
[
sinh(θ+φ)=sinhθcoshφ
−coshθsinhφ
]- tan2A=
2tanA
1 −tan^2 A
[
tanh2θ=
2tanhθ
1 +tanh^2 θ]