Higher Engineering Mathematics

THE RELATIONSHIP BETWEEN TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS 175

B

cosθ−cosφ=−2 sin

(

θ+φ

2

)

sin

(

θ−φ

2

)

(see Chapter 18, page 184)

thus cosjA−cosjB

=−2 sinj

(

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

)

Butj^2 =−1, hence

coshA−coshB=2 sinh

(

A+B

2

)

sinh

(

A−B

2

)

Problem 5. Develop the hyperbolic identity

corresponding to sin 3θ=3 sinθ−4 sin^3 θby

writingjAforθ.

SubstitutingjAforθgives:

sin 3jA=3 sinjA−4 sin^3 jA

and since from equation (6),

sinjA=jsinhA,

jsinh 3A= 3 jsinhA− 4 j^3 sinh^3 A

Dividing throughout byjgives:

sinh 3A=3 sinhA−j^2 4 sinh^3 A

But j^2 =−1, hence

sinh 3A=3 sinhA+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 θis 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 44].

Now try the following exercise.

Exercise 79 Further problems on hyper-

bolic identities

In Problems 1 to 9, use the substitutionA=jθ

(andB=jφ) 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)=cosAcosB−sinAsinB
   cosh (θ+φ)
   =coshθcoshφ+sinhθsinhφ

]

3. sin (A−B)=[sinAcosB−cosAsinB
   sinh (θ+φ)=sinhθcoshφ
   −coshθsinhφ

]

4. tan 2A=

2 tanA

1 −tan^2 A[

tanh 2θ=

2 tanhθ

1 +tanh^2 θ

]

5. cosAsinB=

1

2

[ sin (A+B)−sin (A−B)]

⎡

⎣coshθcoshφ=

1

2

[sinh(θ+φ)

−sinh(θ−φ)]

⎤

⎦

6. sin^3 A=

3

4

sinA−

1

4

sin 3A

[

sinh^3 θ=

1

4

sinh 3θ−

3

4

sinhθ

]

7. cot^2 A(sec^2 A−1)= 1
   [coth^2 θ(1−sech^2 θ)=1]