Higher Engineering Mathematics

174 GEOMETRY AND TRIGONOMETRY

=−

2

j

(

sinjθ

j

)

( cosjθ)

=2 sinjAcosjA sincej^2 =− 1

i.e. sinj 2 A=2 sinjAcosjA

Now try the following exercise.

Exercise 78 Further problems on the rela-

tionship between trigonometric and hyper-

bolic functions

Verify the following identities by expressing in

exponential form.

1. sinj(A+B)=sinjAcosjB+cosjAsinjB


2. cosj(A−B)=cosjAcosjB+sinjAsinjB


3. cosj 2 A= 1 −2 sin^2 jA


4. sinjAcosjB=^12 [ sinj(A+B)+sinj(A−B)]


5. sinjA−sinjB

=2 cosj

(

A+B

2

)

sinj

(

A−B

2

)

17.2 Hyperbolic identities

From Chapter 5, coshθ=^12 (eθ+e−θ)

Substitutingjθforθgives:

coshjθ=^12 (ejθ+e−jθ)=cosθ, from equation (3),

i.e. coshjθ=cosθ (7)

Similarly, from Chapter 5,

sinhθ=^12 (eθ−e−θ)

Substitutingjθforθgives:

sinhjθ=^12 (ejθ−e−jθ)=jsinθ, from equation (4).

Hence sinhjθ= jsinθ (8)

tanjθ=

sinjθ

coshjθ

From equations (5) and (6),

sinjθ

cosjθ

=

jsinhθ

coshθ

=jtanhθ

Hence tanjθ=jtanhθ (9)

Similarly, tanhjθ=

sinhjθ

coshjθ

From equations (7) and (8),

sinhjθ

coshjθ

=

jsinθ

cosθ

=jtanθ

Hence tanhjθ=jtanθ (10)

Two methods are commonly used to verify hyper-

bolic identities. These are (a) by substitutingjθ(and

jφ) in the corresponding trigonometric identity and

using the relationships given in equations (5) to (10)

(see Problems 3 to 5) and (b) by applying Osborne’s

rule given in Chapter 5, page 44.

Problem 3. By writingjAforθin

cot^2 θ+ 1 =cosec^2 θ, determine the

corresponding hyperbolic identity.

SubstitutingjAforθgives:

cot^2 jA+ 1 =cosec^2 jA,

i.e.

cos^2 jA

sin^2 jA

+ 1 =

1

sin^2 jA

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 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 jAand jBforθ

andφrespectively in the trigonometric identity

for cosθ−cosφ, show that

coshA−coshB

=2 sinh

(

A+B

2

)

sinh

(

A−B

2

)