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.
- sinj(A+B)=sinjAcosjB+cosjAsinjB
- cosj(A−B)=cosjAcosjB+sinjAsinjB
- cosj 2 A= 1 −2 sin^2 jA
- sinjAcosjB=^12 [ sinj(A+B)+sinj(A−B)]
- 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
)