160 Higher Engineering Mathematics
But from Chapter 5, Problem 6,
cosh
2
θ − sinh
2
θ = 1,
hence cos
2 j θ + sin
2 j θ = 1
Problem 2. Verify that sin j 2 A = 2 sin jA cos jA.
From equation (6), writing 2 A for θ, sin j 2 A= j sinh 2 A,
and from Chapter 5, Table 5.1, page 45, sinh 2A =
2 sinh A cosh A.
Hence,
sin j 2 A = j (2 sinh A cosh A)
But, sinh A =
1
2 (e A − e − A ) and cosh A =
1
2 (e A + e − A )
Hence, sin j 2 A = j 2
e A − e − A
2
e A + e − A
2
= −
2
j
e A − e − A
2
e A + e − A
2
= −
2
j
sin j θ
j
(cos j θ)
= 2 sin jA cos jA since j
2
= −1
i.e.
sin j 2A = 2 sin jAcos jA
Now try the following exercise
Exercise 70 Further problems on the
relationship between trigonometric and
hyperbolic functions
Verify the following identities by expressing in
exponential form.
1. sin j (A + B) = sin jA cos jB + cos j Asin jB
2. cos j (A − B) = cos jA cos jB + sin j Asin jB
3. cos j 2 A =1 − 2 sin 2 jA
4. sin jA cos jB =
1
2 [sin j (A + B) + sin j (A − B)]
5. sin jA − sin jB
= 2 cos j
A + B
2
sin j
A − B
2
16.2 Hyperbolic identities
From Chapter 5, cosh θ =
1
2 (e θ + e −θ )
Substituting j θ for θ gives:
cosh j θ =
1
2 (e j θ + e − j θ ) = cos θ, from equation (3),
i.e. cosh jθ = cos θ
(7)
Similarly, from Chapter 5,
sinh θ =
1
2 (e
θ
− e
−θ
)
Substituting j θ for θ gives:
sinh j θ =
1
2 (e j θ − e − j θ ) = j sin θ, from equation (4).
Hence sinh jθ = j sin θ
(8)
tan j θ =
sin j θ
cosh j θ
From equations (5) and (6),
sin j θ
cos j θ
=
j sinh θ
cosh θ
= j tanh θ
Hence tan jθ = j tanh θ
(9)
Similarly, tanh j θ =
sinh j θ
cosh j θ
From equations (7) and (8),
sinh j θ
cosh j θ
=
j sin θ
cos θ
= j tan θ
Hence tanh jθ = j tan θ
(10)
Two methods are commonly used to verify hyperbolic
identities. These are (a) by substituting j θ (and j φ) 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 45.
Problem 3. By writing jA for θ in cot 2 θ + 1 =
cosec 2 θ, determine the corresponding hyperbolic
identity.
Substituting jA for θ gives:
cot
2 jA + 1 = cosec
2 jA,
i.e.
cos 2 jA
sin 2 jA
+ 1 =
1
sin 2 jA
But from Chapter 5, Problem 6,
cosh
2
θ − sinh
2
θ = 1,
hence cos
2 j θ + sin
2 j θ = 1
Problem 2. Verify that sin j 2 A = 2 sin jA cos jA.
From equation (6), writing 2 A for θ, sin j 2 A= j sinh 2 A,
and from Chapter 5, Table 5.1, page 45, sinh 2A =
2 sinh A cosh A.
Hence,
sin j 2 A = j (2 sinh A cosh A)
But, sinh A =
1
2 (e A − e − A ) and cosh A =
1
2 (e A + e − A )
Hence, sin j 2 A = j 2
e A − e − A
2
e A + e − A
2
= −
2
j
e A − e − A
2
e A + e − A
2
= −
2
j
sin j θ
j
(cos j θ)
= 2 sin jA cos jA since j
2
= −1
i.e.
sin j 2A = 2 sin jAcos jA
Now try the following exercise
Exercise 70 Further problems on the
relationship between trigonometric and
hyperbolic functions
Verify the following identities by expressing in
exponential form.
1. sin j (A + B) = sin jA cos jB + cos j Asin jB
2. cos j (A − B) = cos jA cos jB + sin j Asin jB
3. cos j 2 A =1 − 2 sin 2 jA
4. sin jA cos jB =
1
2 [sin j (A + B) + sin j (A − B)]
5. sin jA − sin jB
= 2 cos j
A + B
2
sin j
A − B
2
16.2 Hyperbolic identities
From Chapter 5, cosh θ =
1
2 (e θ + e −θ )
Substituting j θ for θ gives:
cosh j θ =
1
2 (e j θ + e − j θ ) = cos θ, from equation (3),
i.e. cosh jθ = cos θ
(7)
Similarly, from Chapter 5,
sinh θ =
1
2 (e
θ
− e
−θ
)
Substituting j θ for θ gives:
sinh j θ =
1
2 (e j θ − e − j θ ) = j sin θ, from equation (4).
Hence sinh jθ = j sin θ
(8)
tan j θ =
sin j θ
cosh j θ
From equations (5) and (6),
sin j θ
cos j θ
=
j sinh θ
cosh θ
= j tanh θ
Hence tan jθ = j tanh θ
(9)
Similarly, tanh j θ =
sinh j θ
cosh j θ
From equations (7) and (8),
sinh j θ
cosh j θ
=
j sin θ
cos θ
= j tan θ
Hence tanh jθ = j tan θ
(10)
Two methods are commonly used to verify hyperbolic
identities. These are (a) by substituting j θ (and j φ) 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 45.
Problem 3. By writing jA for θ in cot 2 θ + 1 =
cosec 2 θ, determine the corresponding hyperbolic
identity.
Substituting jA for θ gives:
cot
2 jA + 1 = cosec
2 jA,
i.e.
cos 2 jA
sin 2 jA
+ 1 =
1
sin 2 jA