Higher Engineering Mathematics

B

Geometry and trigonometry

17

The relationship between trigonometric

and hyperbolic functions

17.1 The relationship between
trigonometric and hyperbolic
functions

In Chapter 24, it is shown that

cosθ+jsinθ=ejθ (1)

and cosθ−jsinθ=e−jθ (2)

Adding equations (1) and (2) gives:

cosθ=

1

2

(ejθ+e−jθ) (3)

Subtracting equation (2) from equation (1) gives:

sinθ=

1

2 j

(ejθ−e−jθ) (4)

Substitutingjθforθin equations (3) and (4) gives:

cosjθ=

1

2

(ej(jθ)+e−j(jθ))

and sinjθ=

1

2 j

(ej(jθ)−e−j(jθ))

Sincej^2 =−1, cosjθ=^12 (e−θ+eθ)=^12 (eθ+e−θ)

Hence from Chapter 5,cosjθ=coshθ (5)

Similarly, sinjθ=

1

2 j

(e−θ−eθ)=−

1

2 j

(eθ−e−θ)

=

− 1

j

[

1

2

(eθ−e−θ)

]

=−

1

j

sinhθ (see Chapter 5)

But −

1

j

=−

1

j

×

j

j

=−

j

j^2

=j,

hence sinjθ=jsinhθ (6)

Equations (5) and (6) may be used to verify that in all

standard trigonometric identities,jθmay be written

forθand the identity still remains true.

Problem 1. Verify that cos^2 jθ+sin^2 jθ=1.

From equation (5), cosjθ=coshθ, and from equa-

tion (6), sinjθ=jsinhθ.

Thus, cos^2 jθ+sin^2 jθ=cosh^2 θ+j^2 sinh^2 θ, and

sincej^2 =−1,

cos^2 jθ+sin^2 jθ=cosh^2 θ−sinh^2 θ

But from Chapter 5, Problem 6,

cosh^2 θ−sinh^2 θ=1,

hence cos^2 jθ+sin^2 jθ = 1

Problem 2. Verify that sinj 2 A=2 sinjAcosjA.

From equation (6), writing 2Aforθ,

sinj 2 A=jsinh 2A, and from Chapter 5, Table 5.1,

page 45, sinh 2A=2 sinhAcoshA.

Hence, sinj 2 A= j(2 sinhAcoshA)

But, sinhA=^12 (eA−e−A) and coshA=^12 (eA+e−A)

Hence, sinj 2 A=j 2

(

eA−e−A

2

)(

eA+e−A

2

)

=−

2

j

(

eA−e−A

2

)(

eA+e−A

2

)