Chapter 16

The relationship between

trigonometric and

hyperbolic functions

16.1 The relationship between

trigonometric and hyperbolic

functions

In Chapter 21, 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

2j

(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))

Since j^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 for

θand 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