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