MATH6501 - Autumn 2016
Solutions to Problem Sheet 3
1. Squaring the two equations
cosh x + sinh x ≡ ex
cosh x − sinh x ≡ e−x
yields
cosh2 x + 2 sinh x cosh x + sinh2 x ≡ e2x
2
2
2x
cosh x − 2 sinh x cosh x + sinh x ≡ e .
Then we can add Equations (1) and (2) to obtain
2 cosh2 x + 2 sinh2 x ≡ e2x + e−2x .
Finally, divide both sides by 2 to give
cosh2 x + sinh2 x ≡ cosh 2x,
which is the desired identity.
2. (a) We will need the definition
ex − e−x
,
2
which tells us that the LHS is
sinh x =
sinh 3x =
(ex )3 − (e−x )3
e3x − e−3x
=
2
2
and
3
ex − e−x
2
x
3
(e ) − 3ex + 3e−x − (e−x )3
.
8
3
sinh x =
=
(1)
(2)
MATH6501 - Autumn 2016
Then
(ex )3 − 3ex + 3e−x − (e−x )3
4 sinh x = 4
8
x
3
x
−x
(e ) − 3e + 3e − (e−x )3
=
2
(ex )3 − (e−x )3
ex − e−x
=
−3
,
2
2
3
and the RHS is
3 sinh x + 4 sinh3 x =
(ex )3 − (e−x )3
,
2
therefore LHS≡RHS, and so
sinh 3x ≡ 3 sinh x + 4 sinh3 x.
Remark: This is analogous with the trigonometric
identity
sin 3x ≡ 3 sin x − 4 sin3 x,
except that the sign in the RHS has been flipped.
This is predicted by Osborne’s rule, since the term
4 sin3 x contains a product of two sines.
(b) Again, start with the definition
cosh x =
ex + e−x
,
2
then consider the LHS, i.e.
3x
e + e−3x
(ex )3 + (e−x )3
cosh 3x =
=
.
2
2
MATH6501 - Autumn 2016
Meanwhile
3
ex + e−x
2
x
3
(e ) + 3ex + 3e−x + (e−x )3
,
8
3
cosh x =
=
thus
(ex )3 + 3ex + 3e−x + (e−x )3
4 cosh x = 4
8
x
3
x
−x
(e ) + 3e + 3e + (e−x )3
=
2
ex + e−x
(ex )3 + (e−x )3
+3
,
=
2
2
3
and the RHS is
4 cosh3 x − 3 cosh x =
(ex )3 + (e−x )3
,
2
so LHS≡RHS. Thus
cosh 3x ≡ 4 cosh3 x − 3 cosh x.
Remark: The analogous trigonometric identity is
cos 3x ≡ 4 cos3 x − 3 cos x.
Note that since there are no products of sines, all
of the signs match up perfectly, as predicted by
Osborne’s Rule.
MATH6501 - Autumn 2016
3. (a) We begin by rewriting y = coth−1 x as
x = coth y =
ey + e−y
.
ey − e−y
Next,
x(ey − e−y )
= (ey + e−y )
⇒
x(e2y − 1)
= (e2y + 1)
⇒
e2y (x − 1)
=x+1
⇒
(ey )2 (x
− 1) = x + 1,
which rearranges to give
(ey )2 (x − 1) − (x + 1) = 0,
which is a quadratic equation for ey (with a = (x−1),
b = 0, c = −(x + 1). Solving this for ey gives
r
x+1
y
e =±
,
x−1
but ey > 0, so we choose the positive square root
and
!
r
r
x+1
x+1
y
⇒
y = ln
,
e =
x−1
x−1
i.e.
"
y = ln
x+1
x−1
1 #
2
⇒
1
y = ln
2
x+1
x−1
,
which is the desired result. Note that any other
valid method which gives this result will also be
accepted.
MATH6501 - Autumn 2016
(b) Using Equation (1) from part (a), we see that for
x = 54 ,
−1 5
y = coth
4
!
5
+
1
1
= ln 45
2
4 −1
1
9/4
= ln
2
1/4
1
= ln 9
2
= ln 3.
4. Here is a sketch of the scenario. . .
Figure 1: A plot of the shape of the live cable; this is given by
the blue curve. Also, if you thought the pavement (shown in
orange) was on the x-axis, think again! The y-coordinate of
the pavement is 20 cosh(1) − 16.5 ≈ 14.3 feet, not zero.
MATH6501 - Autumn 2016
If a = 20 then the equation becomes
x
.
y(x) = 20 cosh
20
The poles are at x = ±20, where
y(±20) = 20 cosh(1) ≈ 30.8
(remembering that cosh(x) is an even function, so cosh(−x) =
cosh(x).)
The lowest point of the curve is at the minimum, where
dy
dx = 0:
0 = y 0 (x)
x
20
=
sinh
,
20
20
i.e. when x = 0. This is clear anyway from thinking
about the shape of the cosh graph.
When x = 0,
y(0) = 20 cosh(0) = 20.
So the difference in height between the top of the cable
and the bottom of the cable is 30.8 − 20 = 10.8 feet
00
= 100 9 12 .
The cable hangs from the tops of poles that are 160 600 tall,
00
00
so the lowest point of the cable is 160 600 − 100 9 12 = 50 8 12
00
off the ground. So Carl, at 60 1 12 , will definitely get
00
fried, while Amanda, at 50 3 12 , will survive (but she will
probably be shocked emotionally, even if not electrically!)