A Course in Mathematical Methods for Physicists – Russell L. Herman (1st Edition, 9781466584679) – Complete Solutions Manual

SOLUTIONS MANUAL

A COURSE IN MATHEMATICAL METHODS FOR PHYSICISTS



CHAPTER NO. 01: INTRODUCTION AND REVIEW
1. Prove the following identities using only the definitions of the trigono-
metric functions, the Pythagorean identity, or the identities for sines and
cosines of sums of angles.

a. cos 2x = 2 cos2 x − 1.

cos 2x = cos2 x − sin2 x
= cos2 x − (1 − cos2 x )
= 2 cos2 x − 1.

b. sin 3x = A sin3 x + B sin x, for what values of A and B?

sin 3x = sin x cos 2x + sin 2x cos x
= sin x (cos2 x − sin2 x ) + 2 sin x cos2 x
= 3 sin x (1 − sin2 x ) − sin3 x
= 3 sin x − 4 sin3 x.

So, A = −4, B = 3.
 
θ π
c. sec θ + tan θ = tan + .
2 4
 
θ π

θ π
 sin 2 + 4
tan + =  
2 4 cos θ
+ π
2 4

sin 2θ cos π4 + sin π4 cos 2θ
=
cos 2θ cos π4 − sin π4 sin 2θ
sin 2θ + cos 2θ
=
cos 2θ − sin 2θ
!
sin 2θ + cos 2θ cos 2θ + sin 2θ
=
cos 2θ − sin 2θ cos 2θ + sin 2θ
 2
cos 2θ + sin 2θ
=
cos2 θ
2 − sin2 θ
2
cos2 θ
2 + sin2 2θ + 2 sin 2θ cos 2θ
=
cos θ

, 1 + sin θ
=
cos θ
= sec θ + tan θ.

2. Determine the exact values of
π
a. sin .
8

π 1 π
sin2 = 1 − cos
8 2 4
√ !
1 2
= 1−
2 2
1  √ 
= 2− 2 .
4
π 1
q √
Therefore, sin = 2− 2.
8 2
b. tan 15o .


tan 15o = tan(60o − 45o )
tan 60o − tan 45o
=
1 + tan 60o tan 45o
√
3−1
= √
1+ 3
√ √ !
3−1 1− 3
= √ √
1+ 3 1− 3
√
= 2 − 3.

One can get the same answer using
√
2 osin2 15o 1 − cos 30o 2− 3 √ 2
tan 15 = = = √ = ( 2 − 3) .
cos2 15o 1 + cos 30o 2+ 3

c. cos 105o .


1
cos2 105o (1 + cos 210o )
=
2
1
= (1 − cos 30o )
2
1 √
= (2 − 3).
4
p √
Therefore, cos 105o = − 21 2 − 3. One could also use

1
cos2 105o = sin2 15o = (1 − cos 30o ) .
2
Note that this answer can be denested:
1
q √ 1
q √
− 2− 3 = − 8−4 3
2 4

, 1
q √ √
= − 6−2 2 6+2
4q
1 √ √
= − ( 6 − 2)2
4
1 √ √
= ( 2 − 6).
4
3. Denest the following if possible.
p √
a. 3 − 2 2.
p √ √
Assume that 3 − 2 2 = a + b 2. Then,
√ √
3 − 2 2 = ( a + b 2)2
√
= a2 + 2b2 + 2ab 2.

This would hold if a2 + 2b2 = 3 and 2ab = −2. These simultaneous
equations have
p the√solutions a = −b = ±1. Therefore, the positive
√
solution is 3 − 2 2 = 2 − 1.
p √
b. 1 + 2.
Following the reasoning from the last problem, one tries to solve
the system a2 + 2b2 = 1 and 2ab = 1. Solving the second equation
for b = 1/2a and inserting this into the first equation, we have

2a4 + a2 − 2 = 0.

Solving, we obtain √
2 17 − 1
a = .
4
p √
Therefore, 1 + 2 cannot be denested.
p √
c. 5 + 2 6.
√ √ √ √ √
Note that 5 + 2 6 = 2 + 3 + 2 2 3 = ( 2 + 3)2 . Therefore,
p √ √ √
5 + 2 6 = 2 + 3.
p
3
√ p3
√
d. 5+2− 5 − 2.
p
3
√ √
We first guess the form 5 ± 2 = a + b 5. Cubing both sides of
this equation, we find
√ √ √
5 ± 2 = a3 + 3a2 b 5 + 15ab2 + 5b3 5.

This leads to the system of equations

a3 + 15ab2 = ±2
2 3
3a b + 5b = 1.

Solving the second equation for a, we have
p
3b(1 − 5b3 )
a=± .
3b
Substituting this into the first equation of the system, we obtain
p
3b(1 − 5b3 )(40b3 + 1)
= ±2.
9b2

, Now, square both sides of the equation. Rearranging and factoring,
we have
(8b3 − 1)(1000b6 − 25b3 + 1) = 0.
The only real solutions come from b3 = 1/8. Thus, b = 1/2.
Inserting this result into the expression for a,
p
3/2(1 − 5/8) 3/4 1
a=± =± =± .
3/2 3/2 2
√ √
Therefore, 5 ± 2 = 12 ( 5 ± 1).
Using this result, we can proceed to solve the problem:
q√ q√
3 3 1 √ 1 √
5+2− 5 − 2 = ( 5 + 1) − ( 5 − 1) = 1.
2 2
√
e. Find the roots of x2 + 6x − 4 5 = 0 in simplified form.
p √ p √
From the quadratic formula, x = −3 ± 9 + 4 5. Setting 9 + 4 5 =
√
a + b 5, we need to solve the equations

a2 + 5b2 = 9
2ab = 4.

Solving for b = 2a , and substituting into the first equation, we have

a4 − 9a2 + 20 = 0.

Solving for a2 ,
9±1
a2 = = 4, 5.
2
√ √ p √
So, a = 2, 5 and, therefore, b = 2/ 5, 1. Both cases give 9 + 4 5 =
√
2 + 5.
Using this result, we find the solutions of the problem, x = −1 +
√ √
5, x = −5 − 5.
5
4. Determine the exact values of
4
 
3
a. sin cos−1 .
θ 5
3 3 3
Figure 1.1: Triangle for Problem 4a.
This problem takes the form sin θ for θ = cos−1 , or cos θ = .
5 5
One can draw a triangle with base of length 3 and hypotenuse
4
of length 5 as shown in Figure 1.1. Then, sin θ = . This is also
5
obtainable from the Pythagorean identity,
 2
7 3
+ sin2 θ = 1.
5
x  x
b. tan sin−1 .
√
θ 7
49 − x2 Similar to the last problem one can use the triangle in Figure 1.2.
x
Figure 1.2: Triangle for Problem 4b. Let sin θ = . Then, tan θ = 49−x x2 .
7