Contents
195
Chapter 4 Trigonometry for Combinations of Angles
4.1 Sums and Differences of Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
4.2 Multiples of Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
4.3 From Sums to Products and Back . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
Appendices
1001
A Algebra Review I: Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1001
B Algebra Review II: Matrix Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . 1011
i
Chapter 4
Trigonometry for Combinations of Angles
In this Chapter we learn how to handle combinations of angles, such as sums, differences,
doubles and halves.
4.1 Sums and Differences of Angles
In this section we introduce several key formulas that allow us to compute the trigonometric
functions of the sum and the difference of two angles.
 Before you begin this section, please read Appendix B, which reviews matrix arithmetic.
Rotation Matrices
The most important matrices we use in Trigonometry are the rotation matrices, which we have
already encountered several times (see Sections 2.1, 2.2, and 3.1). These special matrices are those
of the form:
cos τ −sin τ
Rτ =
.
(4.1.1)
sin τ cos τ
Their main feature, upon which everything learned about trigonometric functions builds, is the
following.
Rotation Principle
→
−
Given
a rotation measure τ , the vector obtained by applying a τ -rotation to a vector v =
x
is given by the product:
y
cos τ −sin τ
x
xcos τ − ysin τ
→
−
Rτ v =
·
=
.
sin τ cos τ
xsin τ + ycos τ
y
cos τ
→
In particular, the unit vector −
uτ =
, which represents the first column of the
sin τ
→
−
→
−
rotation matrix Rτ , and
is given by u τ = Rτ u 0 , is the result of applying a τ -rotation to
1
→
the vector −
u0 =
.
0
Since applying a rotation is the same as multiplying by a rotation matrix, we clearly see that rotation matrices satisfy the following important fact, which first appeared as formula (??) in Section
2.2.
195
196
4.1. SUMS AND DIFFERENCES OF ANGLES
Product Rule for Rotation Matrices
The product of two rotation matrices is again a rotation matrix, namely
Rα+β = Rα · Rβ .
(4.1.2)
x
→
−
by the product Rα · Rβ , we can also
C LARIFICATION. When we multiply some v =
y
write
→
→
(Rα · Rβ )−
v = Rα · (Rβ −
v ),
→
which can be understood as a two-step transformation of −
v : first we apply a β-rotation, then we
apply an α-rotation (to what we got from the first step). Clearly, this is the same as applying an
→
(α+β)-rotation, that is, multiplying directly −
v by Rα+β , thus proving the equality (4.1.2).
The Angle Sum Formulas for Sine and Cosine
When we write down the Product Rule (4.1.2) explicitly (by identifying the coefficients of all
three matrices involved in it), we get:
cos(α+β) −sin(α+β)
cos α −sin α
cos β −sin β
=
·
(4.1.3)
sin(α+β) cos(α+β)
sin α cos α
sin β cos β
When we take a closer look at the matrix product Rα · Rβ that appears in the right-hand side
of (4.1.3), especially to the entries that sit in the first column of Rα · Rβ , we immediately get:
The Angle Sum Formulas for Sine and Cosine
cos(α+β) = cos αcos β−sin αsin β;
sin(α+β) = sin αcos β+cos αsin β;
(4.1.4)
(4.1.5)
 As we see, the cosine and sine of a sum of two angles are given by complicated formulas.
Equalities like “cos(α + β) = cos α + cos β” or “sin(α + β) = sin α + sin β” are WRONG!
(They are what we call false identities!)
The best way to memorize the Angle Sum Formulas is to remember the Product Rule (4.1.2)
and the definition (4.1.1) of rotation matrices.
C LARIFICATION. What we see from (4.1.4) and (4.1.5) is that: in order to be able to compute
the trigonometric functions of an angle sum α + β, we need to know all four values of cos α, sin α,
cos β, and sin β.
Example 4.1.1. Suppose we are given the following information:
3
(a) α is in Quadrant III, and has tan α = .
4
12
(b) β is in Quadrant II, and has sin β = .
13
Given this information, we wish to find sine and cosine of α + β, as well as the quadrant where
α + β sits in.
Based on the Clarification above, our “wish list” is:
cos α = ?
cos β = ?
sin α = ?
12
sin β = .
13
CHAPTER 4. TRIGONOMETRY FOR COMBINATIONS OF ANGLES
197
We will fill in our “wish list” using the techniques introduced in Section 2.2. (When we first dealt
with these type problems, we learned of two techniques: the algebraic method and the coordinate
method. For the benefit of the reader, we will illustrate both.)
To find sine and cosine of α, we use the coordinate method. We build α as an angle in standard
position, and we pick a point P (x, y) on the terminal side of α, which we know that must satisfy
y
= tan α. Based on the given information on α, a correct choice would be x = −4 and y = −3,
x
p
p
√
√
which will then give us r = x2 + y 2 = (−4)2 + (−3)2 = 16 + 9 = 25 = 5, from which
x
4
y
3
we quickly get: cos α = = − and sin α = = − .
r
5
r
5
As for β, we already have its sine, so we can use the algebraic method for finding its cosine:
s
r
r
2
q
144
12
169 144
2
cos β = ± 1 − sin β = ± 1 −
=± 1−
=±
−
=
13
169
169 169
r
r
169 − 144
25
5
=±
=±
=± .
169
169
13
Since β is in Quadrant II, where cosine is negative, the correct value is: cos β = −
5
.
13
If we go back to our “wish list,” we now have all we need:
4
5
5
cos β = −
13
cos α = −
3
5
12
sin β = ,
13
sin α = −
so using the angle sum formulas we can compute:
4
5
3
12
cos(α+β) = cos α cos β − sin α sin β = −
− −
=
−
5
13
5
13
20 36
20 + 36
56
=
+
=
= ;
65 65
65
65 3
5
4
12
sin(α+β) = sin α cos β + cos α sin β = −
−
+ −
=
5
13
5
13
15 48
15 − 48
33
=
−
=
=− .
65 65
65
65
Since cos(α+β) is positive, and sin(α+β) is negative, it follows that α+β is in Quadrant IV.
The Angle Difference Formulas for Sine and Cosine
Based on the Angle Sum Formulas, we can also deal with differences. After all, we can always
write α−β = α+(−β). Using the formulas for negatives, we know that cos(−β) = cos β and
sin(−β) = −sin β, and therefore, the cosine and sine of the difference are given by:
198
4.1. SUMS AND DIFFERENCES OF ANGLES
The Angle Difference Formulas for Sine and Cosine
cos(α−β) = cos αcos β+sin αsin β;
sin(α−β) = sin αcos β−cos αsin β;
(4.1.6)
(4.1.7)
Example 4.1.2. Suppose we want to compute the cosine and sine of the angle
8
ψ = arcsin −
17
24
− arccos
,
25
and find the quadrant ψ sits in.
8
24
We start off by naming the first angle arcsin −
= α, and the second angle arccos
=
17
25
β, so the angle we are interested in is ψ = α − β.
Using the formulas from Section 2.6 (see Trigonometric Functions of Inverses), we can quickly
compute:
8
8
sin α = sin arcsin −
=− ;
17
17
r
s
2 r
8
8
64
225
15
cos α = cos arcsin −
= 1− −
= 1−
=
= ;
17
17
289
289
17
24
24
cos β = cos arccos
= ;
25
25
s
r
2 r
24
24
576
49
7
sin β = sin arccos
= 1−
= 1−
=
= .
25
25
625
625
25
With our “wish list” filled, using the angle differnece formulas we can compute:
15
24
8
7
cos ψ = cos(α−β) = cos α cos β + sin α sin β =
+ −
=
17
25
17
25
360
56
360 − 56
304
=
−
=
=
;
425 425
425
425 8
24
15
7
sin ψ = sin(α−β) = sin α cos β − cos α sin β = −
−
=
17
25
17
25
−192 − 105
192 105
297
=−
−
=
=−
.
425 425
425
425
Since cos ψ is positive, and sin ψ is negative, it follows that ψ is in Quadrant IV.
Application: Expressions of the Form E(x) = a cos kx + b sin kx
Using the Angle Difference Formula for Cosine, there is a neat way to simplify expressions of
the form:
E(x) = acos kx + bsin kx.
CHAPTER 4. TRIGONOMETRY FOR COMBINATIONS OF ANGLES
199
The main idea is to make √
such expressions match the cosine of a difference. Of course, this is only
possible, if we divide by a2 + b2 , so a better expression we should first consider is:
√
a
b
cos kx + √
sin kx.
a2 + b2
a2 + b2
We know, from Section 3.3, that we can always find an angle τ , such that

a

 cos τ = √ 2
a + b2
b

 sin τ = √
a2 + b2
−
v =
for instance we can take τ to be standard direction angle of the vector →
by the formula:
τ=
a
, which is given
b

a


(sign of b) arccos √
, if b 6= 0



a2 + b2


a


, if b = 0
 arccos √ 2
a + b2
With this choice of τ , we can write
√
(4.1.8)
a
b
cos kx + √
sin kx = cos kx cos τ + sin kx cos τ = cos(kx − τ ),
2
2
+b
a + b2
a2
so the original expression we wanted to simplify can now be written as follows.
With τ is given by (4.1.8), the expression E(x) = acos kx + bsin kx can also be presented
as:
√
E(x) = acos kx + bsin kx = a2 + b2 cos(kx − τ ).
(4.1.9)
Example 4.1.3. Suppose we are given the function f (x) = 3cos 5x − 2sin 5x, and we are
asked to find its maximum value, as well as the x-values where f (x) attains its maximum.
The idea is to re-write f (x) as in (4.1.9). We start off by identifying the coefficients: a = 3,
b = −2. For future use, we also compute at this time the value:
p
√
√
a2 + b2 = 32 + (−2)2 = 13.
Since b is negative, the angle τ given by (4.1.8) is:
a
3
τ = − arccos √
= − arccos √
.
a2 + b2
13
We defer the calculation of τ until later. However, for the record, we
point
point
that, by construca
3
tion (remember that τ is a standard direction angle of the vector
=
) we already
b
−2
know that
−π ≤ τ ≤ π.
(4.1.10)
200
4.1. SUMS AND DIFFERENCES OF ANGLES
With τ more-or-less in hand, using (4.1.9) we can now write our given function as
f (x) =
√
13 cos(5x − τ ).
(4.1.11)
√
With this alternative presentation of f (x), it is now clear that its maximum value is 13. Furthermore, the x-values where this maximum is attained are the solutions of the equation
cos(5x − τ ) = 1.
Using the substitution 5x − τ = y, this equation becomes cos y = 1, and all its solutions are1
y = 2nπ, n integer, so when we go back to the substitution, we get 5x −√τ = 2nπ, which yields
5x = τ + 2nπ, thus the x-values where f (x) attains its maximum value ( 13) are:
x=
2nπ
τ
+
, n integer.
5
5
(4.1.12)
It should be pointed out here that, when we plug in n = 0, we get the special value
x=
τ
≃ −.117600521,
5
√
which is the closest x-value to zero, at which f (x) attains its maximum value ( 13). This is due
π π
τ
to the fact that, by dividing all sides in (4.1.10) by 5, it follows that is in the interval − , ,
5
5 5
2π
so adding any non-zero multiple of
will result in a number outside this interval.
5
Using presentations like (4.1.9), we can also develop a method for solving equations of the
form
acos kx + bsin kx = c.
(4.1.13)
Assume both a and b are non-zero. If we set τ to be given by (4.1.8), then the solutions of
the equation
√(4.1.13) are as follows.
√
I. If c > √a2 + b2 , or c < − a2 + b2 , then there are no solutions.
II. If c = a2 + b2 , then the solutions are:
x=
1
τ + 2nπ , n integer.
k
√
III. If c = − a2 + b2 , then the solutions are:
x=
1
1
τ + π + 2nπ , n integer.
k
There is no need to use arccos for this equation! We know that the solutions are the “peaks” of cosine!
CHAPTER 4. TRIGONOMETRY FOR COMBINATIONS OF ANGLES
201
√
√
III. If − a2 + b2 < c < a2 + b2 , then the solutions are:
1
c
x=
τ ± arccos √
+ 2nπ , n integer.
k
a2 + b2
C LARIFICATION . With τ given by (4.1.8), the equation (4.1.13) is equivalent to:
cos(kx − τ ) = √
which after the substitution kx − τ = u becomes
cos u = √
c
,
a2 + b2
c
.
a2 + b2
Depending of the four possibilities listed, we immediately get:
c
c
I. If √
> 1, or √
< −1, then there are no solutions.
a2 + b2
a2 + b2
c
II. If √
= 1, then:
a2 + b2
u = 2nπ, n integer.
c
= −1, then:
III. If √
a2 + b2
u = π + 2nπ, n integer.
c
< 1, then:
III. If −1 < √
2
a + b2
c
u = ± arccos √
+ 2nπ, n integer.
a2 + b2
Going back to the substitution, we clearly have kx = τ + u, and then all the given formulas are
clear, by dividing everything by k.
Example 4.1.4. Let us solve the equation cos 3t + sin 3t = 1.
π
Our angle τ can be easily guessed here. Since a = b = 1, we clearly have τ = , so we can
4
write
√
√ h
1
1
π
πi √
π
cos t+sin t = 2 √ cos 3t + √ sin 3t = 2 cos 3t cos + sin 3t sin
= 2 cos 3t −
.
4
4
4
2
2
√
π
Our equation becomes 2 cos 3t −
= 1, which can be rewritten as
4
π
1
cos 3t −
=√ .
4
2
Using the substitution 3t −
π
= u, this equation reduces to
4
1
cos u = √ ,
2
202
4.1. SUMS AND DIFFERENCES OF ANGLES
which yields: u = ±arccos
we now get
1
π
√ +2nπ = ± +2nπ, n integer. Going back to our substitution,
4
2
3t −
so when we add
π
π
= ± + 2nπ,
4
4
π
to both sides we get
4
3t =
π π
± + 2nπ,
4
4
which after dividing everything by 3 yields
1 π π
t=
± + 2nπ , n integer.
3 4
4
If we wish, wecan break thesesolutions into two lists:
1 π π
π 2nπ
(a) t =
+ + 2nπ = +
, n integer;
3 4
4
6
3
2nπ
1 π π
(b) t =
− + 2nπ =
, n integer.
3 4
4
3
Example 4.1.5. Suppose we are asked to find all solutions of the equation
2 cos 2x + sin 2x = 1,
that are in the interval [−π, π].
We will transform the left-hand side of our equation in a manner similar to what we did in
Example 4.1.3. We start off by identifying the coefficients: a = 2, b = 1. For future use, we also
compute at this time the value:
√
√
√
a2 + b2 = 22 + 12 = 5.
Since b is positive, the angle τ given by (4.1.8) is:
a
2
τ = arccos √
= arccos √ .
a2 + b2
5
We defer the calculation of√τ until later. With τ more-or-less in hand, using
√(4.1.9) we can now
write 2cos 2x + sin 2x = 5 cos(2x − τ ), so now our equation becomes: 5 cos(2x − τ ) = 1,
which we can also write as:
1
cos(2x − τ ) = √ .
5
Using the substitution 2x − τ = w, our equation reads:
1
cos w = √ ,
5
which has as its solutions:
w = ± arccos
1
√
5
+ 2nπ, n integer.
CHAPTER 4. TRIGONOMETRY FOR COMBINATIONS OF ANGLES
203
1
As we did with τ , we will defer the calculation of arccos √ . Let us just give it a name for
5
now: call it φ. If we go back to our substitution, we get 2x − τ = ± φ + 2nπ, which yields
2x = τ ± φ + 2nπ, and then we get:
x=
1
τ ±φ
[τ ± φ + 2nπ] =
+ nπ, n integer.
2
2
We can now start our computations, and split the solutions into the following two lists (depending
on the ± sign):
τ +φ
+ nπ ≃ 0.785398163 + nπ, n integer. From this list, it is clear that only the
(a) x =
2
values given by n = 0 and n = −1 will yield numbers in [−π, π]:
x1 ≃ 0.785398163;
x2 ≃ 0.785398163 − π ≃ −2.35619449.
τ −φ
+ nπ ≃ −0.321750054 + nπ, n integer. From this list, it is clear that only the
2
values given by n = 0 and n = 1 will yield numbers in [−π, π]:
(b) x =
x3 ≃ −0.321750054;
x4 ≃ −0.321750054 + π ≃ 2.819842099.
The Tangent Formulas
If we pay special attention to the Angle Sum Formulas, we see that we can write
tan(α+β) =
sin(α+β)
sin α cos β + cos α sin β
=
.
cos(α+β)
cos α cos β − sin α sin β
(4.1.14)
We can push this a little bit by force-factoring cos α cos β both upstairs and downstairs, so we can
write
sin α cos β
cos α sin β
Numerator = cos α cos β
+
= cos α cos β [tan α + tan β] ;
cos α cos β cos α cos β
cos α cos β
sin α sin β
−
= cos α cos β [1 − tan α tan β] .
Denominator = cos α cos β
cos α cos β cos α cos β
This way, going back to (4.1.14), we arrive to the following two interesting formulas (the one for
differences is obtained the same way):
204
4.1. SUMS AND DIFFERENCES OF ANGLES
Angle Sum/Difference Formulas for Tangent
tan α+tan β
;
1−tan α tan β
tan α−tan β
tan(α−β) =
.
1+tan α tan β
tan(α+β) =
What is nice about these two formulas is the fact that, if we only want to know tangent of a
sum or a difference of two angles, we do not need all four values of sines and cosines.
Example 4.1.6. Suppose we want to compute tan [arctan(10) + arctan(7)]. If we set up
arctan(10) = α and arctan(7) = β, then tan α = 10 and tan β = 7, so we can immediately
compute
tan [arctan(10) + arctan(7)] = tan(α + β) =
tan α + tan β
10 + 7
17
=
=− .
1 − tan α tan β
1 − 10 · 7
69
In this particular example, we can also locate the quadrant of α + β, as follows. We know that,
π π
since both α and β are in − ,
and their tangents are positive, these two angles belong in fact
2 2
π
π
to the interval 0, , thus 0 < α, β < , so if we add them, we get 0 < α + β < π, so α + β can
2
2
only be in Quadrant I or II. Since tan(α + β) is negative, it follows that α + β is in Quandrant II.
Exercises
In Exercises 1-8 you are asked to prove several well known identities, using the addition and
subtraction formulas.
π
= cos t.
1. sin t +
2
2. sin t −
3. cos t +
4. cos t −
π
= − cos t.
2
π
= − sin t.
2
π
= sin t.
2
5. sin(t + π) = − sin t.
6. sin(t − π) = − sin t.
7. cos(t + π) = − cos t.
8. cos(t − π) = − cos t.
CHAPTER 4. TRIGONOMETRY FOR COMBINATIONS OF ANGLES
205
In Exercises 9-13 you are asked to find the exact values of sine, cosine, and tangent, for several
◦
◦
◦
◦
angles, denoted by α, that can be easily computed out of the “familiar” angles 30
√ , 45 , 60 and 90 .
3 1
1 1
For instance, the exact value of cos 75◦ is: cos 30◦ cos 45◦ − sin 30◦ sin 45◦ =
·√ − ·√ =
2
2 2
2
√
3−1
√ .
2 2
9. α = 75◦ . (Above we found cosine; find also sine and tangent.)
10. α = 15◦ .
11. α = 105◦.
12. α = 165◦.
13. α = 195◦.
In Exercises 14-17 you are asked to find the exact values of sine, cosine, and tangent, for a sum
or difference of two angles, based on some information.
3
12
14. Given α in Quadrant II, with sin α = , and β in Quadrant III, with tan β = , find sine,
5
5
cosine and tangent of the angle γ = α + β, as well as the quadrant of γ.
15. Same problem as above, except that γ = α − β.
7
8
16. Given α, β in Quadrant II, with tan α = − , and cot β = − , find sine, cosine and
24
15
tangent of the angle γ = α + β, as well as the quadrant of γ.
17. Same problem as above, except that γ = α − β.
"
1
18. Find the exact value of sin arcsin
+ arcsin
2
√ !#
2
.
2
1
2
19. Find the exact value of sin arcsin
− arccos −
.
3
3
3
12
20. Find the exact value of cos arcsin −
+ arctan −
.
5
13
21. Find the exact value of tan [arctan(5) + arctan(−2)].
1
22. Find the exact value of tan arctan(4) − arctan
.
4
√
23. Find all solutions of the equation cos t + 3 sin t = 1. Use exact value.
√
24. Find all solutions of the equation cos t − 3 sin t = −1. Use exact value.
206
4.2. MULTIPLES OF ANGLES
1
25. Find all solutions of the equation cos t + sin t = √ . Use exact value.
2
1
26. Find all solutions of the equation − cos t + sin t = √ . Use exact value.
2
4.2 Multiples of Angles
In this section we use the Angle Addition and Subtraction Formulas to obtain several formulas
that are very useful for computing multiples and fractions of angles.
The Double Angle Formulas
The formulas for double angles are obtained by specializing the Angle Addition Formulas
sin(α+β) = sin αcos β+cos αsin β,
cos(α+β) = cos αcos β−sin αsin β,
tan α+tan β
tan(α+β) =
,
1−tan α tan β
(4.2.1)
(4.2.2)
(4.2.3)
to the case when β = α. The package below summarizes all the possible versions of the resulting
identities.
Double Angle Formulas
sin 2α = 2 sin α cos α =
2 tan α
1 + tan2 α
1−tan2 α
cos 2α = cos2 α−sin2 α = 2 cos2 α−1 = 1 − 2 sin2 α =
1 + tan2 α
2 tan α
tan 2α =
1−tan2 α
(4.2.4)
(4.2.5)
(4.2.6)
C LARIFICATIONS. The first equalities in (4.2.4), (4.2.5) and (4.2.6) follow directly from the
Angle Addition Formulas (4.2.1), (4.2.2) and (4.2.3) shown above.
The second and third equalities in (4.2.5) follow by replacing sin2 α = 1 − cos2 α, or replacing
cos2 α = 1 − sin2 α.
1
As for the third equalities in (4.2.4) and (4.2.5), they follow from the identity
=
1 + tan2 α
1
= cos2 α, so if we compute the final right-hand sides of both (4.2.5) and (4.2.6), we get
sec2 α
2 tan α
1
2 sin α
= 2 tan α ·
=
· cos2 α = 2 sin α cos α = sin 2α;
2
2
1 + tan α
1 + tan α
cos α
1 − tan2 α
1
sin2 α
2
= (1 − tan α) ·
= 1−
· cos2 α = cos2 α − sin2 α = cos 2α.
2
2
2
1 + tan α
1 + tan α
cos α
A DDITIONAL C LARIFICATIONS. What we see from (4.2.5), (4.2.4) and (4.2.6) is that, in order
to be able to compute the trigonometric functions of a double angle 2α, then depending on what
function we need to compute, we need to complete one of the following “wish lists”
CHAPTER 4. TRIGONOMETRY FOR COMBINATIONS OF ANGLES
207
( A ) Find tan α: this will easily give us all trigonometric functions of the double angle 2α.
( B ) If we only want to compute cos 2α, then one of the values sin α or cos α suffices.
( C ) If need to find sin 2α, then as an alternative to ( A ), we may also use the first equality in
(4.2.4), if we know both sin α and cos α.
5
Example 4.2.1. Suppose we know that α is an angle in Quadrant III with cos α = − , and
13
we are asked to compute sine and cosine of 2 α, as well as the quadrant 2 α sits in.
The cosine of 2 α can be easily computed solely out of cos α, using (4.2.5):
2
5
25
50
169
119
cos 2α = 2cos α − 1 = 2 −
−1 =2·
−1=
−
=−
.
13
169
169 169
169
2
As for the sine of 2 α, we either need tan α, or sin α. Of course, knowing one of these values will
give us the other one, so it is up to us to decide which one we want to compute. We opt here to
compute the sine:
s
r
2
√
5
25
2
=
sin α = ± 1 − cos α = ± 1 − −
=± 1−
13
169
r
r
169
25
144
12
=±
−
=±
=± .
169 169
169
13
12
Since α is in Quadrant III, it follows that sin α is negative, so the correct value is: sin α = − ,
13
and then using the first formula from (4.2.4), we get:
12
5
120
−
=
.
sin 2α = 2 sin α cos α = 2 −
13
13
169
Since cos 2α is negative, and sin 2α is positive, it follows that 2α is in Quadrant II.
Example 4.2.2. Suppose we know that tan α = −2, and we are asked to compute sine and
cosine of 2 α, as well as the quadrant 2 α sits in. (Notice that we are not given any information
about the Quadrant of α!)
In this case, our knowledge of tan α allows us to compute both sine and cosine of 2 α directly:
2 tan α
2(−2)
−4
4
=
=
=
−
;
2
1 + tan α
1 + (−2)2
1+4
5
1−tan2 α 1 − (−2)2
1−4
3
cos 2α =
=
=− .
2
2
1 + tan α 1 + (−2)
1+4
5
sin 2α =
Since both sin 2α and cos 2α are negative, it follows that 2α is in Quadrant III.
The Triple Angle Formulas
The Double Angle Formulas can be combined with the Angle Addition Formulas to yield the
following interesting formulas for triple angles.
208
4.2. MULTIPLES OF ANGLES
Triple Angle Formulas
sin 3α = 3 sin α − 4 sin3 α
cos 3α = 4 cos3 α − 3 sin α
3 tan α − tan3 α
tan 3α =
1 − 3 tan2 α
(4.2.7)
(4.2.8)
(4.2.9)
C LARIFICATION. Formula (4.2.7) follows from (4.2.1), (4.2.4) and (4.2.5):
sin 3α = sin(2α + α) = sin 2α cos α + cos 2α sin α =
= 2 sin α cos2 α + (1 − 2 sin2 α) sin α =
= 2 sin α (1 − sin2 α) + (1 − 2 sin2 α) sin α =
= 2 sin α − 2sin3 α + sin α − 2 sin3 α =
= 3 sin α − 4 sin3 α.
Formula (4.2.8) follows from (4.2.2), (4.2.4) and (4.2.5):
cos 3α = cos(2α + α) = cos 2α cos α − sin 2α sin α =
= (2 cos2 α − 1) cos α − 2 sin2 α cos α =
= 2 cos3 α − cos α) − 2(1 − cos2 α) cos α =
= 2 cos3 α − cos α) − 2 cos α + 2 cos3 α =
= 4 cos3 α − 3 cosα.
Formula (4.2.9) follows from (4.2.3) and (4.2.6):
2 tan α
+ tan α
1 − tan2 α
=
2 tan α
1−
· tan α
1 − tan2 α
2 tan α
tan α(1 − tan2 α)
3 tan α − tan3 α
+
2
1 − tan2 α
1 − tan2 α
= 1 − tan α 2
=
=
1 − tan α
2 tan2 α
1 − 3 tan2 α
−
1 − tan2 α 1 − tan2 α
1 − tan2 α
3 tan α − tan3 α 1 − tan2 α
3 tan α − tan3 α
=
·
=
.
1 − tan2 α
1 − 3 tan2 α
1 − 3 tan2 α
tan 2α + tan α
tan 3α = tan(2α + α) =
=
1 − tan 2α · tan α
Below is a very interesting application of the Triple Angle Formulas, which illustrates how we
can use them to obtain the exact values of a new angle.
π
Example 4.2.3. Consider the angle α = 36◦ = angle. Since 3α + 2α = 5α = 180◦ = π, it
5
follows that cos 3α = cos(π − 2α) = − cos 2α, that is,
cos 3α + cos 2α = 0,
(4.2.10)
so if we let cos α = x, the using (4.2.8) and (4.2.5), then the equality (4.2.10) reads:
4x3 − 3x + 2x2 − 1 = 0.
(4.2.11)
CHAPTER 4. TRIGONOMETRY FOR COMBINATIONS OF ANGLES
209
3π
3π
3π
+2·
= 3π, it follows that the angle α =
(= 108◦ ) satisfies cos 3α =
5
5
5
cos(3π − 2α) = cos(π − 2α) = − cos 2α, so α also satisfies (4.2.10). Clearly, α = π(= 180◦ )
also satisfies (4.2.10), so the values x1 = cos 36◦ , x2 = cos 108◦ and x3 = cos 180◦ = −1 are all
the solutions of the cubic equation. (4.2.11). This means that the polynomial on the left-hand side
of (4.2.11) factors as
Since we also have 3·
4x3 − 3x + 2x2 − 1 = 4(x − cos 36◦ )(x − cos 108◦ )(x+1),
which means that
(x − cos 36◦ )(x − cos 108◦ ) =
4x3 − 3x + 2x2 − 1
= 4x2 − 2x − 1.
x+1
This means that the numbers cos 36◦ and cos 108◦ are precisely the two solutions of the quadratic
equation
4x2 − 2x − 1 = 0.
The two solutions of these equations are easily obtained using the Quadratic Formula:
x=
2±
p
√
√
√
√
(−2)− 4(4)(−1)
2 ± 20
2±2 5
2(1 ± 5)
1± 5
=
=
=
=
.
2·4
8
8
8
4
< 0, it follows that√the correct way to match these valSince x1 = cos 36◦ > 0 and x2 = cos 108◦√
√
1+ 5
1− 5
ues with the above solutions is: x1 =
and x2 =
. Since sin 36◦ = 1 − cos2 36◦ ,
4
4
the two exact values we are interested in are:

√

1+ 5

◦


 cos 36 = v 4
u
√ !2 p
√
u
1
+
5
10
−
2
5



sin 36◦ = t1 −
=


4
4
The Half Angle Formulas
We now take another look at the Double Angle Formulas, especially at the last equality in
(4.2.4) and the second through fourth equalities in (4.2.5).
For instance, if we consider the second and the third equalities in (4.2.5), they give us the
equalities
2 cos2 α = 1 + cos 2α and 2 sin2 α = 1 − cos 2α,
which allow us to conclude that:
1 + cos 2α
;
2
1 − cos 2α
sin2 α =
.
2
cos2 α =
(4.2.12)
(4.2.13)
210
4.2. MULTIPLES OF ANGLES
Another “game we can play” is to compute the two numerators in right-hand sides in (4.2.12)
and (4.2.13) using the last equality in (4.2.5):
1 − tan2 α
1 + tan2 α 1 − tan2 α
2
=
+
=
;
2
2
2
1 + tan α
1 + tan α 1 + tan α
1 + tan2 α
1 − tan2 α
1 + tan2 α 1 − tan2 α
2tan2 α
1 − cos 2α = 1 −
=
−
=
;.
1 + tan2 α
1 + tan2 α 1 + tan2 α
1 + tan2 α
1 + cos 2α = 1 +
(4.2.14)
(4.2.15)
When we combine these two identities with the last equality in (4.2.4), we obtain
2 tan α
2
sin 2α
2 tan α
= 1 + tan α =
2
1 + cos 2α
1 + tan2 α
1 + tan2 α
2 tan2 α
2
2 tan2 α
1 − cos 2α
= 1 + tan α =
2 tan α
sin 2α
1 + tan2 α
1 + tan2 α
·
1 + tan2 α
= tanα;
2
(4.2.16)
·
1 + tan2 α
= tanα.
2 tan α
(4.2.17)
If we start now with some angle θ, and we set up α = θ/2, then the above formulas yield:
Half Angle Formulas
r
1 − cos θ
;
2
r
1 + cos θ
cos(θ/2) = ±
;
2
r
1 − cos θ
sin θ
1 − cos θ
tan(θ/2) = ±
=
=
.
1 + cos θ
1 + cos θ
sin θ
sin(θ/2) = ±
(4.2.18)
(4.2.19)
(4.2.20)
C LARIFICATIONS. Formulas (4.2.18) and (4.2.19) follow directly from (4.2.12) and (4.2.13).
sin(θ/2)
By taking the ratio
, the first equality in (4.2.20) follows immediately. The other two
cos(θ/2)
equalities from (4.2.20) follow from (4.2.16) and (4.2.17).
 If we need to compute the sine or cosine of a half angle, we must resolve the ± sign in each
of the formulas (4.2.18) and (4.2.18). This meas that our “wish list” must include the quadrant
information on θ/2. However, when we only asked to compute the tangent of a half-angle, we have
the option of using the other formulas from (4.2.20), which do not require this information.
Example 4.2.4. Suppose we hand to compute sine and cosine of θ/2, given that π < θ < 2π
3
and cos θ = − .
5
Since we are given cos θ, all we need is the quadrant information on θ/2. Based on the given
inequalities involving θ, which we can divide by 2, we get:
π
θ
< < π,
2
2
CHAPTER 4. TRIGONOMETRY FOR COMBINATIONS OF ANGLES
211
which tells us, among other things, that θ/2 is in Quadrant II, so using (4.2.18) and (4.2.19), with
the correct signs, we get
v
u
u
3
u1 − −
t
5
v
v
u
u8
r
u1 + 3
u
t
t
1 − cos θ
5 =
5 = 4 = √2 ;
sin(θ/2) = +
=
=
2
2
2
2
5
5
v
v
v
u
u
u2
u
3
r
r
u1 − 3
u
u1 + −
t
t
t
1 + cos θ
1
1
5
5
5
cos(θ/2) = −
=−
=−
=−
=−
= −√ .
2
2
2
2
5
5
r
Example 4.2.5. Suppose we hand to compute sine and cosine of θ/2, given that 270◦ < θ <
5
360◦ and sin θ = − .
13
Our “wish list” is a little longer here, because we need
(i) Quadrant information on θ/2;
(ii) cos θ; since we are only given sin θ, we also need
(iii) Quadrant information on θ !!!
Start off with (iii). Based on the given information, we immediately get that θ sits in Quadrant IV,
so cos θ is positive, so we have
cos θ = +
p
s
r
r
2
5
25
144
12
=+ 1−
=+
=+ .
1 − sin2 θ = + 1 − −
13
169
169
13
Based on the given inequalities for θ, we get (upon dividing by 2) the inequalities
360◦
, that is,
2
θ
135◦ < < 180◦ ,
2
270◦
θ
< <
2
2
which clearly tells us that θ/2 is in Quadrant II, so using (4.2.18) and (4.2.19), with the correct
signs, we get
v
v
u
u 1
r
u 1 − 12
u
t
t
1 − cos θ
1
1
13 =
13 =
sin(θ/2) = +
=
=√ ;
2
2
2
26
26
v
v
u
u 25
r
r
u 1 + 12
u
t
t
1 + cos θ
13 = − 13 = − 25 = − √5 .
cos(θ/2) = −
=−
2
2
2
26
26
r
Example 4.2.6. Using Half Angle Formulas, we can compute the exact value of halves of the
π
familiar angles. For instance, we can consider the angle = 22.5◦ , which is in Quadrant I, so its
8
212
4.2. MULTIPLES OF ANGLES
sine and cosine will be:
sin
π
8
cos
π
8
v
u
u 1 − √1
t
2
=+
=
;
2
2
v
v
u
u
u 1 + √1
u 1 + cos π
t
t
2
4 =
=+
.
2
2
v
u
u 1 − cos π
t
4
Even though the above values do not look “nice” at all, they are nevertheless exact.
Trigonometric Equations Involving Multiple Angles
Using the Double (or Triple) Angle Formulas it is possible to solve more complicated equations, as seen in the following Example.
Example 4.2.7. Suppose we want to sole the equation
sin 2t − sin t = 0.
Using the Double Angle, we can replace sin 2t = 2 sin t cos t, so the above equation becomes
2 sin t cos t − sin t = 0.
Now we can factor the left-hand side, so our equation becomes
sin t(2 cos t − 1) = 0,
and then we can split into two equations:
( A ) sin t = 0, which has as solutions: t = nπ, n integer;
1
( B ) 2 cos t − 1 = 0, which is equivalent to cos t = , so it has has as solutions (see Section
2
2.7):
π
1
+ 2nπ = ± + 2nπ, n integer.
t = ± arccos
2
3
Exercises
In Exercises 1-7 you are asked to find the sine, cosine and tangent of the double angle 2θ, as
well as the quadrant 2θ sits in, based on the given information on θ. Use exact values.
8
1. cos θ = , θ in Quadrant IV.
17
2. sin θ = −
7
, θ in Quadrant III.
25
3. tan θ = −7.
4. θ = arccos −
5. θ = arcsin −
12 13
35 37
CHAPTER 4. TRIGONOMETRY FOR COMBINATIONS OF ANGLES
213
6. cot θ = −3.
7. θ = arctan(−4).
In Exercises 8-12 you are asked to find the sine, cosine and tangent of the half angle θ/2, based
on the given information on θ. Use exact values.
3
8. sin θ = − , π < θ < 2π.
5
1
9. sin θ = − , 3π < θ < 5π.
4
10. tan θ = −
24
, −3π < θ < −2π.
25
1
11. cos θ = − , 180◦ < θ < 360◦.
2
5
13
In Exercises 13-19 you are asked to find all solutions of the given trigonometric equation
√
13. sin 2x = 3 sin x. Use exact values.
12. θ = arccos −
14. sin x + cos 2x = 1. Use exact values.
1
15. cos 2x + cos x = √ . Use exact values.
2
16. sin 3x + sin x = 0. Use exact values.
17. cos 2x − tan x = 1. Use exact values.
18. sin 2t − tan x = 0. Use exact values.
19. tan 2x = tan x Use exact values.
In Exercises 20-26 you are asked to verify the given identity.
20. sin 4t = 4 sin t cos t cos 2t.
21. (sin u + cos u)2 = 1 + sin 2u.
22. cos4 x − sin4 x = cos 2x.
23.
sec2 t
= sec 2t.
2 − sec2 t
24. cos 4t = 8 cos4 t − 8 cos2 t + 1.
25. 2 csc 2u = csc u sec u.
26. tan(t/2) = csc t − cot t.
214
4.3. FROM SUMS TO PRODUCTS AND BACK
4.3 From Sums to Products and Back
In this section we build on the Angle Addition and Subtraction Formulas to several useful
identities that involve sums and products of sines and cosines.
The Product-to-Sum Identities
Let us start off with the Angle Addition and Subtraction Formula for cosines, and put them side
by side:
cos(α−β) = cos αcos β+sin αsin β,
cos(α+β) = cos αcos β−sin αsin β.
(4.3.1)
(4.3.2)
When we add these two identities, the product sin αsin β cancels out, so we simply get:
cos(α−β) + cos(α+β) = 2 cos αcos β.
(4.3.3)
Likewise, when we subtract (4.3.2) from (4.3.1), the the product cos αcos β cancels out, so now
we get:
cos(α−β) − cos(α+β) = 2 sin αsin β.
(4.3.4)
When we put the Angle Addition and Subtraction Formulas for sines side by side:
sin(α+β) = sin αcos β+cos αsin β,
sin(α−β) = sin αcos β−cos αsin β,
(4.3.5)
(4.3.6)
and we add them, the product cos αsin β cancels out, so now we get:
sin(α+β) + sin(α−β) = 2 sin αcos β.
(4.3.7)
So now, when we read each one of the formulas (4.3.3), (4.3.4), (4.3.7) from right to left, then
after dividing each one of them by 2, we obtain the following formula package.
Product-to-Sum Identities
cos αcos β =
sin αsin β =
sin αcos β =
1
cos(α−β) + cos(α+β) ;
2
1
cos(α−β) − cos(α+β) ;
2
1
sin(α+β) + sin(α−β) .
2
(4.3.8)
(4.3.9)
(4.3.10)
C LARIFICATIONS. At a first glance, the formulas (4.3.8) and (4.3.9) may appear to be unbalanced. What happens if we flip α and β? The left-hand sides of (4.3.8) and (4.3.9) will be
unchanged, but what happens in the right-hand side? As it turns out, since cos is even, we can
always replace
cos(β−α) = cos(α−β),
so the right-hand sides of (4.3.8) and (4.3.9) will not change either.
As for the formula (4.3.10), when we flip α and β, we should take into account the fact that sin
is odd, so the correct replacement is
sin(β−α) = − sin(α−β),
CHAPTER 4. TRIGONOMETRY FOR COMBINATIONS OF ANGLES
215
so in addition to (4.3.10), we also have another possible expansion:
cos αsin β = 12 sin(α+β) − sin(α−β) .
(4.3.11)
Deciding, whether to use (4.3.10) or (4.3.11) to expand of product of one sine and one cosine, is
up to you.
Application: Fourier Expansion
The main application of the Product-to-Sum Formulas is concerned with certain types of
trigonometric sums described as follows.
For any integer N ≥ 0, a Fourier function of order N is a sum of the form:
F(x) = a0 + a1 cos x + a2 cos 2x + a3 cos 3x + . . . + aN cos Nx +
+ b1 sin x + b2 sin 2x + b3 sin 2x + . . . + bN sin Nx,
(4.3.12)
where a0 , a1 , a2 , . . . , aN , b1 , b2 , . . . , bN are constants, with either aN or bN non-zero.
What is not at all obvious is the fact that, whenever a function F(x) can be written as a sum as
above, then the integer N, as well as the coefficients a0 , a1 , a2 , . . . , aN , b1 , b2 , . . . , bN , are uniquely
determined. With this uniqueness in mind (which will be clarified in Section 5.2), the sum that
appears in the right-hand side of (4.3.12) is what we call the Fourier expansion of F.
Right now, the only thing we are concerned with is finding Fourier expansions for various
functions.
Example 4.3.1. Suppose we are asked to find the Fourier expansion of
f (t) = (1 + sin 3t)sin2 t.
We start off by reducing sin2 t using formula (4.3.9):
sin2 t =
1
1 1
1
cos(t − t) − cos(t + t) = cos 0 − cos 2t = − cos 2t.
2
2
2 2
(Here we also used the obvious identity cos 0 = 1.) Now we can replace the factor sin2 t back in
f (t), and obtain (by “folding”):
1 1
1 1
1
1
f (t) = (1 + sin 3t)
− cos 2t = − cos 2t + sin 3t − sin 3t cos 2t,
(4.3.13)
2 2
2 2
2
2
1
and all we have to take care of is the last term. Using (4.3.11), and leaving − aside for a moment,
2
we can write:
sin 3t cos 2t =
1
1
1
1
sin(2t + 3t) + sin(3t − 2t) = sin 5t + sin t = sin 5t + sin t,
2
2
2
2
so when we go back to (4.3.13), we get:
1 1
1
1 1
1
1 1
1
1
1
f (t) = − cos 2t + sin 3t −
sin 5t + sin t = − cos 2t + sin 3t − sin 5t − sin t.
2 2
2
2 2
2
2 2
2
4
4
216
4.3. FROM SUMS TO PRODUCTS AND BACK
Of course, when we want to arrange the Fourier expansion more neatly, we can (and should) write
1 1
1
1
1
f (t) = − cos 2t − sin t + sin 3t − sin 5t,
(4.3.14)
2 2
4
2
4
so our function f (t) is a Fourier function of order 5.
C LARIFICATION : (R EAD THIS ONLY IF YOU ARE AN ENGINEERING STUDENT !). Why does
one care about Fourier expansions? Consider for instance the function f from Example 4.3.1, and
think of it as representing the wave function of some sound produced by several “instruments.”
(Think of these “instruments” for instance, as musical instruments in a band. You can also think
of them as keys on a piano keyboard.) Assume the pitch of “instrument k” is the function pk (t) =
sin kt, where k is some positive integer. Besides these “instruments” we also have to account for
“constant sounds,” which are constant functions. When we look closely at the Fourier expansion
1
(4.3.14), we see that our sound is put together using a “constant sound” c = , and “instruments”
2
that use k = 1, 2, 3, 5. To see the volume of each “instrument” that contributes to (4.3.14), we need
to rearrange all terms with negative coefficients, as well as those that have cosine functions. We do
π
π
this using shifting formulas like − sin α = sin(α − π) and cos α = sin α +
= − sin α − .
2
2
Using such identities, the Fourier expansion (4.3.14) can be rearranged as:
1 1
1
1
1
f (t) = − cos 2t − sin t + sin 3t − sin 5t =
2 2
4
2
4
π 1
1
1
1 1
= + sin 2t −
+ sin(t − π) + sin 3t + sin(5t − π) =
2 2
2
4
2
4
π 1
1
1 π 1 1 = + sin 2 t −
+ sin(t − π) + sin 3t + sin 5 t −
=
2 2
4
4
2
4
5
1 1
π 1
1
1
π
= + p2 t −
+ p1 (t − π) + p3 (t) + p5 t − .
2 2
4
4
2
4
5
Based on this calculation, we now see that the “sound” f (t) is assembled using:
1
• the “constant sound” at volume c = , together with
2
1
• “instrument 1” at volume , shifted by π;
4
1
π
• “instrument 2” at volume , shifted by ;
2
4
1
• “instrument 3” at volume ;
2
1
π
• “instrument 5” at volume , shifted by .
4
5
The Sum-to-Product Identities
Suppose we “play” with the Product-to-Sum Formulas (4.3.8), (4.3.9), (4.3.10) and (4.3.11) by
multiplying all identities by 2. When reading what we get from right to left, we get the following
identities
cos(α−β) + cos(α+β) = 2 cos αcos β;
cos(α−β) − cos(α+β) = 2 sin αsin β;
sin(α+β) + sin(α−β) = 2 sin αcos β;
sin(α+β) − sin(α−β) = 2 sin β cos α.
(4.3.15)
(4.3.16)
(4.3.17)
(4.3.18)
CHAPTER 4. TRIGONOMETRY FOR COMBINATIONS OF ANGLES
217
Suppose now we use the substitution
α+β =u
α−β =v
which can be solved as a system of equations in α and β to give:

u+w


 α = 2


 β = u−v
2
Using these computations, when we go back to the identities (4.3.15), (4.3.16), (4.3.17) and
(4.3.18), we now get the following formula package.
Sum-to-Product Identities
u+v
u−v
cos u+cos v = 2 cos
;
cos
2
2
u+v
v−u
cos u−cos v = 2 sin
sin
;
2
2
u+v
u−v
sin u+sin v = 2 sin
cos
;
2
2
u−v
u+v
sin u−sin v = 2 sin
.
cos
2
2
(4.3.19)
(4.3.20)
(4.3.21)
(4.3.22)
Application: Equations
C LARIFICATION. The Sum-to-Product Identities are very useful tools for factoring sums or
differences of sines or cosines. This is particularly useful when we want to solve certain equations, as seen in the Examples below, which shows how certain equations can be reduced (after
substitutions) to x-intercept equations, which are those of the form sin ? = 0, or cos ? = 0.
 When solving x-intercept equations of the form sin ? = 0, or cos ? = 0 we do not need to
use inverse trigonometric functions as we did in Section 2.7! The solutions are much simpler!
I. The solutions of sin ? = 0 are: nπ, n integer.
π
II. The solutions of cos ? = 0 are: + nπ, n integer.
2
Example 4.3.2. Suppose we want to find all solutions of the equation
sin 3x = sin x.
(4.3.23)
Subtract the right-hand side, so our equation is the same as:
sin 3x − sin x = 0.
(4.3.24)
Using (4.3.22) we can factor
sin 3x − sin x = 2 sin
3x − x
2
cos
3x + x
2
= 2 sin x cos 2x,
218
4.3. FROM SUMS TO PRODUCTS AND BACK
so our equation becomes
2 sin x cos 2x = 0.
By setting each factor equal to zero (of course, we cannot have 2 = 0), the above equation breaks
into two possibilities:
( A ) sin x = 0, which yields x = nπ, n integer.
π
( B ) cos 2x = 0, which yields 2x = + nπ, thus
2
π nπ
1 π
+ nπ = +
, n integer.
x=
2 2
4
2
Example 4.3.3. Suppose we want to find all solutions of the equation
sin 3x + cos 5x = 0.
(4.3.25)
There are many ways to solve this equation, based on transforming the left-hand side of
(4.3.25) either into a sum/difference of sines, or into a sum/difference of cosines. For instance,
π
using the well-known identity sin α = cos
− α , our equation can be presented as:
2
π
− 3x + cos 5x = 0.
(4.3.26)
cos
2
Using (4.3.19) we can factor
cos
π
π
− 3x + cos 5x = 2 cos  2
2
π
= 2 cos  2
so our equation becomes
− 3x + 5x
2
+ 2x
2


π
− 3x − 5x
− 8x

 cos  2
π
 cos  2
2
2

=
 = 2 cos π + x cos π − 4x ,
4
4
π
+ x cos
− 4x = 0.
4
4
By setting each factor equal to zero (of course, we cannot have 2 = 0), the above equation breaks
into two possibilities:
π
π
π
( A ) cos
+ x , which yields + x = + nπ, thus
4
4
2
2 cos
π
x=−
π
π π
π
+ + nπ = + nπ, n integer.
4
2
4
π
π
π
( B ) cos
− 4x = 0, which yields − 4x = + nπ, which is the same as 4x = −
4
2
4
π 4 π π
π
+ nπ = − − nπ = − − nπ thus
2
4
2
4
1 π
π
nπ
x=
− − nπ = − −
, n integer.
4
4
16
4
CHAPTER 4. TRIGONOMETRY FOR COMBINATIONS OF ANGLES
219
Application: Identities
Another application of the Sum-to-Product Identities concerns certain trigonometric identities
which involve ratios, as illustrated in the following Example.
Example 4.3.4. Suppose we are asked to prove the identity:
sin 5x + sin 3x
= tan 4x.
cos 5x + cos 3x
We start off with the left-hand side, and use the Sum-to-Product Identities to factor both the
numerator and the denominator:
5x − 3x
5x + 3x
2 sin
cos
sin 5x + sin 3x
2 sin 4x cos x
2
2
=
LHS =
=
.
5x − 3x
5x + 3x
cos 5x + cos 3x
2 cos 4x cos x
2 cos
cos
2
2
Now we can finish everything by simplifying 2 cos x:
LHS =
sin 4x
2 sin 4x cos x
=
= tan 4x = RHS.
2 cos 4x cos x
cos 4x
Exercises
In Exercises 1-7 you are asked to find the Fourier expansion of the given function.
1. f (t) = sin3 t.
2. f (t) = cos3 t.
3. f (t) = sin4 t.
4. f (t) = cos4 t.
5. f (t) = sin2 t + cos 2t.
6. f (t) = (1 + sin t)3 .
7. f (t) = (2 sin t − cos t)2 .
In Exercises 8-12 you are asked to find all solutions of the given trigonometric equation.
8. sin 4x − sin 2x = 0. Use exact values.
9. sin 4x − cos 2x = 0. Use exact values.
10. cos x + cos 3x = 0. Use exact values.
11. cos 5x − cos 3x = 0. Use exact values.
220
4.3. FROM SUMS TO PRODUCTS AND BACK
12*. sin x + sin 5x = sin 3x. Use exact values. (H INT: Factor LHS, then subtract RHS, then
factor again.)
In Exercises 13-18 you are asked to prove the given identity.
sin 4t + sin 6t
13.
= cot t.
cos 4t − cos 6t
14.
sin t + sin 4t + sin 7t
= tan 4t.
cos t + cos 4t + cos 7t
u + v
sin u + sin v
2
15.
=
u
−
v .
sin u − sin v
tan
2
tan
16.
cos u − cos v
u + v
v − u
= tan
tan
.
cos u + cos v
2
2
17. sin 2x + sin 4x + sin 6x = 4 cos x cos 2x sin 3x.
18*.
sin x + sin 2x + sin 3x + sin 4x + sin 5x
= tan 3x.
cos x + cos 2x + cos 3x + cos 4x + cos 5x
Appendices
A Algebra Review I: Equations
This Appendix provides a quick review of the various types of equations frequently used in
Trigonometry. We understand equations as equalities between expressions that involve unknown
quantities. The simplest examples of equations are those the contain only one unknown, for example
(A.1)
x2 + 2x = 2x2 − 3,
where the unknown quantity (a.k.a. variable) is denoted by x. To solve an equation in one variable
simply means to find the values for the variable which satisfy the equality given in the equation.
Such values are called solutions of the equations.
If you know how to solve the equation (A.1), do it now! (Otherwise, wait until we discuss
Quadratic Equations later in this section.)
When comparing two equations, we often use the following terminology.
◮
Two equations are equivalent, if they have the same solutions.
With the above definition in mind, the “game” we play when we want to solve an equation is
to transform them into equivalent ones, so we change the equation:
transformation
original (old) equation −−−−−−−−−−−−→ transformed (new) equation.
A “perfect” transformation is one that has the old and the new equations equivalent. A “safe”
(but possibly “imperfect”) transformation is one for which the new equation does not loose any
solutions of the old one.
The two most popular “perfect” transformations are:
I. Add or subtract a common expression from both sides of the equation.
II. Multiply or divide both sides of the equation by a non-zero quantity.
Example A.1. Solve: 8x − 2 = 5x − 8.
Solution.
original equation:
8x − 2 = 5x − 8;
add 2, subtract 5x:
8x − 2 + 2 − 5x = 5x − 8 + 2 − 5x;
combine like terms:
3x = −6;
−6
divide by 3(6= 0!):
x=
= −2.
3
Example A.2. If we start with the equation x2 = x and we think of dividing by x, we are
incorrect! We are not exactly sure if we are dividing by a non-zero number! After dividing by x
the new equation looks like x = 1, so it is already solved. However, in the process of dividing by x
we eliminated the possibility that x = 0, which in fact is another solution of the original equation.
Thus dividing by x made us loose a solution.
1001
1002
A. ALGEBRA REVIEW I: EQUATIONS
Linear Equations
A basic linear equation, is one of the form:2
number1 · ? = number2 ,
(A.2)
One such equation showed up in Example A.1, where we saw how to solve it. The rules for solving
such equations are:
Solutions of basic linear equations
Depending on whether one or both coefficients vanish in a basic linear equation of the form
(A.2), the solutions of the equation are as follows.
number2
• If number1 6= 0, there is only one solution: ? =
, so its solution set has only one
number1
element.
• If number1 = 0 and number2 6= 0, then the equation has no solutions, so its solution set
is ∅ – the empty set.
• If number1 6= 0 and number2 = 0, then the equation has all real numbers as solutions,
so its solution set is R – the set of all real numbers.
A slightly larger class of equations are the so-called general linear equations, which look like:
♥ · ? + ♠ = ♦ · ? + ♣.
(A.3)
As we have already seen in Example A.1, the following principle is clearly true in general.
Any general linear equation of the form (A.3) can be “perfectly” transformed into a basic
linear equation of the form (A.2).
Example A.3. Let us consider the following equation in two variables:
xy = x + 2y.
(A.4)
We can think of (A.4) as a linear equation in two ways:
(i) The unknown is x, so we think y as a known quantity, thus the equation looks like:
♥ · x = x + ♦,
(A.5)
with both ♥(= y) and ♦(= 2y) treated as numbers.
(ii) The unknown is y, so we think x as a known quantity, thus the equation looks like:
♣ · y = ♠ + 2y
(A.6)
with both ♣(= x) and ♠(= x) treated as numbers.
2
At times, when we don’t want to use specific letters, we use “kindergarten” notation, such as the one shown in
(A.2). This textbook also uses “templates” in which various symbols like ♥, ♣, ♠, ♦, #, ♯, ♮, etc. denote place-holders
for known or unknown (most often numerical) quantities.
APPENDICES
1003
This means that when we are asked to solve (A.4) for x, we must think as in (i); but when we are
asked to solve (A.4) for y, we must think as in (ii). With these considerations in mind the solutions
are as follows. (C AUTION : The solutions given in (A.8) and (A.10) are sloppy! We will revisit
them in the note that follows this Example.)
(i). To solve for x we make a plan on how to solve (A.5): we would subtract x, then group like
terms to make the equation look like a basic linear one: number1 · x = number2 , and finally we
would solve by division. With this plan in mind, our steps are as follows:
original equation:
subtract x:
combine like terms:
divide by (y − 1):
xy = x + 2y;
xy − x = 2y;
(y − 1)x = 2y;
2y
x=
.
y−1
(A.7)
(A.8)
(ii). To solve for y, our plan is similar: subtract 2y, then group like terms to make the equation
look like a basic linear one: number1 · y = number2 , and finally we would solve by division.
original equation:
subtract 2y:
combine like terms:
divide by (y − 1):
xy = x + 2y;
xy − 2y = x;
(x − 2)y = x;
x
y=
.
x−2
(A.9)
(A.10)
 Both answers given in the preceding Example are incomplete. The transformations of the
original equation (A.4) into both (A.7) and (A.9) are “perfect,” but the final answers given in (A.8)
and (A.10) are sloppy, because we did not care that much about dividing by a number which might
in fact be equal to zero: (i) in (A.8) we must account for the possibility that y − 1 = 0, that is,
y = 1; (ii) in (A.10) we must account for the possibility that x − 2 = 0, that is, x = 2. Therefore
the complete answers should look like:

 x
2y



, if x 6= 2
,
if
y
=
6
1


x−2
y−1
x=
y=




 no solution, if y = 1
no solution, if x = 2
Proportions
In elementary school we learned that a proportion is an equality of fractions (ratios), so it is
presented as
numerator2
numerator1
=
.
(A.11)
denominator1
denominator2
Of course, since we deal with fractions, the denominators are assumed to be non-zero. Since the
division is the inverse operation to multiplication, proportions can always be transformed according
to the following scheme.
1004
A. ALGEBRA REVIEW I: EQUATIONS
Proportions in fraction-less form
I. A proportion of the form (A.11) yields an equality of two products, obtained by crossmultiplication
numerator1 × denominator2 = numerator2 × denominator1 .
II. Conversely, an equality of products, in which each product has a non-zero factor
♠ · ♥ = ♣ · ♦, ♥, ♦ =
6 0,
yields a proportion obtained by cross-division by the two non-zero factors:
♠
♣
= .
♦
♥
Going back and forth between two forms of a proportion is a useful technique for solving basic
proportion equations, which are proportions in which one element is unknown, while the other
?
♣
♣
♠
= , or like
= . As long as the given denominators
three are given, so they look like
♦
♥
?
♥
in a proportion equation are non-zero, we can “perfectly” transform the equation using crossmultiplication into a basic linear equation, which we can then solve by division. We summarize
this method using the phrase “cross-multiply, then divide.”
3
7
Example A.4. Consider the equation: = . After cross-multiplication (which is a “perfect”
t
9
transformation, because the only given denominator is 9 6= 0), we get the basic linear equation
27
7 t = 3 · 9 = 27, which by division (which is OK, since we divide by 7 6= 0) yields: t =
=
7
3 67 = 2.857142857142 · · · = 2.857142.
Equations Vs. Identities
In Algebra we often use identities (or formulas), which make our life a little easier, especially
when we need to solve equations, or when we want to simplify certain algebraic expressions. When
comparing identities with equations, the bottom line is:
An identity is a “trivial” equation, than is, an equality which is satisfied by all possible
values of the variables.
Example A.5. On the one hand, the equation (x + 1)2 = x2 + 2x + 1 is an identity: no matter
what value of x we plug in, the equality will be satisfied.
On the other hand, the equation (x + 1)2 = x2 + 1 is not an identity: there are values for x, for
instance x = 1, for which the equality fails.
Among the many of identities used in Algebra, the package below is the most popular one:
Quadratic Product Identities
(♥ + ♠)2 = ♥2 + 2♥♠ + ♠2 ;
(♥ − ♠)2 = ♥2 − 2♥♠ + ♠2 ;
(♥ + ♠)(♥ − ♠) = ♥2 − ♠2 .
APPENDICES
1005
Equations which are “almost trivial,” so they hold for instance for all values in an interval
(or a union of intervals) are referred to as conditional identities. We write them using the format
“equation, condition on the variable(s),” for example
x2 − 1
= x + 1,
x−1
x 6= 1.
Power Equations with Positive Integer Exponents
Besides linear equations, the power equations are the next easiest to solve. A power equation
is one of the form:
?♥ = number.
(A.12)
with number and ♥ real numbers. The number ♥ is called the exponent of the power equation. At
this point we only discuss the easiest case, when the exponent is a positive integer3 , so the left-hand
of (A.12) is the product:
?♥ = ?
| · ?{z· · · ?} .
♥ factors
Of course, when the exponent is equal to 1, the equation is already solved, so we can in fact assume
that it is at least 2.
 Unless we are in some very special cases (as in Example A.6 ) the way we solve power
equations is not an honest one. What we are about to do is to make a statement and cook up a
notation that will help us cheat a little bit.
Power Equation Facts and Radical Notation
Assume the exponent ♥ is an integer ≥ 2.
FACT 1. Whenever the equation (A.12) has real solutions, exactly one of its real solutions
has
q the same sign as number (the right-hand side). This special solution is denoted by
♥
q
♥
number. However, if (A.12) has no real solution (see Fact 2 below), then the quantity
number is not defined!
FACT 2. Depending on the parity of of the exponent ♥, the complete solutions of the power
equation (A.12) are as follows.
q
I. If ♥ is odd, then the equation has exactly one real solution: ? =
♥
number.
II. If ♥ is even, then according to the sign of number (the right-hand side), we have the
following three possibilities.
(a) If number > 0, the equation has
q two real solutions:
• one positive solution: ? =
3
♥
number, and
The case when the exponent is an arbitrary real number will be treated in Section 0.2.
1006
A. ALGEBRA REVIEW I: EQUATIONS
q
• one negative solution: ? = − ♥ number;
√
(b) If number = 0, the equation has exactly one real solution: ? = ♥ 0 = 0.
(c) If number < 0, the equation has no real solution.
q
For cases (a) and (b), we can also use the “lazy” notation: ? = ± ♥ number.
interest, particularly in Geometry
and Trigonometry, is when
C ONVENTION. A special case of p
p
2
the exponent is ♥ = 2. In this case
is simply denoted by
, and we call it the square root
operation.
√
Example A.6. 16 =?
According to the above definition, we must look at the equation ?2 = 16, solve it, and pick the
solution
which is positive. It is clear that the particular solution we are looking for is 4, so we have
√
16 = 4. As for the equation ?2 = 16, it has two solutions: ? = ±4.
Example A.7. Solve the equation x2 = 2.
√
According to Fact 2.II above, the solutions of this equation are: x = ± 2. On the one hand,
our use the square root√
notation is clearly a way to “cheat” this problem, because we still don’t
know how big or
√ small 2 really
√ is. On the other hand, one can also make a claim that, in fact we
do know what 2 quite well: 2 is the unique positive real number whose square is equal to 2.
When doing calculations, this implicit description is good enough, for example
√
2)10 = 32,
√
√
and this is why we say that “ 2 is an exact value” notation. If we want to get an idea how 2
“feels” as a number, then we will √
have to use a calculator (or do a hand computation as shown in
Appendix A), which reveals that: 2 = 1.4142135623730950488016887242 . . . . Unlike what √
we
4
saw in Example A.4, there is no repeating√pattern of decimals here, and this is evidence that 2
is irrational. One can actually prove that 2 is indeed irrational.
As we understand them, radicals are quantities described implicitly, so our preceding definition
can be restated as follows.
For any real number, and any integer ♥ ≥ 2,
 ♥
q
 ? = number,
♥
and
number =? means:

? is a real number which has same sign as number
If such a quantity exists, it is unique. If such a quantity does not exist, then
not defined.
√
♥
number is
The following sign features are always helpful when dealing with radicals:
In Example A.4 we encountered the rational number 27
7 , whose decimal expansion was 2.857142, meaning that
after the decimal point, the string “857142” repeats itself. This happens with any rational number.
4
APPENDICES
Sign Rules for Radicals
q
any
0 = 0;
q
odd
negative = negative;
1007
q
any
q
even
positive = positive;
negative = undefined.
 All identities included in the formula package shown below are conditional! (Besides the
main condition, other restrictions are shown next to the identity.)
Radical Identities
Assume all radicals below are defined:
p
♥
♥
number = number.
(Undoing Identity I)
(
p
number, if ♥ is odd;
♥
number♥ = (Undoing Identity II)
number, if ♥ is even.
q
q
♥ √
♠
♥·♠
number =
number
(“Cascading” Formula)
p
p
p
♥
number1 · number2 = ♥ number1 · ♥ number2 .
(Product Formula)
r
√
♥
numerator
numerator
♥
= √
, if denominator 6= 0.
(Quotient Formula)
♥
denominator
denominator
C LARIFICATION. In the even case of undoing formula II, one uses the absolute value of
number, which is defined as
number =
number,
−number,
if number≥ 0
if number< 0
In other words |number| is pretty much the same as number, except that it removes the “−” sign,
if number happens to be negative. For instance |3| = | − 3| = 3, so when we only care about the
absolute values, the numbers 3 and −3 become indistinguishable.
 Undoing√ even radicals without the use of absolute value is incorrect!
In particular, an
2 = x” is not a true identity. In works for x ≥ 0, but fails for x < 0, for instance:
equality
like
“
x
p
√
(−2)2 = 4 = 2 6= −2.
Example A.8. We can use the Product and Quotient Formulas to pull out squares from under
square roots, for instance:
r
√
√
√ √
√
18
18
9·2
9 2
3 2
=√
=√
=√ √ = √ .
175
175
25 · 7
25 7
5 7
1008
A. ALGEBRA REVIEW I: EQUATIONS
The final answer in the above calculation can be simplified to “look a little nicer” in two ways:
√
√
3 2
3( 2)2
3·2
6
√ = √ √ = √
= √ ; or
(A.13)
5 7
5 7 2
5 7·2
5 14
√
√ √
√
√
3 14
3 2
3 2 7
3 2·7
√ = √
=
=
.
(A.14)
5·7
35
5 7
5( 7)2
Simplifying as we did in (A.13) is referred to as rationalizing the numerator, whereas what we did
in (A.14) is referred to as rationalizing the denominator.
Quadratic Expressions
Besides linear and power equations, the so-called quadratic equations are the next easiest to
solve. Concerning the algebraic expressions involved in such equations, we use the following
terminology.
A quadratic expression in a variable v is an algebraic expression of the form
leader · v 2 + middle · v + trailer.
(A.15)
As far as v is concerned, we understand all three quantities leader, middle, trailer as numbers (or “constants”), which we call coefficients, and we refer to them as follows:
• leader is called the leading coefficient: it is always assumed to be non-zero.
• middle is called the linear coefficient.
• trailer is called the constant term.
For any quadratic expression as above, we also define its discriminant to be the quantity
Discriminant = (middle)2 − 4 · leader · trailer.
The discriminant of a quadratic expression is important to us in several ways. Its first application, although “mysterious” and often overlooked, deals with the transformation of any quadratic
expression into one that (after a suitable change of variables) has no middle term:
Completed Square Identity
middle 2
Discriminant
leader · v + middle · v + trailer = leader · v +
−
. (A.16)
2 · leader
4 · leader
2
C LARIFICATION. How does the above identity come about this way?
Let us set this question aside for the moment, and consider the following “game.” Assume we
have a simpler quadratic expression, of the form v 2 +#·v, and we want to complete it to a square,
so we want to add some quantity ? to make it a square:
v 2 + # · v + ? = (v + ♦)2 .
(A.17)
Using the first Quadratic Product Identity, we know that, whatever ♦ is, the right hand side in
(A.17) must have the form v 2 + 2♦ · v + ♦2 , so (A.17) must look like:
v 2 + # · v + ? = v 2 + 2♦ · v + ♦2 .
APPENDICES
1009
The only way to get a match is when # = 2♦ and ? = ♦2 . This then gives ♦ = #/2 and
? = #2 /4, so the correct way to complete the square in (A.17) is:
#2
v +#·v+
=
4
2
2
#
v+
.
2
Equivalently, we can also write:
2
#
#2
v +#·v = v+
−
.
2
4
2
(A.18)
How about a slightly more general expression, of the form ℓv 2 + mv? All we have to do is to
match it with the identity obtained by multiplying (A.18) by ℓ:
2
#
ℓ#2
ℓv + ℓ# · v = ℓ v +
−
.
2
4
2
(A.19)
Matching the left-hand side of this identity with ℓv 2 + mv is the same as saying that ℓ# = m,
that is, # = m/ℓ. With this replacement, the identity (A.19) becomes:5
“Easy” Completed Square Identity
m 2 m2
ℓv + mv = ℓ v +
−
.
2ℓ
4ℓ
2
(A.20)
Finally, for the most general quadratic expression ℓv 2 + mv + t, we use (A.20) and add the
constant term t to it:
m 2 m2
m 2 m2 4ℓt
ℓv 2 + mv + t = ℓ v +
−
+t=ℓ v+
−
+
=
2ℓ
4ℓ
2ℓ
4ℓ
4ℓ
m 2 4ℓt − m2
m 2 m2 − 4ℓt
m 2 D
=ℓ v+
+
=ℓ v+
−
=ℓ v+
− ,
2ℓ
4ℓ
2ℓ
4ℓ
2ℓ
4ℓ
where D is the discriminant.
Example A.9. Complete the square in: 2x2 − 12x − 5.
Solution. The leader is ℓ = 2, the middle is m = −12 and the trailer is t = −5, so the discrimant is D = (−12)2 − 4 · 2 · (−5) = 144 + 40 = 184. Thus, after completing the square, our
expression is:
2
184
−12
2x − 12x − 5 = 2 x +
−
= 2(x − 3)2 − 23.
2·2
4·2
2
Example A.10. Complete the squares in: x2 + 14x − 4y 2 + 8y − 7.
5
Upon replacing # = m/ℓ, the last term
last term in (A.20).
ℓ#2
4
in (A.19) becomes:
ℓ(m2 /ℓ2 )
4
=
m2 /ℓ
4
=
m2
4ℓ ,
which is exactly the
1010
A. ALGEBRA REVIEW I: EQUATIONS
Solution. This is a quadratic expression in two variables. It will be better in this case to split
our expression as a sum of two separate expressions plus a constant:
2
2
x2 + 14x − 4y 2 + 8y − 7 = x
(A.21)
+ 8y − 7,
| +
{z14x} + −4y
| {z }
(I)
( II )
and to work on each of the two expressions individually, using the easy square completion (A.20).
In ( I ), the leader is 1, the middle is 14 (and trailer is 0), so after completing the square, this
expression is
2
14
142
2
x + 14x = x +
−
= (x + 7)2 − 49.
(A.22)
2·1
4·1
In ( II ), the leader is −4, the middle is 8 (and trailer is 0), so after completing the square, this
expression is
2
8
82
2
−4y + 8y = (−4) y +
= −4(y − 1)2 + 4.
(A.23)
−
2 · (−4)
4 · (−4)
Now we go back to (A.21) and replace ( I ) and ( II ) using the above two identities:
2
2
2
x2 + 14x − 4y 2 + 8y − 7 = |x2 +
{z14x} + |−4(y −{z1) + 4} − 7 = (x + 7) − 4(y − 1) − 52.
(I)
( II )
Quadratic Equations
A basic quadratic equation (with unknown v) is one of the form;
ℓv 2 + mv + t,
(A.24)
where the leading coefficient ℓ is not equal to zero.
We (should) always solve these equations using the following.
Quadratic Formula
Assuming we have a quadratic equation (A.24), with real coefficients, the real solutions are
determined by the sign of the discriminant D = m2 − 4ℓt, as follows:
( A ) If D < 0, the equation has no real solutions.
m
( B ) If D = 0, the equation has exactly one real solution: v = − .
2ℓ
( C ) If D > 0, the equation has two distinct real
√ solutions:
−m ± D
v=
.
(A.25)
2ℓ
C LARIFICATION. Using square completion, we can re-write our equation (A.24) as
m 2 D
ℓ v+
−
= 0,
2ℓ
4ℓ
which upon multiplying everything by 4ℓ, can be re-written as:
m 2
4ℓ2 v +
− D = 0.
2ℓ
APPENDICES
1011
m 2 m 2
Since the first term can be simplified as 4ℓ2 v +
= 2ℓ v +
= (2ℓv + m)2 , upon
2ℓ
2ℓ
adding D, our equation now reads:
(2ℓv + m)2 = D.
2
We can treat this
√ as a power equation ? = D, with unknown ? = 2ℓv + m, so when we solve it
we get ? = ± D, so we now have
√
2ℓv + m = ± D,
and the desired formula (A.25) follows immediately, by subtracting m, then by dividing by 2ℓ.
It should be pointed out
√ here that the Quadratic Formula (A.25) also works in case ( B ), but in
that case we will have ± D = 0.
B Algebra Review II: Matrix Arithmetic
In this Appendix we review the basic facts concerning matrix arithmetic. We think a matrix
simply as a rectangular table filled with numbers. The size of a matrix is always be specified in the
form
number of rows × number of columns,
so for example a 2 × 2 matrix will be presented as
A=
♣ ♦
♥ ♠
.
We “navigate” the entries (a.k.a. coefficients) in our matrix using notations like aij , where the
first index (i) denotes the row number, and the second index (j) denotes the column number. For
example, the coefficients of the matrix A given above are: a11 = ♣, a12 = ♦, a21 = ♥, and
a22 = ♠. Two matrices are equal, if they have same size, and their coefficients match according to
their position.
 The main limitation of addition and subtraction of matrices is the fact that matrices can be
Matrix Addition, Subtraction, and Scalar Multiplication
added/subtracted, only if they have the same size.
Within this limitation, addition and subtraction of matrices of same size is carried on entry-byentry, so the result has the same size as the terms in the sum/difference. Specifically, if we look at
two m × n matrices




b11 b12 b13 . . . b1n
a11 a12 a13 . . . a1n
 a21 a22 a23 . . . a2n 
 b21 b22 b23 . . . b2n 




B =  ..
A =  ..
..
..
..  ,
..
..
..  ,
 .
 .
.
.
. 
.
.
. 
bm1 bm2 b13 . . . bmn
am1 am2 a13 . . . amn
1012
B. ALGEBRA REVIEW II: MATRIX ARITHMETIC
their sum/difference will be again two m × n matrices:



A+B=

a11 + b11
a21 + b21
..
.
a13 + b13
a23 + b23
..
.
...
...
a1n + b1n
a2n + b2n
..
.
am1 + bm1 am2 + bm2 a13 + bm3 . . . amn + bmn



A−B=

a12 + b12
a22 + b22
..
.
a11 − b11
a21 − b21
..
.
a12 − b12
a22 − b22
..
.
a13 − b13
a23 − b23
..
.
...
...
a1n − b1n
a2n − b2n
..
.
am1 − bm1 am2 − bm2 a13 − bm3 . . . amn − bmn



;






Example B.1.
1 2 3
−2 1 −3
1 + (−2)
2+1
3 + (−3)
11 3 0
+
=
=
.
4 5 6
7 −1 4
4+7
5 + (−1)
6+4
11 4 10
Another easy operation involving matrices is scalar multiplication, which is carried on by
multiplying each entry in the matrix by the given number. Specifically, if we start with some real
number t and an m × n matrix



A=

a11
a21
..
.
a12
a22
..
.
a13 . . .
a23 . . .
..
.
a1n
a2n
..
.
am1 am2 a13 . . . amn



,

then the scalar product tA is again an m × n matrix



tA = 

Example B.2.
ta11
ta21
..
.
ta12
ta22
..
.
ta13 . . .
ta23 . . .
..
.
ta1n
ta2n
..
.
tam1 tam2 ta13 . . . tamn



.


 
 

1 2
(−2)1 (−2)2
−2 −4
(−2)  3 4  =  (−2)3 (−2)4  =  −6 −8 
5 6
(−2)5 (−2)6
−10 −12
It is fairly obvious that matrix subtraction can also be understood using addition and scalar
multiplication by −1, that is,
A − B = A + (−1)B.
In many respects, addition and scalar multiplication for matrices behaves very much like real number arithmetic.
APPENDICES
1013
If A, B and C are matrices of same size m × n, then the following identities hold.
I. Associativity of Addition: (A + B) + C = A + (B + C).
II. Commutativity of Addition: A + B = B + A.
III. Opposite Property: (−A) + A = A + (−A) = 0m×n ; here −A = (−1)A and 0m×n
is the m × n zero matrix, which has all entries equal to zero.
IV. Easy Scalar Multiplications: 0A = 0m×n ; 1A = A; t0m×n = 0m×n .
V. Associativity of Scalar Multiplication: s(tA) = t(sA) = (st)A.
VI. Distributivity over Scalar Addition: (s + t)A = sA + tA.
VII. Distributivity over Matrix Addition: t(A + B) = tA + tB.
Matrix Multiplication
The product of matrices is built up starting with its simplest version, which is described as
follows. Given a row matrix R and a column matrix C


y1
 y2 


R = [x1 x2 . . . xn ], C =  ..  ,
 . 
yn
both having the same number of entries, the row-times-column product RC is the number:
R · C = x1 y1 + x2 y2 + . . . + xn yn .
More generally, given an m × n matrix A, and and n × k matrix B, the matrix product P = AB
is an m × k matrix, constructed by the the following rules:
(i) P has as many rows as A has, and as many columns as B has. Thus A is an m × k matrix.
(ii) The entry pij in P (that sits in row i and column j) is the row-times-column product
pij = [ith row of A] · [j th column of B].
 The matrix product P = AB is defined only if the following match occurs:
number of columns in A = number of rows in B.
In general, the matrix product is not commutative! This means that, even in the cases when both
the matrix products AB and BA are defined, they may be different. For example,
1 2
1 0
1 4
1 0
1 2
1 2
·
=
, but
·
=
.
3 4
0 2
3 8
0 2
3 4
6 8
Example B.3. In Trigonometry, we only use 2 × 2 and 2 × 1 matrices, for which the only
possible products we are dealing with are covered by the following formulas
♣ ♦
s
♣s + ♦t
·
=
;
(B.1)
♥ ♠
t
♥s + ♠t
♣ ♦
s u
♣s + ♦t ♣u + ♦v
·
=
.
(B.2)
♥ ♠
t v
♥s + ♠t ♥u + ♠v
Even though matrix multiplication is not commutative, it still has some nice properties which
again resemble some from real number arithmetic.
1014
B. ALGEBRA REVIEW II: MATRIX ARITHMETIC
I. Associativity of Matrix Multiplication: If AB and BC are defined then the following
products are defined and are equal: (AB)C = A(BC).
II. Distributivity on the Right over Matrix Addition: If AB and AC are defined, then so is
A(B + C), and furthermore: A(B + C) = AB + AC.
III. Distributivity on the Left over Matrix Addition: If AC and BC are defined, then so is
(A + B)C, and furthermore: (A + B)C = AC + BC.
One special type of matrices, which appear very naturally when dealing with matrix products,
are the identity matrices In , which are n×n (thus square) matrices, which have 1’s on the diagonal,
and zeros everywhere else. For example,
I2 =
1 0
0 1
;


1 0 0
I3 =  0 1 0  .
0 0 1
The most important property of such matrices is the following “easy” multiplication rule:
If A is an m × n matrix, then:
Im A = AIn = A.
(B.3)
Invertible Matrices
When dealing with matrix multiplications, one can also ask if there is any notion of division.
After all, when doing number arithmetic, we always understand the division operation as the “undoing” of a multiplication, or equivalently, multiplication by reciprocals. With these considerations
in mind, one introduces the following definition.
A square n × n matrix A is said to be invertible, if one can find another square n × n matrix
B, such that
AB = In .
C LARIFICATIONS. If a matrix B as above exists, then it is in fact unique, and it also satisfies
the equality
BA = In .
For this, and many other reasons, the matrix B is denoted by A−1 , and is referred to as the inverse
of A.
The inverse matrices (if they exist) are very useful for solving systems of linear equations. Any
system with n equations and n unknowns, of the form


a11 x1 + a12 x2 + · · · + a1n xn = c1


 a21 x1 + a22 x2 + · · · + a2n xn = c2
(B.4)
..

.


 a x +a x +···+a x = c
n2 2
nn n
n
n1 1
APPENDICES
1015
can also be presented in matrix form as a multiplication equation AX = C, where:



A=

a11 a12 . . . a1n
a21 a22 . . . a2n
..
..
..
.
.
.
an1 an2 . . . ann



;




X=

x1
x2
..
.
xn




;



C=

c1
c2
..
.
cn



.

The point is that, if A is invertible, then the equation AX = C can be easily solved by multiplying
both sides on the left by’ A−1 , to give X = A−1 C. In turn, this means that, if we have a way to
compute the inverse matrix, by finding its coefficients, say
A
−1



=

b11 b12 . . . b1n
b21 b22 . . . b2n
..
..
..
.
.
.
bn1 bn2 . . . bnn



,

then the given system (B.4) is immediately solved in matrix form:





x1
x2
..
.
xn


 
 
=
 
b11 b12 . . . b1n
b21 b22 . . . b2n
..
..
..
.
.
.
bn1 bn2 . . . bnn
 
 
 
·
 
c1
c2
..
.
cn



,

which when we write everything down reads:


x1 = b11 c1 + b12 c2 + · · · + b1n cn


 x2 = b21 c1 + b22 c2 + · · · + b2n cn
..

.


 x = b c +b c +···+b c
n
n1 1
n2 2
nn n
In the C OLLEGE A LGEBRA course you learned how to decide if a matrix is invertible and (if
it is) how to compute its inverse, both “by hand” (using reduced row-echelon forms) and with the
help of a scientific calculator. Fortunately, in the 2 × 2 case (the only one needed in Trigonometry),
this problem is quite easy to solve, using the following scheme.
Invertibility Test & Formula for 2 × 2 Matrices
♣ ♦
A matrix A =
is invertible, if and only if the quantity det(A) = ♣♠ − ♦♥
♥ ♠
(which is referred to as the determinant of A) is non-zero. Moreover, if this is the case, then:
1016
B. ALGEBRA REVIEW II: MATRIX ARITHMETIC
1
♠ −♦
A =
.
(B.5)
det(A) −♥ ♣
1 2
Example B.4. Consider the matrix A =
, and let us find its inverse (if it has one).
3 4
By definition of the determinant, we have det(A) = 1 · 4 − 2 · 3 = −2 6= 0, so A is indeed
invertible, and using (B.5) its inverse is:
1
4 −2
−2 1
−1
A =
=
.
3
− 12
−2 −3 1
2
−1