Download Trigonometry

Chapter 1
Trigonometry
1.1
Angular Measure
Consider the circle in the plane of radius 1, centre the origin. Henceforth this will be known as
the unit circle. Let A be the point (1,0) and let P be any point on the circle.
Degree Measure
P
θ
O
A
Figure 3.1.1
Then we shall say that ∠AOP has degree measure θ◦ if
lenth of arc AP
θ◦
=
,
circumference
360
i.e.
length of arc AP
.
(1.1)
circumference
In this definition please note that the length of arc AP is measured in an anti-clockwise direction
from A. This is, of course, nothing but the usual definition of the degree measure of an angle
used in secondary schools. Thus if P = (0, 1) then
θ◦ = 360 ×
length of arc AP =
2π
π
circumference
=
= .
4
4
2
Hence
π
360◦
360◦
× =
= 90◦ .
2π
2
4
Evidently if P = (−1, 0) then θ◦ = 180◦ , while if P = (0, −1) then θ◦ = 270◦ (see Figure 3.1.2).
θ◦ =
1
2
CHAPTER 1. TRIGONOMETRY
270◦
O
z
A
Figure 3.1.2
In this way any point P on the unit circle determines a degree measure, and if we take any
degree measure from 0◦ to 360◦ we shall obtain a point P on the unit circle. But why stop
there? Suppose we wish to consider angles of more than 360◦ . For example take 380◦ . Starting
at A = (1, 0), go once around the unit circle, giving 360◦ , and then a further 20◦ (Figure 3.1.3).
380◦
O
6
A
Figure 3.1.3
Notice that we end up at the same point P as is obtained by considering 20◦ .
THIS DOES NOT MEAN 380◦ = 20◦
As a second example consider 1437◦ . In this case go three times around the circle, using up
3 × 360◦ = 1080◦ . There remains 357◦ . So we proceed a further 357◦ , almost getting back to
where we started.
Now we can see what θ◦ means for any non-negative θ. Can we not attach some meaning
to negative angles? What could −45◦ mean? (We could define −45◦ to mean anything we
choose, but we wish to make a sensible and useful definition.) What we should like is that
45◦ + (−45◦ ) = 0◦ . Now 45◦ is going to mean moving to a point P anti-clockwise from A
through 1/8 of a circumference. 0◦ is going to mean remaining at A. Thus it seems sense to
define −45◦ to mean moving to a point Q, say, from A clockwise through 1/8 of a circumference
(Figure 3.1.4).
O
A
−45◦
Q
Figure 3.1.4
In general if θ◦ is any angle, −θ◦ is the same angle, but the rotation is in the opposite direction.
It may have occurred to you that the number 360 in formula (1.1) was a little curious. Why
360 and not 240 or any other number? The reason for the choice of 360 is historical and dates
back to Babylonian times. Clearly the number 360 is quite arbitrary and (1.1) would be simpler
without it. This will give a different kind of angular measure.
1.1. ANGULAR MEASURE
3
Radian Measure
We shall say that ∠AOP has radian measure θ if
θ = length of arc AP
(1.2)
P
k θ
θ
O
?
A
Figure 3.1.5
In Figure 3.1.5 the arc length AP is θ, so ∠AOP has radian measure θ. Notice that if ∠AOP
is a right angle then
circumference
π
θ=
= .
4
2
Thus
90◦ =
π
radians,
2
and
1 radian =
180◦ = π radians,
180◦
≈ 57.296◦ .
π
Indeed if θ = 1 radian, then arc AP will have length 1.
P
Y
1
O
1
1
A
?
Figure 3.1.6
The following are useful to remember:
30◦ = π/6 radians,
45◦ = π/4 radians,
180◦ = π radians,
60◦ = π/3 radians, 90◦ = π/2 radians,
360◦ = 2π radians.
Of course you should not learn these off by heart. Any one of them will do. Practice will
yield familiarity.
Figure 3.1.7 may help to illustrate the relationship between degrees and radians. Radian
measures are written in the interior of the circle, degrees on the outside.
4
CHAPTER 1. TRIGONOMETRY
90◦
60◦
π/2
45◦
2π/3
π/3
π/4
30◦
π/6
5π/6
120◦
150◦
0 0◦
180◦ π
4π/3
240
7π/4
◦
315◦
Figure 3.1.7
Exercise 3.1
1. Sketch the points obtained on the unit circle by the following angles:
(a) 45◦
(b) 210◦
(c) −80◦
2. Convert to radians:
(a) 45◦
(b) 180◦
◦
(f) 210
(g) 240◦
◦
(k) −60
(l) −45◦
◦
(p) −315
(q) −225◦
◦
(u) 100
(v) 5◦
(d) 530◦
(c) 60◦
(h) 225◦
(m) −75◦
(r) −150◦
(w) 37◦
(e) −390◦
(d) 270◦
(i) 315◦
(n) 520◦
(s) −135◦
(x) 247◦
(f) 1800◦ .
(e) 135◦
(j) −30◦
(o) −405◦
(t) −130◦
(y) −543◦
3. Convert to degrees:
(a) 3π/2 rad.
(e) 3 rad.
(i) 7π/15 rad.
(m) 6.2 rad.
1.2
(b) π/10 rad.
(f) −5π/8 rad.
(j) 7 rad.
(c) 9π/2 rad.
(g) −11π/12 rad.
(k) −11π/15 rad.
(d) −3π/4 rad.
(h) −3π/8 rad.
(l) −34π/12 rad.
Trigonometr1c Functions
Let P (x, y) be any point on the unit circle and let A = (1, 0) and O = (0, 0). Let ∠AOP = θ
and the perpendicular from P to the x-axis meets that axis at K. Then P K = y and OK = x.
. P (x, y)
.......
.
.
.. ..
... ...
.
.
.. θ ..
O
K A
Figure 3.2.1
1.2. TRIGONOMETR1C FUNCTIONS
5
We define:
1
= sec θ = secant θ (x 6= 0)
x
1
= csc θ = cosecant θ (y 6= 0)
y
x
= cot θ = cotangent θ (y 6= 0).
y
x = cos θ = cosine θ
y = sin θ = sine θ
y
= tan θ = tangent θ (x 6= 0)
x
These six functions of θ are called trigonometric functions.
A large number of facts are immediately clear.
(a) Since −1 ≤ x ≤ 1, i.e. |x| ≤ 1, then | cos θ| ≤ 1. Similarly | sin θ| ≤ 1.
(b) It follows directly from the definitions that
tan θ =
Furthermore
cot θ =
cos θ
,
sin θ
sin θ
,
cos θ
sec θ =
cos θ 6= 0
(1.3)
1
1
, csc θ =
.
cos θ
sin θ
(c) Since x2 + y 2 = 12 , by Pythagoras’ Theorem, then
sin2 θ + cos2 θ = 1.
(1.4)
Note that sin2 θ means (sin θ)(sin θ) = (sin θ)2 .
This is quite different from sin(θ2 ).
(d) Dividing (1.4) by sin2 θ (when sin θ 6= 0), we obtain
1+
cos2 θ
1
=
sin2 θ
sin2 θ
whence
csc2 θ = 1 + cot2 θ
(1.5)
Similarly, dividing by cos2 θ, we have
sec2 θ = 1 + tan2 θ.
(1.6)
(e) If φ = θ + 2π, then the point P 0 on the unit circle determined by φ will be the same as the
point P determined by θ. In fact φ is the same as θ with one rotation added on. Then
sin φ = sin θ (since P 0 = P ), cos φ = cos θ etc. Indeed if n is any integer, then
sin(θ + 2nπ) = sin θ, cos(θ + 2nπ) = cos θ
and the same for the other four trigonometric functions.
(f) Directly from the definitions,
sin 0 = 0,
sin π/2 = 1,
cos π/2 = 0,
sin π = 0,
sin 3π/2 = −1,
tan 0 = 0,
tan π = 0,
cos 3π/2 = 0,
cos 0 = 1.
tan π/2 is undefined
cos π = −1
tan 3π/2 is undefined.
6
CHAPTER 1. TRIGONOMETRY
(g) If P is in the first quadrant, sin θ > 0, cos θ > 0, tan θ > 0.
If P is in the second quadrant, sin θ > 0, cos θ < 0, tan θ < 0.
If P is in the third quadrant then sin θ < 0, cos θ < 0, tan θ > 0.
If P is in the fourth quadrant then sin θ < 0, cos θ > 0, tan θ < 0.
It follows that if 0 < θ < π then sin θ > 0 while if π < θ < 2π then sin θ < 0.
Also if −π/2 < θ < π/2, then cos θ > 0, while if π/2 < θ < 3π/2 then cos θ < 0.
Finally, if 0 < θ < π/2 or π < θ < 3π/2 then tan θ > 0, while if π/2 < θ < π or
3π/2 < θ < 2π then tan θ < 0. All this information is summarized in Figure 3.2.2.
II
I
sin θ > 0
cos θ < 0
tan θ < 0
sin θ > 0
cos θ > 0
tan θ > 0
sin θ < 0
cos θ < 0
tan θ > 0
sin θ < 0
cos θ > 0
tan θ < 0
III
IV
Figure 3.2.2
(h) If θ = π/4, then x = y = sin θ = cos θ.
Since
sin2 (π/4) + cos2 (π/4) = 1
by (1.4), then
so
2 sin2 (π/4) = 1,
1
sin π/4 = cos π/4 = √ , and tan π/4 = 1.
2
............. P
√ 6 1 ....
..
1/ 2
..
? π/4 ...
A
O¾ √1/ 2
Figure 3.2.3
(i) If θ =
π
(= 60◦ ), then ∠OP B = 30◦ = π/6 (see Figure 3.2.4). Hence
3
OB =
1
1
OP = .
2
2
1.2. TRIGONOMETR1C FUNCTIONS
7
Also
P B 2 = OP 2 − OB 2 = 1 −
Thus
1
3
= .
4
4
√
PB =
i.e.
3
,
2
√
sin π/3 =
3
,
2
cos π/3 =
1
,
2
..........P
..
6
.
√
1 ...
..
3/2
...
?π/3 ..
O¾ 1/2
tan π/3 =
√
3.
A
Figure 3.2.4
By a similar argument,
sin
π
1
= ,
6
2
cos
√
π
3
=
,
6
2
tan
π
1
=√ .
6
3
These results may be set out as follows, in a table of standard angles:
θ
0
π/6
sin θ
0
cos θ
1
tan θ
0
1/2
√
3/2
√
1/ 3
π/4
√
1/ 2
√
1/ 2
1
π/3
√
3/2
π/2
π
3π/2
1
0
−1
1/2
√
3
0
−1
0
undef
0
undef
(j) Let P be the point on the unit circle with coordinates (x, y) and let ∠AOP = θ,
i.e. x = cos θ, y = sin θ. Suppose Q is the point with coordinates (x, −y). Evidently
∠AOQ = −θ.
. P (x, y)
.......
.
.
.. ..
... ...
.
.
..
.
..... θ ...
..
...
O −θ
A
... ...
... ...
... ..
...
Q
Figure 3.2.5
8
CHAPTER 1. TRIGONOMETRY
Hence
cos(−θ) = x = cos θ,
sin(−θ) = −y = − sin θ,
tan(−θ) = −y/x = − tan θ.
In summary,
cos(−θ) = cos θ,
sin(−θ) = − sin θ,
tan(−θ) = − tan θ.
Thus for example, sin(−π/6) = − sin π/6 = −1/2; cos(−π/6) = cos π/6 =
(1.7)
√
3/2, etc.
Once the values of any trigonometric function are known from θ = 0 to θ = π/2, all other
values can be calculated.
Example 1.2.1
Find : (a) sin
3π
4
(b) tan
11π
6
(c) cos
7π
.
6
(a) sin 3π/4 is the value of the y coordinate of P when ∠AOP = 3π/4.
P (x, y) ......................
...
...
.
.
3π/4 .....
.
O
A
Figure 3.2.6
√
From Figure 3.2.6 it is clear that sin 3π/4 = sin π/4 = 1/ 2.
(b) It is easy to see from Figure 3.2.7 that
√
tan = 11π/6 = tan(−π/6) = − tan π/6 = −1/ 3.
11π/6
O
−π/6
A
P (x, y)
Figure 3.2.7
(c) From Figure 3.2.8,
√
7π
π
3
cos
= − cos = −
.
6
6
2
1.2. TRIGONOMETR1C FUNCTIONS
9
π/6
7π/6
A
O
P (x, y)
Figure 3.2.8
Notice now that we can regard any of the trigonometric functions
√ as either depending on
an angle or just on a real number. Thus when we say sin π/4 = 1/ 2, we may have in mind
an angle of 45◦ or the real number π/4 ≈ 0.7854.
We conclude this section with a simple identity.
Example 1.2.2
Prove the identity (cot θ − csc θ)2 =
1 − cos θ
.
1 + cos θ
The way to proceed is to start with one side and show it equals the other. Letting LHS
and RHS stand for left-hand side and right-hand side respectively, we have
LHS =
=
=
=
=
³ cos θ
−
1 ´2
sin θ
sin θ
1
(cos θ − 1)2
sin2 θ
(cos θ − 1)2
1 − cos2 θ
(cos θ − 1)2
(1 − cos θ)(1 + cos θ)
cos θ − 1
−
= RHS.
cos θ + 1
Exercise 3.2
1. Find, where possible, the exact
values of θ:
(a) 0
(b) 4π/3
(f) π
(g) 3π/2
(k) −3π/4
(l) −7π/4
(p)−11π/6
(q) 10π/3
(u) −13π/3
(v) 3π
values of all six trigonometric functions for the following
(c) 4π/3
(h) 7π/2
(m) −π/3
(r) 13π/6
(w)103π
(d) 2π/3
(i) 5π/6
(n) −π/6
(s) 23π/6
(x) 2780π
(e) π/2
(j) 7π/6
(o) −7π/2
(t) 101π/4
(y) 53π − π/4
2. Find, where possible, the exact values of all six trigonometric functions for the following
values of θ:
(a) 30◦
(b) 120◦
(c) 210◦
(d) 135◦
(e) 240◦
(f) 225◦
(g) 315◦
(h) −45◦
(i) −135◦
(j) 420◦
(k) 765◦
(l) −660◦
(m) 1440◦
10
CHAPTER 1. TRIGONOMETRY
3. Simplify:
³ 103π ´
³ 103π ´
(a) sin
+ sin −
9
9
q
(b)
sin2 (π/15 + cos2 π/15
(c) sec2 (−π/8) − tan2 (−π/8)
(d) csc2 (π/7) − cot2 (π/7)
(e) − tan2 θ + sec2 θ − sin2 θ
(f) sin4 θ + 2 sin2 θ cos2 θ + cos4 θ.
4. In what quadrant must θ lie in order that the following be satisfied?
(a) cos θ > 0 and sin θ < 0
(c) sin θ > 0 and cot θ < 0
(e) tan θ > 0 and csc θ < 0
(b) cos θ > 0 and tan θ < 0
(d) csc θ > 0 and sec θ < 0
(f) sec θ < 0 and csc θ < 0.
5. Prove the following identities:
(a) (1 − sin2 θ)(1 + tan2 θ) = 1
(b) (tan θ − sec θ)2 =
(c)
csc θ − 1
csc θ + 1
1 − sec θ
= − csc θ
tan θ − sin θ
1 + cos2 x
= 2 csc2 x − 1.
sin2 x
cot x − tan x
(e)
= csc x − sec x
sin x + cos x
(d)
(f) csc4 x − cot4 x = (1 − cot2 x) csc2 x
r
1 − cos x
1 − cos x
(g)
=
1 + cos x
| sin x|
(h) sin3 x + cos3 x = (1 − sin x cos x)(sin x + cos x).
6. Let C1 be the circle passing through the origin with centre (2, 0) and C2 be the circle
passing through the origin with centre (0, 4). The line y = x tan θ cuts C1 again at A and
C2 again at B. Let M be the mid-point of AB. Show that M has coordinates
(2 cos2 θ + 4 sin θ cos θ, 2 sin θ cos θ + 4 sin2 θ)
and hence deduce that M lies on the circle x2 + y 2 − 2x − 4y = 0.
1.3
Graphs of Trigonometric Functions
In this section we shall discuss the graphs of the elementary trigonometric functions f (θ) =
sin θ, f (θ) = cos θ and f (θ) = tan θ as well as some related functions.
(a) f (θ) = sin θ.
Recall that sin θ is the y coordinate of the point P (x, y) on the unit circle (Figure 3.3.1).
First note that f (0) = 0. As θ moves from 0 to π/2 , P moves from A to B, y moves from 0 to
1 and so f (θ) increases from 0 to 1. Then as θ increases from π/2 to π, P moves from B to C,
y moves from 1 back to 0 and thus f (θ) decreases back to 0.
1.3. GRAPHS OF TRIGONOMETRIC FUNCTIONS
B
6
sin θ
? θ
O
C
.. P
..
..
..
..
..
..
..
11
A
Figure 3.3.1
While θ moves from π to 3π/2, y becomes negative and f (θ) continues to decrease, reaching
−1 when θ = 3π/2. Finally, as θ increases to 2π, sin θ increases back to 0 again. After that,
the whole process repeats itself, coming back to where it started every time θ increases by 2π.
(Recall that f (θ + 2nπ) = sin(θ + 2nπ) = sin θ = f (θ).) The same thing happens when θ is
negative.
Graph of y = sin θ
y
1
y = sin θ
π
−π/2
−π
π/2
3π/2
2π
5π/2 θ
−1
Figure 3.3.2
Notice that the graph lies entirely between −1 and 1, i.e. range(f ) = [−1, 1]. Also each wave
(counting a peak and a trough) has length 2π. We say that f is periodic of period 2π.
Once again θ is regarded either as a real number or as an angle measured in radians (and
not degrees).
(b) g(θ) = cos θ
Here the process is almost the same. Recall that cos θ is the x-coordinate of the point
P (x, y) (see Figure 3.3.3). When θ = 0, P lies on A, x = 1 and so g(0) = 1. After that g(θ)
decreases to 0 as θ reaches π/2 and further to −1 when θ is π. Then it increases back to 1
when θ goes up to 2π.
12
CHAPTER 1. TRIGONOMETRY
B
O ¾
C
θ
cos θ
.. P
..
..
..
..
- A
Figure 3.3.3
Graph of y = cos θ
y
1
−π/2
y = cos θ
π
π/2
3π/2
θ
2π
−1
Figure 3.3.4
Notice that the shape is the same as that of the graph of sin θ but it has been moved to
the left by π/2, i.e. sin θ is obtained by moving the graph of cos θ to the right by π/2. Thus
cos(θ − π/2) = sin θ. Hence by (1.7)
cos(π/2 − θ) = sin θ.
(1.8)
Also, if we move the graph of sin θ to the right by π/2 we shall obtain the same graph as that
of cos θ only “turned over”, that is with the sign changed. Hence
sin(θ − π/2) = − cos θ,
i.e.
− sin(π/2 − θ) = − cos θ
by (1.7). Hence
sin(π/2 − θ) = cos θ.
(1.9)
Notice that π/2 − θ is the complement of θ. The first two letters of the word “complement”
are used in the word “cosine”. So the cosine is the sine of the complement. In the same way
tan(π/2 − θ) = cot θ
cot(π/2 − θ) = tan θ
sec(π/2 − θ) = csc θ
csc(π/2 − θ) = sec θ.
1.3. GRAPHS OF TRIGONOMETRIC FUNCTIONS
13
(c) h(θ) = tan θ.
In Figure 3.3.5 tan θ = y/x, which is just the gradient of OP .
B
C
.................. P
..
6
..
y
..
..
θ
?
¾
O
A
x
Figure 3.3.5
Thus as P moves from A and θ increases to π/2, the slope of OP increases from 0 to become
larger and larger as P approaches π/2, at which point it is undefined. Then as θ increases
to π the slope changes from very large and negative up to zero, and in the next quadrant it
becomes larger and larger (and positive), becoming undefined at 3π/2. Between 3π/2 and 2π
the slope is negative, moving from very large and negative back to zero. The whole process
clearly repeats itself every time θ increases by 2π.
Graph of y = tan θ
y
y = tan θ
π
−π
−π/2
π/2
2π
3π/2
5π/2
θ
Figure 3.3.6
In many practical applications wave forms such as those exhibited by sin θ and cos θ arise.
However, the amplitude (how high the graph is), the period (how long the waves are) and the
value at θ = 0 may be changed.
Example 1.3.1
Sketch the graph of y = 2 sin θ.
This is just like sin θ except that the y values are twice as large, so that the graph will
vary between y = −2 and y = 2.
14
CHAPTER 1. TRIGONOMETRY
y
2
1
π
π/2
y = sin θ
−1
y = 2 sin θ
−2
θ
2π
Figure 3.3.7
Example 1.3.2
Sketch the graph of y = −3 cos θ.
This is just like cos θ, except the values will vary from −3 to 3 and the graph has been
turned over.
3
y = −3 cos θ
1
−π/2
π/2
π
2π
θ
y = cos θ
−3
Figure 3.3.8
Example 1.3.3
Sketch y = sin 2θ.
This is similar to sin θ, but now when θ increases from 0 to π/2, y increases to sin 2π/4 =
sin π/2 = 1. Then as θ moves from π/4 to π/2, y returns to 0. This means everything is
happening twice as fast and we get back to where we started when θ reaches π.
1.3. GRAPHS OF TRIGONOMETRIC FUNCTIONS
15
y
y = sin θ
1
π
π/2
3π/2
y = sin 2θ
θ
2π
−1
Figure 3.3.9
Example 1.3.4
Sketch the graph of y = cos(θ/2).
This is very similar to the previous example, but things are now happening only half as
fast as for y = cos θ.
1
y
y = cos θ
−π/2
π/2
π
2π
y = cos(θ/2)
−1
Figure 3.3.10
Example 1.3.5
Sketch y = sin(θ + 3π/2).
This is just sin θ shifted to the left by 3π/2.
y
1
y = sin θ
π/2
π
3π/2
2π
θ
y = sin(θ + 3π/2)
−1
Figure 3.3.11
It is interesting to note that the graph we obtain is the same as that obtained by shifting
sin θ to the right by π/2, i.e.
sin(θ + 3π/2) = sin(θ − π/2).
Also it is evident that the graph is just cos θ “turned over”, i.e.
sin(θ + 3π/2) = − cos θ.
16
CHAPTER 1. TRIGONOMETRY
We can combine all three ideas of the last three examples together.
Example 1.3.6
Sketch y = 3 sin(2θ + π/2).
Firstly the amplitude is 3, i.e. y ∈ [−3, 3]. Secondly the graph repeats itself every
time θ increases by π, i.e. the wavelength or period is only π. Finally, when θ = 0,
y = 3 sin(π/2) = 3.
3
y = 3 sin(2θ + π/2)
π/2
−π/4
π/4
2π
3π/4
5π/4
−3
Figure 3.3.12
Finally let us sketch the graph of y = csc θ.
First recall that csc θ = 1/ sin θ. Thus when sin θ = 0, csc θ is undefined. When θ is small
and positive, sin θ is also small and positive and so csc θ is large and positive (for example if
sin θ = 1/100, then csc θ = 100). Now as θ increases to 1 at π/2, csc θ decreases to 1. Note
that if θ ∈ (0, π/2), then sin θ ∈ (0, 1) and so csc θ ∈ (1, ∞). In the second quadrant the reverse
happens. Thus sin θ decreases, so csc θ increases. In the third and fourth quadrant the same
sort of thing happens, but the signs are reversed. Figure 3.3.13 shows the graphs of y = sin θ
and csc θ on the same axes.
y = csc θ
1
−π
y = sin θ
−1
π
2π
Figure 3.3.13
The graphs of y = sec θ and y = cot θ can be drawn in a similar way.
1.4. ADDITION AND DOUBLE ANGLE FORMULAE
17
Exercise 3.3
1. Sketch the graphs of:
(a) y = sin 3θ
(d) y = cos 4θ
(g) y = 3 cos 2θ
(j) y = 3 cos(2θ − pi)
(m) y = 1/(1 + sin θ)
(p) y = − tan θ
(s) y = −2 tan 3θ
(v) y = sec 2θ
(y) y = 4 cot(θ − 3π/2)
1.4
(b) y = 2 cos θ
(e) y = 3 cos θ
(h) y = 1 + sin θ
(k) y = 4 cos(θ/2)
(n) y = | sin θ|
(q) y = 1 + tanθ
(t) y = sec θ
(w) y = sec θ/2
(z) y = −2 sec(2π − θ)
(c) y = 2 sin θ
(f) y = − sin 4θ
(i) y = −4 sin(θ − π/2)
(l) y = 3sin(π/6 + θ/2)
(o) y = tan 2θ
(r) y = tan(θ + π/2)
(u) y = cot θ
(x) y = csc(θ + π)
Addition and Double Angle Formulae
It is useful in many applications to have formulae for the values of such expressions as sin 2θ,
cos(θ + φ), etc. It is tempting to think that
sin 2θ = 2 sin θ,
cos(θ + φ) = cos θ + cos φ
etc.
These are totally WRONG.
The situation is more complicated. For example if we take θ = π/4, then 2θ = π/2. Hence
sin 2θ = sin π/2 = 1,
2
but 2 sin θ = 2 sin π/2 = √ .
2
Further, taking θ = φ = 0,
cos(θ + φ) = cos 0 = 1,
but
cos θ + cos φ = cos 0 + cos 0 = 2.
Writing f (θ) = sin θ or cos θ or tan θ it is NOT TRUE that
f (θ + φ) = f (θ) + f (φ)
or that
f (kθ) = kf (θ).
(1.10)
In fact, the only functions for which the relations (1.10) hold are straight lines through the
origin. THEY DO NOT HOLD FOR ANY TRIGONOMETRIC FUNCTIONS.
Now let P be any point on the unit circle such that ∠AOP = θ, let Q be a point on the unit
circle such that ∠AOQ = θ + φ, and let R be a point on the unit circle such that ∠AOR = −φ,
where as usual, A is the point (1,0) (see Figure 3.4.1).
Notice then that
∠AOQ = ∠ROP = θ + φ.
Hence
chord AQ = chord RP.
Furthermore,
P = (cos θ, sin θ),
Q = (cos(θ + φ), sin(θ + φ)),
and
R = (cos(−φ), sin(−φ)) = (cos φ, − sin φ).
Now by (1.11),
AQ = RP.
(1.11)
18
CHAPTER 1. TRIGONOMETRY
Q
O
Y θ+φ
φ
θ ?
−φ
P
A
R
Figure 3.4.1
Using the distance formula, we have
AQ2
= [cos θ + φ) − 1]2 + [sin(θ + φ)]2
= cos2 (θ + φ) − 2 cos(θ + φ) + 1 + sin2 (θ + φ)
= 2 − 2 cos(θ + φ)
and
RP 2
= [cos θ − cos φ]2 + [sin θ + sin φ]2
= cos2 θ − 2 cos θ cos φ + cos2 φ + sin2 θ + 2 sin θ sin φ + sin2 φ
= 2 + 2 sin θ sin φ − 2 cos θ cos φ.
Since AQ2 = RP 2 , we obtain
cos(θ + φ) = cos θ cos φ − sin θ sin φ.
(1.12)
Also, since cos(−φ) = cos φ and sin(−φ) = − sin φ, changing φ to −φ in (1.12) yields
cos(θ − φ) = cos θ cos φ + sin θ sin φ.
(1.13)
Equations (1.12) and (1.13) may be summarized in the crucial addition formula below.
Addition Formula for Cosine
cos(θ ± φ) = cos θ cos φ ∓ sin θ sin φ
(1.14)
Now let us consider sin(θ + φ). By (1.8) and (1.9),
sin(π/2 − x) = cos x,
and
cos(π/2 − x) = sin x.
Thus
sin(θ + φ) =
cos(π/2 − (θ + φ))
= cos((π/2 − θ) − φ)
= cos(π/2 − θ) cos φ + sin(π/2 − θ) sin φ
(by (1.14))
= sin θ cos φ + cos θ sin φ (by (1.8) and (1.9)).
Similarly,
sin(θ − φ) = sin θ cos φ − cos θ sin φ,
and we obtain the fundamental formula below.
1.4. ADDITION AND DOUBLE ANGLE FORMULAE
19
Addition Formula for Sine
sin(θ ± φ) = sin θ cos φ ± cos θ sin φ
(1.15)
Finally let us consider tan(θ + φ). By (1.3) we have
tan(θ + φ)
=
=
sin(θ + φ)
cos(θ + φ)
sin θ cos φ + cos θ sin φ
cos θ cos φ − sin θ sin φ
(by (1.14) and (1.15)). Now dividing the numerator and denominator by cos θ cos φ, we obtain
tan(θ + φ) =
tan θ + tan φ
1 − tan θ tan φ
A similar formula may be derived for tan(θ − φ).
Addition Formula for Tangent
tan(θ ± φ) =
tan θ ± tan φ
1 ∓ tan θ tan φ
Example 1.4.1
π
7π
and cos
.
12
12
π
π π
7π
π π
Now
= − , while
= + .
12
3
4
12
3
4
Thus (by 1.15)
Evaluate sin
sin π/12 =
=
sin(π/3 − π/4)
sin π/3 cos π/4 − cos π/3 sin π/4
√
3 1
1 1
√ − √
=
2
2 2 2
√ √
√
√
√
3 2
2
6− 2
=
−
=
.
4
4
4
Similarly
cos
7π
12
=
cos(π/3 + π/4)
=
cos π/3 cos π/4 − sin π/3 sin π/4
√
1 1
3 1
√ −
√
2 2
2
2
√
√
√
1− 3
2− 6
√
=
.
4
2 2
=
=
The following formulae are also very useful. Setting θ = φ in (1.14), we have
cos 2θ = cos2 θ − sin2 θ.
(1.16)
20
CHAPTER 1. TRIGONOMETRY
Now bearing in mind that cos2 θ + sin2 θ = 1, it follows that
cos2 θ = (1 − sin2 θ) − sin2 θ = 1 − 2 sin2 θ,
and also
cos 2θ = cos2 θ − (1 − cos2 θ) = 2 cos2 θ − 1.
Double Angle Formulae for Cosine
In summary we obtain the double-angle formulae:

 cos2 θ − sin2 θ
cos 2θ =
1 − 2 sinθ

2 cos2 θ − 1
(1.17)
Double Angle Formula for Sine
Also setting φ = θ in (1.15), we obtain
sin 2θ = 2 sin θ cos θ
(1.18)
while from (1.16) we have a similar formula for tan 2θ.
Double Angle Formula for Tangent
tan 2θ =
2 tan θ
1 − tanθ
(1.19)
It is obviously a nearly impossible task to memorize all of these formulae – and there are
more to come! Some however are essential. In particular (1.3), (1.4), (1.14) and (1.15) should
be learned. It should be known how to derive all the others. If a large number of practice
exercises are done, many of the others will stick in ones memory without conscious effort.
Example 1.4.2
Prove that 2 sin2 x = tan x sin 2x.
Now
tan x sin 2x =
2 tan x sin x cos x by (1.18)
sin x
= 2
sin x cos x by (1.3)
cos x
2
= 2 sin x.
Example 1.4.3
Prove that (sin x + cos x)2 = 1 + sin 2x.
Multiplying out the left-hand side, we have
(sin x + cos x)2
= sin2 x + 2 sin x cos x + cos2 x
= 1 + 2 sin x cos x (by (1.4))
= 1 + sin 2x,
by (1.18).
1.4. ADDITION AND DOUBLE ANGLE FORMULAE
21
Here is an incorrect “proof” of the above.
(sin x + cos x)2 = 1 + sin 2x,
so
sin2 x + 2 sin x cos x + cos2 x = 1 + 2 sin x cos x
or
1 + 2 sin x cos x = 1 + 2 sin x cos x,
i.e.
0 = 0.
Note that in this case what was required has NOT been proved. What has been proved is that
if (sin x + cos x)2 = 1 + sin 2x, then 0 = 0. This is quite useless. Everybody knows that 0 = 0
anyway.
To emphasize the point consider the following “proof” that 2 = 1.
2 = 1.
So
0.2 = 0.1
or
0 = 0.
Exercise 3.4
1. (a) Prove that cos 3θ = 4 cos3 θ − 3 cos θ. [Hint : cos 3θ = cos(2θ + θ).]
(b) Derive a similar formula for sin 3θ.
(c) Use (a) and (b) to deduce an expression for tan 3θ in terms of tan θ.
2. By using the addition formulae, evaluate exactly:
(a) sin 105◦
(b) cos 75◦
(c) tan 15◦ .
θ
θ
θ
= 2 sin cos .
2
2
2
(b) Deduce formulae for cos θ and tan θ in terms of θ/2.
3. (a) Show that sin
(c) Evaluate cos π/8, sin π/8 and tan π/8.
4. Suppose a = sin 18◦ . Express the following in terms of a:
(a) sin(−18◦ )
(b) cos2 18◦
(c) cos 108◦
(d) tan 162◦
(e) cos 288◦ .
5. Verify that x = 18◦ satisfies the equation sin 2x = cos 3x. Using the double angle formula
and the formula for cos 3x given above in question 1, obtain an equation for sin 18◦ , and
hence find sin 18◦ .
6. Prove the (very easy) identity 2 cos 2x = 1 + cos 2x. Hence sketch the curve y = 2 cos2 x.
7. Generalise formulae (1.14) and (1.15) to obtain formulae for sin(x+y+z) and cos(x+y+z).
8. Obtain formulae for cot(x + y) in terms of cot x and cot y, and sec(x + y) in terms of
secants and tangents.
9. Obtain formulae for cot 2x and sec 2x in terms of cot x and sec x respectively.
22
CHAPTER 1. TRIGONOMETRY
10. Prove the following identities:
(a) sin(π − θ) = sin θ
(c) sin 4θ = 4 sin θ cos θ cos 2θ
(e) tan θ + cot θ = sec θ csc θ
(g) (tan θ + sec θ)2 =
1 + sin θ
1 − sin θ
(b) cos(π + θ) = − cos θ
sin 2θ
(d) tan θ =
1 + cos 2θ
1 + cot θ
sin θ + cos θ
(f)
=
tan θ
sin θ − cos θ
(h) 8 cos4 θ = 3 + 4 cos 2θ + cos 4θ.
(i) (sin θ − cos θ)2 = 1 − sin 2θ
(j) cot θ − tan θ = 2 cot 2θ
(k) tan θ + cot θ = 2 csc 2θ
(l) sin(x + y) sin(x − y) = sin2 x − sin2 y
(m) cos(x + y) cos(x − y) = cos2 x − sin2 y
11. Use the identities for cos(A − B) and cos(A + B) to prove that
1 + cos(A − B) cos(A + B) = cos2 A + cos2 B.
Hence show that 1 + cos 2x = 2 cos2 x and
3+cos(x−y) cos(x+y)+cos(y−z) cos(y+z)+cos(z−x) cos(z+x) = 2(cos2 x+cos2 y+cos2 z).
12. Suppose θ + φ = π/4.
1 − tan θ
. (b) Hence prove that (1 + tan θ)(1 + tan φ) = 2.
1 + tan θ
√
(c)Deduce that tan π/8 = 2 − 1.
(a) Show that tan φ =
13. Show that:
(a) tan 2A =
2 tan A
1 − tan2 A
From (a) deduce that tan θ =
cos θ and sin θ in terms of t .
1.5
(b) tan 3A =
3 tan A − tan3 A
.
1 − 3 tan2 A
2t
, where t = tan θ/2. Derive similar expressions for
1 − t2
Applications to Triangles etc.
The treatment so far may seem very far removed from secondary school trigonometry. This
difference is, in fact, very slight. We shall then apply the theory to the solution of triangles as
well as to the derivation of a number of other simple facts.
y
C
K
θ
O
L B
Figure 3.5.1
x
1.5. APPLICATIONS TO TRIANGLES ETC.
23
Suppose then we have a triangle OBC in which ∠OBC is a right angle. Fix coordinates so
that the origin is O and the positive x-axis lies along OB. Suppose OC = r and ∠BOC = θ.
With centre O and radius 1 draw another circle. This will cut OC (or OC produced) in K.
Let L be the foot of the perpendicular from K to OB (see Figure 3.5.1).
Then by definition
sin θ
=
=
=
KL
KL
OK
CB
OC
(since OK = 1)
by similar triangles.
In the same way cos θ = OL = OL/OK = OB/OC and tan θ = CB/OB.
C
θ
O
B
Figure 3.5.2
Deleting all the excess information and concentrating only on 4OBC, (Figure 3.5.2), we see
sin θ
=
CB
OP
=
“opposite”
“hypoteneuse”
cos θ
=
OB
OP
=
“adjacent”
“hypoteneuse”
tan θ
=
CB
OB
=
“opposite”
.
“adjacent”
These are just secondary school definitions. Notice, however, that because of our definitions
in Section 3.2, it is now clear why, if θ is obtuse, we should have cos θ < 0 and tan θ < 0. Also
by the very definition of triangles this method will not work unless θ ∈ (0, π), whereas our
more advanced definition from Section 3.2 gives a meaning to the trigonometric functions for
all θ ∈ R.
Example 1.5.1
Let ABC be a triangle with B a right angle, a = 2, and b = 3. (Here a is the side opposite
∠A etc.) Find c and ∠C.
24
CHAPTER 1. TRIGONOMETRY
A
b=3
a=2
C
B
Figure 3.5.3
By Pythagoras’ Theorem
AC 2 = AB 2 + BC 2 ,
so 9 = AB 2 + 4 and
AB =
√
5.
Also BC/AC = cos C. Thus cos C = 2/3 and hence C = 48.19◦ (using a calculator).
(a) Area of a Triangle, Sine Rule.
Suppose ABC is any triangle with ∠BAC = θ. Let D be the foot of the perpendicular from
B to AC (Figure 3.5.4).
c
= θ
A ¾
B
> ...
..
..
..
..
..
..
..
..
..
.. D
b
- C
Figure 3.5.4
Now
sin θ =
BD
,
BA
so
BD = BA sin θ = c sin θ.
Hence
area ABC =
1
1
1
1
base × height = AC.BD = bc sin θ = bc sin A.
2
2
2
2
Area of a Triangle
Thus
area of triangle =
1
bc sin A
2
Similarly
1
1
ab sin C = ac sin B.
2
2
Applications of this formula are well-known from secondary school. It follows that
area =
ab sin C = ac sin B = bc sin A.
(1.20)
1.5. APPLICATIONS TO TRIANGLES ETC.
25
Sine Rule
Dividing by abc, we obtain the sine rule:
sin A
sin B
sin C
=
=
a
b
c
(1.21)
(b) Cosine Rule
Let ABC be a triangle. Arrange axes so that A = (0, 0), B = (c, 0) and C = (x, y). Let the
foot of the perpendicular from C to AB be K (see Figure 3.5.5).
Now
CK
= sin A
CA
and
AK
= cos A,
CA
CK = CA sin A
and
AK = CA cos A.
b
C(x,
.. y)
..
..
..
..
..
..
..
..
..
.. K
.
c
i.e.
A ¾
a
-
B(c, 0)
Figure 3.5.5
So
y = b sin A
and x = b cos A.
By the distance formula
BC 2 = (c − x)2 + y 2 ,
i.e.
a2 = (c − b cos A)2 + (b sin A)2 .
Hence using (1.4),
a2 = b2 cos2 A − 2bc cos A + c2 + b2 sin2 A = b2 + c2 − 2bc cos A.
Cosine Rule
This is the cosine rule:
a2 = b2 + c2 − 2bc cos A
Of course, the results
b2 = a2 + c2 − 2ac cos B
follow in exactly the same way.
and
c2 = a2 + b2 − 2ab cos C
(1.22)
26
CHAPTER 1. TRIGONOMETRY
(c) Arc Length, Segment Area
Suppose we have a circle centre O, radius r. Let P and Q be points on the circle. We wish
to find the length of arc P Q. Let ∠P OQ = θ (see Figure 3.5.6).
Q
O
θ
P
Figure 3.5.6
If the angle at the centre were 2π, the arc length would be the circumference, i.e. 2πr. Hence,
θ
if the angle at the centre is θ, the arc length is
× 2πr = rθ.
2π
Arc Length
We have shown
arc length P Q = rθ
(1.23)
Notice that this formula works only when θ is measured in radians.
Example 1.5.2
Let C be a circle with centre O, radius 3. Find the arc length subtended by an angle of
60◦ at the centre (Figure 3.5.7).
3
O
60◦
Figure 3.5.7
Now 60◦ = π/3 radians, so the arc length is rθ = 3
π
= π.
3
Next suppose C is a circle centre O, radius r. Let P and Q be points on the circumference.
Suppose we wish to find the area of the segment P OQ (see Figure 3.5.8).
1.5. APPLICATIONS TO TRIANGLES ETC.
27
r
O
θ
Figure 3.5.8
If the angle at the centre were 2π, the area described would be πr2 . So if the angle at the
centre is θ (in radians) then the area described is
A=
1 2
πr .
2
A=
1 2
r θ
2
Segment Area
(1.24)
Again this formula works only for θ measured in radians. The simplicity of formulae (1.23)
and (1.24) gives another reason why it is advantageous to work in radians. The reader should
work out what the corresponding formulae are when θ is measured in degrees.
Exercise 3.5
1. Suppose 0 < θ < π/2 and sin θ = 3/4. Draw a right-angled triangle one of whose angles
is θ. Find
(a) cos θ
(b) tan θ
(c) cot θ
(d) sin 2θ
(e) sin(π/2 − θ)
(f) cos(π/2 + θ)
(g) cos(3π/2 − θ)
(h) sin(π/3 + θ)
(i) sec(π + θ)
(j) csc(3π/2 + θ)
(k) tan(53π + θ)
(l) csc(55π/2 − θ)
2. Let C be a circle of radius 4, centre O. Points P and Q lie on C and ∠P OQ = 45◦ .
Calculate the area of the shaded sector shown.
Q
O
45◦
P
3. Solve the following triangles ABC, with the standard notation, i.e. find the unknown
sides and angles. (sin or cos of the angle is enough). Find also the area of the triangles.
(a) a = 5,
(c) a = 3,
(e) a = 3,
(g) a = 3,
b = 6, c = 7
B = 45◦ , C = 30◦
b = 5, c = 7
c = 5, B = 120◦
(b) a = 3, b = 3, C = 30◦
(d) a = 3, b = 4, C = 60◦
(f) cos C = 11/14, b = 7, a = 3
(h) B = 90◦ , b = 3, sin A = 2/3.
28
CHAPTER 1. TRIGONOMETRY
√
4. The triangle ABC is defined by: a = 1 ; b = 3; A = 30◦ . Find sin B. Deduce that there
are 2 triangles which satisfy these conditions, and find the 3 sides and 3 angles of each.
5. Angle A is acute with tan A = 1/2. Angle B is obtuse with tan B = −1/3. Find without
using a calculator the value of cos(A − B).
6. The triangle ABC is defined by: B = 90◦ ; sin A = 1/3; b = 3. Find the perpendicular
distance from B to AC.
7. A parallelogram has sides of 4 and 7 and an included angle of 60◦ . Find the lengths of
the diagonals and the area of the parallelogram.
8. Show that the points P = (a cos θ, b sin θ) and Q = (−a sin θ, b cos θ) lie on the ellipse
x2
y2
+ 2 = 1.
2
a
b
Prove that, as θ varies, the values of OP 2 +OQ2 (where O is the origin) remains constant.
9. (a) By using the cosine rule with the usual notation show that
2ac(1 − cos B) = (b + c − a)(b − c + a),
and obtain a similar expressions for
2ac(1 + cos B).
(b) Using formula (1.20), deduce that
Area(4ABC) =
p
s(s − a)(s − b)(s − c),
where s = (a + b + c)/2.
(c) Find the area of the triangle whose sides are 4, 5 and 7.
1.6
The Solution of Trigonometric Equations
In this section we shall investigate how to solve various trigonometric equations. Usually these
can be reduced to one of the forms cosθ = a, sin θ = a, or tan θ = a. We wish to determine θ.
This is not always easy. In general there may be many solutions and sometimes no solution.
For example the very simple equation
cos θ = 1
(1.25)
is clearly satisfiedby θ = 0; but it is equally satisfied by θ = ±2π, ±4π, . . . The solution is thus
θ = 2nπ, n ∈ Z, or if you prefer the solution set is {2nπ | n ∈ Z}.
By way of contrast, the equation cos θ = 2 has no solutions, since | cos θ| ≤ 1.
The Equation cos θ = a
We shall now suppose that |a| ≤ 1, so that the equation has solutions.
We wish to determine for what values of θ we have cos θ = a. By tables (or calculator,
previous knowledge or whatever) it will be possible to find a value θ1 of θ such that 0 ≤ θ1 ≤ π
and cos θ1 = a (see Figure 3.6.1).
1.6. THE SOLUTION OF TRIGONOMETRIC EQUATIONS
1
y = cos θ
−π/2
y=a
π/2
θ3
θ3 = −θ1
29
π
3π/2
θ1
2π
θ4
θ2
θ2 = θ1 + 2π
−1
Figure 3.6.1
This means θ1 is one solution. Then clearly if θ2 = θ1 + 2π, we have
cos θ2 = cos(θ1 + 2π) = cos θ1 = a.
So θ2 is another solution, and indeed θ1 + 2nπ will be a solution for any integer n. However
this is not all. If θ3 = −θ1 , then
cos θ3 = cos(−θ1 ) = cos θ1 = a,
so that θ3 is also a solution and so is θ3 + 2nπ for each integer n. Hence all solutions are
±θ1 + 2nπ, n ∈ Z.
This can be visualised in another way using the unit circle. We need to find all values of θ
such that the x coordinate of the corresponding point on the circumference is a. Suppose θ1 is
such a value with corresponding point P .
0
θ1
.. P
..
..
..
..
..
..
..
..
.. a
...
..
..
..
..
..
..
.
Q
Figure 3.6.2
Then clearly θ1 + 2nπ, n ∈ Z will generate the same point on the circumference, and hence the
same x-coordinate.
Furthermore, at Q (see Figure 3.6.2) the x-coordinate is still a and the angle is −θ1 . Any
multiple of 2π added on will amount to a number of full rotations added on, taking us back to
where we started, at −θ1 + 2nπ, n ∈ Z and all solutions are given by
±θ1 + 2nπ, n ∈ Z.
30
CHAPTER 1. TRIGONOMETRY
Example 1.6.1
1
.
2
Now cos π/3 = 1/2. So all solutions are θ = ±π/3 + 2nπ, n ∈ Z.
Solve cos θ =
Example 1.6.2
√
3
.
2
√
We firstly note that cos 5π/6 = − 3/2.
Solve the equation cos θ = −
Hence all solutions are θ = 5π/6 + 2nπ, n ∈ Z.
Example 1.6.3
Find all values of θ in degrees between 0◦ and 360◦ for which cos 3θ = 1.
Now noting cos 0 = 1 all solutions are given by
3θ = 0 + 2nπ,
i.e.
θ=
2nπ
,
3
n ∈ Z.
Changing to degrees,
θ = n120◦ ,
◦
n = 0, ±1, ±2, . . .
◦
We want only values between 0 and 360 , so θ = 0◦ , 120◦ , 240◦ or 360◦ , taking n = 0, 1, 2, 3.
The Equation sin θ = a
Again we suppose |a| ≤ 1, so that the equation does have solutions. We wish to determine for
what values of θ we have sin θ = a. Suppose one solution θ1 between −π/2 and π/2 is known,
so that sin θ1 = a.
y = sin θ
1
−π/2
π
π/2
θ1
y=a
3π/2
θ3
θ3 = π − θ1
2π
θ2
θ2 = θ1 + 2π
−1
Figure 3.6.3
Clearly (see Figure 3.6.3), if θ2 = θ1 + 2π, then θ2 is a solution, since
sin θ2 = sin(θ1 + 2π) = sin θ1 = a.
Indeed, θ1 + 2nπ is a solution for any integer n, since
sin(θ1 + 2nπ) = sin θ1 = a.
1.6. THE SOLUTION OF TRIGONOMETRIC EQUATIONS
31
Notice however that θ3 = π − θ1 is also a solution. For
sin θ3
= sin(π − θ1 )
= sin π cos θ1 − cos π sin θ1
= 0 − (−1) sin θ1 = sin θ1 = a.
Hence (π − θ1 ) + 2nπ is also a solution for any n ∈ Z, and the general solution is
θ1 + 2nπ
or
(π − θ1 ) + 2nπ,
n ∈ Z.
Alternatively, using the unit circle, we wish to find all values of θ such that the y-coordinate
of the corresponding point is a (Figure 3.6.4). If θ1 is such a value with corresponding point P
then clearly so are the points θ1 + 2nπ, n ∈ Z.
Q .................a............... P
θ1
0
Figure 3.6.4
However, it is also easy to see that the point Q, on the circumference, corresponding to the
value π − θ1 , also has the same y-coordinate and so will any number of full rotations added to
this. This the complete solution is
θ1 + 2nπ, n ∈ Z or π − θ1 + 2nπ, n ∈ Z.
Example 1.6.4
1
.
2
We know that
Solve sin θ =
sin
π
1
= .
6
2
Hence the general solution is
θ=
π
+ 2nπ
6
i.e.
θ=
³
or
π
+ 2nπ
6
π−
or
π´
+ 2nπ, n ∈ Z,
6
5π
+ 2nπ, n ∈ Z.
6
Example 1.6.5
√
Solve sin θ = −
We know
3
.
2
√
sin(−π/3) = − 32.
32
CHAPTER 1. TRIGONOMETRY
Hence
θ=−
³
π
+ 2nπ
3
i.e.
θ=−
or π −
π
+ 2nπ
3
or
−
π´
+ 2nπ,
3
4π
+ 2nπ,
3
n∈Z
n ∈ Z.
Example 1.6.6
Find all values of θ between 0◦ and 90◦ for which sin 10θ = 0.
Now sin 0◦ = 0, so the general solution is given by
10θ = 0 + 2nπ
or (π − 0) + 2nπ,
n ∈ Z,
i.e.
10θ = nπ = n180◦
Thus
θ = 18n◦ .
We want solutions between 0◦ and 90◦ , hence
θ = 0◦ , 18◦ , 36◦ , 54◦ , 72◦ or 90◦ .
The Equation tan θ = a
y=a
π
−π
θ1
3π/2
θ2
θ2 = θ1 + π
Figure 3.6.5
This is the simplest case. If θ1 is a solution between −π/2 and π/2, then θ2 = θ1 + π is also a
solution, and indeed the general solution is
θ = θ1 + nπ, n ∈ Z,
since
tan(θ1 + nπ) =
tan θ1 + 0
tan θ1 + tan π
=
= tan θ1 = a.
1 − tan θ1 tan π
1−0
Example 1.6.7
Solve the equation tan θ = 1.
We note that tan π/4 = 1. So θ = π/4 + nπ, n ∈ Z.
1.6. THE SOLUTION OF TRIGONOMETRIC EQUATIONS
33
Example 1.6.8
√
Solve the equation cot 2θ = − 3.
√
This is equivalent to tan 2θ = −1/ 3.
Now we know
√
tan(−π/6) = −1/ 3.
Hence
π
+ nπ,
6
n∈Z
π
nπ
+
,
12
2
n ∈ Z.
2θ = −
so
θ=−
Now that we have these preliminaries out of the way, we can tackle some harder examples.
Example 1.6.9
Solve sin x + sin 3x + sin 5x = 0.
We have, using (??),
2 sin 3x cos 2x + sin 3x = 0,
i.e.
sin 3x(2 cos 2x + 1) = 0.
Thus
sin 3x = 0
or
so
3x = nπ
Hence
x=
nπ
3
1
cos 2x = − ,
2
or 2x = ±
or
±
2π
+ 2nπ.
3
π
+ nπ,
3
n ∈ Z.
Example 1.6.10
Solve the equation cos 6θ + cos 4θ + cos 2θ + 1 = 0, for θ between 0◦ and 360◦ .
From (26.8) and (28.8),
2 cos 5θ cos θ + 2 cos2 θ = 0,
i.e.
cos θ(cos 5θ + cos θ) = 0.
Hence
cos θ 2 cos 3θ cos 2θ = 0.
Thus
cos θ = 0 or cos 3θ = 0
So
θ=±
or cos 2θ = 0.
π
π
+ 2nπ or 3θ = ± + 2nπ
2
2
or 2θ = ±
π
+ 2nπ,
2
where n ∈ Z.
Setting n = 0, 1, 2 and 3 and remembering that we wish to have solutions only between
0 and 2π, we obtain
θ=
π 3π π 5π 7π 11π π 3π 5π
7π
,
, ,
,
,
, ,
,
or
,
2 2 6 6 6
6 4 4 4
4
which are easily put into degrees.
34
CHAPTER 1. TRIGONOMETRY
Exercise 3.6
1. Find all solutions to:
(a) sin x = 0
(d) cos x = 1
(g) sec x = 1
(j) sec x = −1
(m) csc x = 2
(p) sec x ≥ 0 √
(s) tan x = − 3
(v) 2 sin2 3x = 1
(y) 3 tan2 x = 1
(b) cos x = 0
(e) tan 3x = 0
(h) cos x = −1
(k) sin x = 1/2
(n) sec x = −2
(q) sin 2x =√1
(t) cot x = 3√
(w) sin x = − 3/2
(z) tan2 x = 1.
(c) cos 4x = 0
(f) tan x = −1
(i) csc x = 1
(l) cos x = −1/2
(o) sin x > 0
(r) cos 2x = −1
(u) 2 sin2 x = 1
(x) cos2 x = 1
2. Find all solutions x, lying in the interval [0, 2π] to the following:
(a) cos x = 0
(d) cos x = −1/2
(g) cos 6x = 0
(j) tan(x/2) = 1
(m) tan x ≤ 0
(p) 4 sin2 x = 1
(b) cos 3x = 1
(e) sin2 x = 1/2
(h) sin 2x < 0
(k) tan2 x = 1
(n) tan(x/2) ≤ 0
(q) tan2 x = 3
(c) sin 2x = 1
(f) sin x < 0
(i) 2 sin2 3x = 1
(l) sec 2x = 2
(o) cot 2x = 1
3. Suppose 0 < α < π/2 and cot α = 4/3.
(a) By drawing a suitable triangle, or otherwise, find cos α and sin α.
(b) Now find in terms of α all solutions to the following equations:
(i) sin x = 3/5
(ii) cos x = −4/5 (iii) cos x = −3/5
(iv) tan 2x = 4/3.
4. Suppose 0 < α < π/2 and sin α = 5/13.
(a) Draw an appropriate triangle and find cos α and cos 2α.
(b) Now find in terms of α all solutions to the following equations:
(i) cot x = 5/12 (ii) sin x = −5/13
(iii) sin 3x = 12/13
(iv) sec 2x = 13/12.
5. By writing the expression in the form C sin(x+α) or C cos(x+α) find the general solution
of the following:
√
√
(a) cos x − sin x = 1/ 2 (b) cos x − sin x = 1 (c) cos x − 3 sin x = 1
√
(d) 3 cos x − sin x = 1.
6. (a) Solve the equation sin x = −1/2.
(b) Sketch the graph of y = 1/2 + sin x, showing points of intersection with the axes.
(c) From your graph or otherwise solve the inequality 0 < 1/2 + sin x.
7. (a) Solve tan x = 1 (for all x).
(b) Sketch the graph of y = −1 + tan x
(c) Solve the inequality tan x ≥ 1.
8. Find the general solutions of the following:
1.6. THE SOLUTION OF TRIGONOMETRIC EQUATIONS
35
(b) sin3 x = sin x
(d) sin2 x = cos2 x
(f) 3 cos2 x + sin2 x + 3 cos x = 3
(h) tan4 x − 4 tan2 x + 3 = 0
(j) sin2 x = 1 − cos x
(l) 2 sin3 x − 3 sin2 x + sin x = 0
(n) sin 3x = sin x
(p) cos2 x = − sin2 x − 2 sin x
(r) x2 (1/2 − cos x) = cos x − 1/2
(t) sin x = 1/2 and tan x < 0
(a) cos3 x = cos x
(c) sin 2x = cos x
(e) sin(x + π/6) = 0
(g) sin x = sin 3x
(i) 4 sin2 x cos x = cos x
(k) cos 3x = 3 cos x
(m) 4 cos 2x = 4 sin x + 5
(o) 2 sin2 x − 5 sin x = −2
(q) 2 cos3 x − 2 cos2 x − cos x = −1
(s) 2 cos 2x > 1
√
(u) cos x = 1/2 and sin x = − 3/2.
9. Find all solutions x lying in the interval [−π, π] to:
(a) 2x sin x = x
(c) sin x + cos x = 0
(e) sin x = cos 2x
(g) sin 2x sin x = cos x
(i) cos 3x = −2 cos x
(k) cos 2x − √
3 cos x + 2 = 0 √
(m) tan x + 3 cot x = 1 + 3
(o) |2 tan x − 1| < 1
(b) 2 sin 2x = 1
(d) sin x + cos x = 1
(f) cos x = 2 sin2 x − 1
(h) tan2 2x = 3
(j) 2 sin 2x sin x + 3 cos 2x = 0
(l) 2 sin2 x + 3 cos x ≤ 3
(n) |2 tan x − 1| = 1
(p) 4 sin3 x − 4 sin2 x − 3 sin x = −3
10. (a) Show that if x ≥ 0, then −x ≤ x sin x ≤ x.
(b) Find where the curve y = x sin x meets the straight lines y = 0, y = −x and y = x.
(c) Sketch the graphs of y = x sin x, y = 0, y = −x and y = x on the same diagram.
11. The triangle ABC is defined by: B = 90◦ , a = 12, c = 5. Find:
(a) b
(b) sin A
(c) cos C
(d) cot A.
Find in terms of A all solutions to the equation tan x = −12/5.
12. The triangle ABC is defined by: B = 90◦ , sin A = 3/4, b = 4. Find:
(a) a
(b) c (c) csc C
(d) cos A.
Now find in terms of A all solutions to the equations
(e) sin x = −3/4
(f) sin 2x = 3/4.
13. The triangle ABC is defined by : B = 90◦ , cos A = 3/4, b = 4. Find:
(a) a (b) c (c) sin(π − A)
(d) cot A
(e) cos(π − A)
(f) tan A.
Find in terms of A all solutions to:
(g) cos x = −3/4
(h) sin 2x = sin 7A.
14. A certain student believes that sin 2x = 2 sin x for any x. Give an example to prove him
wrong, and find all values of x for which he is right. Give a sketch of the curves y = sin 2x
and y = 2 sin x on the same diagram.
15. (a) Solve the equation sin(1/x) = 0.
(b) Sketch the curve y = sin(1/x).
16. By considering sin 3θ = 1/2, and expressing the left-hand side in terms of sin θ, show that
the equation 8x3 − 6x + 1 = 0 has roots sin π/18, sin 5π/18 and − sin 7π/18.
17. Prove that sin 5A − sin A + sin 2A = sin 2A(2 cos 3A + 1). Hence solve the equation
sin 5θ + sin 2θ = sin θ for values of θ in the interval 0◦ < θ < 180◦ .
18. (a) Factorise x3 − 4x2 + 6x − 4
36
CHAPTER 1. TRIGONOMETRY
(b) Show that if
4 − tan θ = 5 sin θ cos θ
3
2
then tan θ is a root of x − 4x + 6x − 4 = 0.
(c) Hence find the values of tan θ for which
4 − tan θ = 5 sin θ cos θ.