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Foundations - Trigonometry
1. Radians
2. Trigonometric Ratios
3. More Ratios
4. Graphs of Trigonometric Functions
5. Trigonometric Equations and Formulae
Last updated September 23, 2015
1
Radians
The radian is the standard unit of angular measure. An angle’s measurement in radians is
equal to the length of a corresponding arc of a circle.
One radian is equal to about 53.7 degrees. The following fact provides an easy way to
remember the conversion between radians and degrees:
360◦ = 2π radians
Thus
1◦ =
π
radians
180
and
1 radian =
180 ◦
.
π
WARNING! Always check on your calculator whether you are working in degrees or
radians.
Exercise 1.1 Convert the following angle measurements to radians.
1) 30◦
2) 120◦
3) 150◦
4) 300◦
5) 270◦
6) 60◦
Exercise 1.2 Convert the following angle measurements to degress.
1)
π
radians
6
4) 7π radians
3π
radians
2
18π
5)
radians
4
2)
1
4π
radians
3
17π
6)
radians
5
3)
2
Trigonometric Ratios
The three most common trigonometric functions - sine, cosine and tangent (sin, cos, tan)
- are defined in terms of the ratios of the sides of a right-angled triangle. Let θ be an acute
angle, i.e. between 0 and π2 radians.
The definitions are as follows.
sin θ =
opposite
AB
=
hypotenuse
BC
cos θ =
adjacent
AC
=
hypotenuse
BC
Notice that
tan θ =
and
sin
π
− θ = cos θ,
tan θ =
opposite
AB
=
adjacent
AC
sin θ
cos θ
cos
π
− θ = sin θ.
2
2
The values of sin, cos and tan of the angles appearing in the following triangles are worth
remembering.
Exercise 2.1 Complete the following table of values of sin, cos and tan.
θ
sin
cos
tan
0 π/6 π/4 π/3
2
π/2
We can now extend the definitions for sin, cos and tan to include all angles. Let θ be an
angle (greater than π/2 radians) obtained by rotating a line starting at OA anticlockwise to
position OP . In this case the angle θ is taken as positive. If the line is rotated anticlockwise,
the angle is taken to be negative.
Let OP = r (always positive). In whatever quadrant OP lies, the trigonometric ratios of
the angle θ are defined in terms of the (x, y)-coordinates of the point P :
cos θ =
x
r
sin θ =
y
r
y
tan θ = .
r
In each case the numerical value (i.e. ignoring the sign) is that of the trig-ratio for the
corresponding acute angle between OP and the x − axis, and the sign is obtained from
the signs of x and y.
In each of quadrants 2, 3, 4 only one of the three ratios sin θ, cos θ, tan θ is positive: sin is
positive in quadrant 2, tan is positive in quadrant 3 and cos is positive in quadrant 4. It is
easy to memorise these:
Note that all three ratios are positive in the first quadrant.
.
Example 2.1 Find the value of tan 2π
3
3
radians lies in the second quadrant. The correMeasured anticlockwise, we find that 2π
3
2π
sponding acute angle is therefore π − 3 = π3 . Also, tan is negative in the second quadrant.
Hence
√
2π
π
tan
= − tan = − 3.
3
3
Example 2.2 Express sin 323◦ in terms of the sine of an acute angle.
The angle 323◦ lies in quadrant 4 and the corresponding acute angle is 360◦ − 323◦ = 37◦ .
Since sin is negative in the fourth quadrant, we have
sin 323◦ = − sin 37◦ .
Exercise 2.2 Find the exact values of the following trig ratios.
4π
3
5π
4) sin
3
5π
6
3π
5) tan
4
1) cos
3
7π
6
7π
6) cos
4
2) tan
3) sin
More Ratios
Three more trigonometric ratios that you may not have encountered are secant, cosecant
and cotangent.
sec θ =
1
cos θ
cosec θ =
1
sin θ
cot θ =
1
tan θ
The following identities hold:
sec
π
cosec
π
2
− θ = cot θ
π
− θ = tan θ.
− θ = cosec θ
tan
− θ = sec θ
cot
2
π
2
2
Example 3.1 Show that
cot θ sec θ = cosec θ.
4
We have
1
1
×
tan θ cos θ
1
cos θ
×
=
sin θ
cos θ
1
=
sin θ
= cosec θ
cot θ sec θ =
as required.
Exercise 3.1 1) Find the exact values of the following ratios.
i) sec
π
6
ii) cot
π
3
iii) cosec
π
4
2) Prove the following identities.
i) sin θ cot θ = cos θ
4
ii) tan φ cosec φ = sec φ
iii) cot θ
cos(π/2 − θ)
sin(π/2 − θ)
=1
Graphs of Trigonometric Functions
It is useful to know the graphs of the functions sin, cos and tan. In particular, they can help
you identify solutions of trigonometric equations.
5
Some facts to note:
1. The graphs of sin and cos have period 2π, i.e. the graphs repeat every 2π-units.
Algebraically, we can express this as sin(x + 2π) = sin(x) and cos(x + 2π) = cos(x).
2. The graph of tan has period π, so tan(x + π) = tan(x).
3. tan x is undefined for all values x =
π
2
+ kπ (k ∈ Z), i.e. ± π2 , ± 3π
, ± 5π
, . . ..
2
2
4. −1 ≤ sin x ≤ 1 and −1 ≤ cos x ≤ 1, while tan x is unbounded.
5. sin x = 0 for all x = kπ and cos x = 0 for all x =
5
π
2
+ kπ (k ∈ Z).
Trigonometric Equations and Formulae
The list of trigonometric formulae is endless. However, there is a core of fundamental identities that are very useful to commit to memory. They are particularly helpful when solving
trigonometric equations and calculating integrals involving trig functions.
(1) sin2 x + cos2 x = 1
(2) sin(x + y) = sin x cos y + sin y cos x
(3) sin(x − y) = sin x cos y − sin y cos x
(4) cos(x + y) = cos x cos y − sin x sin y
(5) cos(x − y) = cos x cos y + sin x sin y
Identities (2)−(5) are called the additional formulae. From these, and identity (1), we derive
more results.
6
(6) sin(2x) = 2 sin x cos x
(7) cos(2x) = cos2 x − sin2 x = 2 cos2 x − 1 = 1 − 2 sin2 x
1
(8) cos2 x = (1 + cos 2x)
2
1
2
(9) sin x = (1 − cos 2x)
2
2
(10) 1 + tan x = sec2 x
The product formulae allow us to write products of trig functions in terms of sums of trig
functions.
1
(sin(x + y) − sin(x − y))
2
1
(12) cos x sin y = (sin(x + y) − sin(x − y))
2
1
(13) cos x cos y = (cos(x + y) + sin(x − y))
2
1
(14) sin x sin y = (cos(x − y) − cos(x + y))
2
(11) sin x cos y =
The ability to solve equations involving trigonometric functions is a useful skill to have. The
above identities, together with knowledge of common equations (e.g. quadratic), will help
you.
Example 5.1 Solve the following equations for the stated range of x.
1) 2 sin x + 1 = 0
0 ≤ x ≤ 2π
2) 3 tan2 x = 1
0 ≤ x ≤ 2π
2
3) 2 cos x − 5 cos x − 3 = 0
x∈R
4) sin x − cos 2x = 0
−π ≤x≤π
1) If 2 sin x + 1 = 0 then sin x = − 12 . For sin x to be negative, it must be the case that x is
either a third or fourth quadrant angle. In order to find these angles, we first compute the
corresponding acute angle, call it φ. Now φ must satisfy
1
sin φ = ,
2
7
so we know that φ = π6 . Therefore
(Quadrant 3)
(Quadrant 4)
π
7π
=
6
6
π
11π
x = 2π − φ = 2π − =
6
6
x=π+φ=π+
2) Given 3 tan2 x = 1 we have tan x = ± √13 . If tan x = √13 then x is either a first or third
quadrant angle. The corresponding acute angle is π6 (since tan π6 = √13 ) and so
x=
π
6
or
x=π+
π
7π
=
.
6
6
For tan x = − √13 either x is a second or a fourth quadrant angle. Thus
x=π−
5π
π
=
6
6
or
x = 2π −
π
11π
=
.
6
6
3) The key to solving this equation is to notice that it is a quadratic equation in terms of
cos x. We proceed by factorising:
(2 cos x + 1)(cos x − 3) = 0.
Therefore cos x = − 21 or cos x = 3. The maximum value of cos x is +1, so there is no such
x that satisfies cos x = 3. For cos x = − 12 the corresponding acute angle is π3 and so
x=π−
π
2π
=
3
3
or
x=π+
π
4π
=
.
3
3
We have to find all x ∈ R that solves the original equation. These are obtained by using the
fact that cos is periodic with period 2π. Thus the complete set of solutions is given by
x=
2π
+ 2kπ
3
or
x=
4π
+ 2kπ
3
for k ∈ Z.
4) We’ll use the identity cos(2x) = 1 − 2 sin2 x to transform the equation into a quadratic
for sin x.
sin x − cos(2x) = 0
⇔ sin x − (1 − 2 sin2 x) = 0
⇔ 2 sin2 x + sin x − 1 = 0
⇔ (2 sin x − 1)(sin x + 1) = 0
1
⇔ sin x =
or
sin x = −1.
2
8
Looking at the graph of sin x for −π ≤ x ≤ π we see that there are only two possible
solutions: x = π6 and x = − π2 .
Exercise 5.1 1) Solve the following equations, stating (i) all possible solutions, and (ii)
solutions in the range 0 ≤ x ≤ 2π.
1
2
2
d) tan x = −3
x
x
f ) sin2
− cos2
=1
2
2
h) 3 − 2 cos2 x = 3 sin x
a) 2 sin(2x) + 1 = 0
b) cos(3x) =
c) sin(2x) = 1
√
e) 3 tan2 x + tan x = 0
√
g) 2 cos2 x + 3 cos x = 0
i) sin2 x − sin x = 2
j) cos2 x + cos x = sin2 x
k) ln(2 − sin2 x) = 0
l) sin x = tan x
2
m) 2 cos x + sin(2x) = 0
n) sin(4x) − 2 cos(2x) = 0
o) tan(2x) − cot(x) = 0
p) sin(2x) cos x − cos(2x) sin x = 0
2) Simplify the following expressions as much as possible.
a) cosec2 x − cot2 x
b) sec2 x − tan2 x
c)
d)
1
1
−
2
1 + tan y 1 + cot2 y
(cos y + sin y)2 − 1
1 − 2 sin2 y
3) Prove the following identities.
a) tan2 x − sin2 x = sin2 x tan2 x
b) cot y + tan y = sec y cosec y
cosec x
d)
= tan x sec x
cot2 x
cot y − tan y
f)
= cos(2y)
cot y + tan y
c) cos(3x) = 4 cos3 x − 3 cos x
e) tan(2y) + cot y =
cot y
cos(2y)
4) Use the addition formulae to find the exact values of (i) cos
9
π
12
and (ii) sin
5π
12
.
6
Answers
Exercise 1.1
1)
π
6
2)
2π
3
3)
5π
6
4)
5π
3
5)
3π
2
6)
π
3
Exercise 1.2
1) 30◦
2) 270◦
3) 240◦
4) 1260◦
5) 810◦
6) 612◦
Exercise 2.1
θ
sin
0 π/6 π/4 π/3
√
1
3
√1
0
2
2
2
cos
1
tan
0
√
3
2
√1
3
√1
2
1
π/2
1
1
√2
3
0
not defined
Exercise 2.2
1
1) −
2
1
2) − √
3
1
3) −
2
4) −
√
3
2
1
6) √
2
5) − 1
Exercise 3.1
1)
2
i) √
3
1
ii) √
3
iii)
√
2
Exercise 5.1
1) Note that only answers to part (i) are given, where k ∈ Z.
11π
7π
+ kπ,
+ kπ
12
12
π
c) x = + kπ
4
5π
+ kπ, kπ
e) x =
6
π
5π
7π
g) x = + kπ,
+ 2kπ,
+ 2kπ
2
6
6
3π
i) x =
+ 2kπ
2
3π
π
+ 2kπ
k) x = + 2kπ,
2
2
π
3π
3π
7π
m) x = + 2kπ,
+ 2kπ,
+ 2kπ,
+ 2kπ
2
4
2
4
π 2kπ π 2kπ
,
+
o) x = +
6
3
2
3
a) x =
10
π 2kπ 5π 2kπ
+
,
+
9
3
9
3
d) No solutions
b) x =
f ) x = π + 2kπ
π
π
5π
+ 2kπ,
+ 2kπ,
+ 2kπ
2
6
6
π
5π
j) x = + 2kπ,
+ 2kπ, π + 2kπ
3
3
h) x =
l) x = kπ
n) x =
π
3π
+ kπ,
+ kπ
4
4
p) x = kπ
2)
a) 1
b) 1
c) cos(2y)
d) tan(2y)
4)
√
1+ 3
(i) cos
= √
12
2 2
π
(ii) sin
11
5π
12
√
1+ 3
= √
2 2