3.5: Trigonometric Functions
Reference Evans 6.1
Consider a right-angled triangle with angle θ and side lengths x, y and h as
shown:
h
θ
y
x
The trigonometric functions sine, cosine and tangent of θ are defined as:
sin θ =
opposite
y
= ,
hypotenuse
h
tan θ =
cos θ =
adjacent
x
=
hypotenuse
h
opposite
y
sin θ
=
=
adjacent
x
cos θ
71
Reciprocal Trigonometric Functions
The reciprocal trigonometric functions secant, cosecant and cotangent are defined as:
sec θ =
1
cos θ
cosec θ =
1
sin θ
cot θ =
1
cos θ
or
tan θ
sin θ
72
Example
13
θ
5
12
For the above right-angled triangle, find:
1. sin θ
2. cos θ
3. tan θ
4. sec θ
5. cosec θ
6. cot θ
You can now attempt Sheet 3 Q19
73
The Unit Circle
We can form right-angled triangles in a unit circle (circle of radius 1).
−
−
→
If θ is the anticlockwise angle between the positive x-axis and the ray OP then for
all θ ∈ R:
y
x
y
sin θ = = y , cos θ = = x and tan θ =
1
1
x
y
y
1
1
y
O
y
1
P(x,y)
1
θ
x 1
x
x
θ
O
O
x
θ
1
1
P(x,y)
1
x
y
y
1
x
negative θ
(clockwise)
P(x,y)
74
Signs of Trigonometric Functions
The signs of the trigonometric functions for different values of θ can be determined
by noting the signs of x and y in the unit circle:
y
1
2nd quadrant
x negative
y positive
1st quadrant
x positive
y positive
y
1
x 1
x
3rd quadrant
x negative
y negative
y
x
4th quadrant
x positive
y negative
75
‘CAST’ Diagram
sin θ =
Since in the unit circle sin θ = y, cos θ = x and tan θ = cos
θ
following ‘CAST’ diagram as a shorthand for remembering signs
functions in different parts of the circle:
y
x , we obtain the
of trigonometric
y
1
ALL sin, cos, tan positive
only SIN positive
S
A
1
T
only TAN positive
x
C
only COS positive
76
Degrees and Radians
In the unit circle, if we measure the length of the arc from A to B in an anticlockwise direction, we have another way of measuring angles.
.
B
1
θ
.A
1
This length is called the radian measure of the angle θ and we can signify it by
the symbol θ c rather than θ o, but usually we don’t show units for radians.
77
Radians on the Unit Circle
Since the circumference of the unit circle is 2πr = 2π · 1 = 2π, a complete
angle around the whole circle is 2π radians. Hence if we divide the circle into four
quarters or quadrants we have radian measures as shown:
π
2
0, 2π
π
3π
2
78
Converting Between Degrees and Radians
π .
To convert from degrees to radians multiply by 180
To convert from radians to degrees multiply by 180
π .
Examples
Convert 360◦, 1◦, 90◦, 135◦ to radians.
7π radians to degrees.
Convert π, 1, π
,
3 6
79
Two Important Triangles
The following triangles help us work out the exact values of sin, cos and tan of
certain special angles.
π π
6 6
2
π
2
3
π3
2
π
1
1
3
π
4
4
1
1
80
Exact Values using Triangles
Using the two important triangles on the previous slide and your knowledge of the
unit circle complete the table with exact values:
θ
sin θ
cos θ
tan θ
0
π
6
π
4
π
3
π
2
π
3π
2
***Important to know these ratios, either the triangles or the table
81
Find the exact values of the following trig ratios:
Examples:
�
1. cos 5π
6
�
2. sin 4π
3
�
�
�
3. tan 3π
4
4. sin 11π
6
5. tan −π
4
�
• quadrant?
�
�
�
Ask yourself:
• angle from x-axis?
�
• ± CAST?
• value of ratio?
�
6. cos −2π
3
�
82
Finding Angles
Here we need to work backwards considering both the value of the ratio and its
sign, as this determines the quadrants of the resultant angles.
Find all values of θ between 0 and 2π satisfying:
1. sin(θ) = 1
2
2. tan(θ) = 1
3. cos(θ) = − √1
2
83
Solving Trigonometric Equations
Solve the following trigonometric equations for x ∈ [0, 2π).
1. cos x + 1
2 =0
2. 2 sin2 x + sin x − 1 = 0
84
3. cos x + sin x = 0
4. 2 tan2 x − 6 = 0
You can now attempt Sheet 3 Q20-21
85
Sine and Cosine Graphs
Reference Evans 6.2
Plotting the values of f (x) = sin x and g(x) = cos x for x ∈ R gives the
following graphs
f �x�
1
�2 Π
�Π
Π
2Π
3Π
4Π
Π
2Π
3Π
4Π
x
�1
g�x�
1
�2 Π
�Π
x
�1
86
Properties of Sine and Cosine Graphs
• The sine and cosine functions repeat themselves after an interval (or period) of
2π units. That is:
. . . sin(x − 2π) = sin x = sin(x + 2π) = sin(x + 4π) . . .,
. . . cos(x − 4π) = cos(x − 2π) = cos x = cos(x + 2π) . . .,
and
for all x ∈ R
These functions are therefore said to be periodic or cyclic, with period 2π.
• The maximum and minimum values of sin x and cos x are 1 and −1 respectively. The graphs of f (x) = sin x and g(x) = cos x are therefore said to have
an amplitude of 1.
87
Graph of Tangent Function
Reference Evans 6.6
sin x is defined for x ∈ R such that cos x �= 0.
The function h(x) = tan x = cos
x
Plotting h(x) = tan x over its domain x ∈ R\{(2k + 1) π2 | k ∈ Z} gives the
following graph
88
Properties of Tan Graph
• The tangent function h(x) = tan x is undefined at x = {(2k + 1) π2 | k ∈ Z}
(this is where cos x = 0).
The graph has vertical asymptotes at these x-values, which are usually indicated
by dotted or dashed vertical lines.
• The tangent function repeats itself after an interval of π units. That is:
π
tan θ = tan(θ + π), for all θ ∈ R\{(2k + 1) | k ∈ Z}
2
This function therefore has period π.
89
Dilations of Trig Graphs
Reference Evans 6.2, 6.3 & 6.6
Sketch the following graphs - the standard sin or cos graph is shown - state the
period and amplitude of the transformed graph in each case.
�
�
2π
Note: a dilation in the x direction (e.g. y = cos bx) changes the period to b
and one in the y direction (e.g. y = a sin x) changes the amplitude (to a)
1.
3.
y = sin 2x
y = −5
2 sin x
2.
4.
y = 3 cos
� x�
x
y = cos 2
90
Translations of Trigonometric Graphs
Sketch the following graphs on the axes below –adding or subtracting a value to
the trig function will translate the graph up or down, while adding or subtracting a
value to the x term will translate the graph to the left or right respectively.
1.
y = tan(x − π4 )
91
2.
y = sin(x) + 3
3.
y = cos(x + π)
92
Multiple Transformations
Sketch the following sequence of graphs on the same set of axes.
�
�
π
y1 = sin x
y2 = −2 sin x
y3 = −2 sin 3x
y4 = −2 sin 3x − 2
You can now attempt Sheet 3 Q22-23
93
A:
Q1
Q2
Q3
Q4
has
has
has
has
angles
angles
angles
angles
from 0 → π2
from π2 → π
from π → 3π
2
3π
from 2 → 2π
Q2
cos π6 =
3
,
2
√
tan π6 =
√1
3
3

6
1

3
2
1

4
1
2
tan 3π
=
2
(sin 3π
)
−1
2
= undefined
3π =
0
(cos 2 )
1
It may help to look at the unit circle above to see these points (and remember that
− π2 and 3π
are just different names for the same point on the circle.
2
Therefore we can find values like:
cos π = −1 or sin − π2 = −1 or
• cos θ is the x coordinate
• sin θ is the y coordinate
sin θ
• tan θ =
cos θ

4
Q4
Q1
Semester 1, 2011
4. For angles that give points on the x or y axes we use the basic definitions:
e.g. sin π6 = 12 ,
(remember SOH-CAH-TOA)
3. Use the special triangles to find
the ratio required
(most students use CAST to remember the signs)
2. Decide if the ratio you need to find (usually sin, cos or tan)
is positive or negative in the quadrant you found in step 1
This is just the first revolution of the unit circle
We can of course find bigger angles by moving Q3
around the circle more than once or negative
angles by going in the opposite direction
•
•
•
•
1. Identify the quadrant that the angle is in:
Finding trig ratios in the Unit Circle
REVISION - TRIGONOMETRY
MAST10012 Introduction to Mathematics
DEPARTMENT OF MATHEMATICS AND STATISTICS
5π
4
is in Q . . . . . .
5π
4
= π + π4 )
(+ or –)
(as
11π
6
is in Q . . . . . .
11π
6
= 2π − π6 )
(+ or –)
(as
8π
3
is in Q . . . . . .
8π
3
= 3π − π3
= π − π3 )
(+ or –)
(as
(x or y)
=⇒ cos 5π =
2
iii. As cos is the x coordinate we know:
ii. It has coordinates (. . . , . . . )
i. 5π is on the . . . . . . axis
(d) cos 5π
=
=⇒ tan 8π
3
iii. From triangle 1 we know that tan π3 = . . . . . .
ii. In Q . . . . . . tan is . . . . . .
i.
(c) tan 8π
3
=
=⇒ cos 11π
6
iii. From triangle 1 we know that cos π6 = . . . . . .
ii. In Q . . . . . . cos is . . . . . .
i.
(b) cos 11π
6
=
=⇒ sin 5π
4
iii. From triangle 2 we know that sin π4 = . . . . . .
ii. In Q . . . . . . sin is . . . . . .
i.
(a) sin 5π
4






1. Find the values of the following trig ratios - the steps mentioned on the previous
page have been spelt out for the first few questions (then you are on your own to do
the rest by following the same process):
PRACTICE EXERCISE A
=⇒
√1
3
tan π6 =
cot
sec
π
6
π
6
cosec
=
=
π
6
√
2
1
3
√2
3
=
(or
=2
and
√
2 3
)
3
(e) cot(6π)
(d) sec ( 3π
)
4
(c) cot( 7π
)
6
(b) sec (− 7π
)
2
(a) cosec ( 5π
)
3
3
3. Find the following trig ratios (you may find your answers to Q2 useful in some cases):
=⇒
=⇒
3
2
1
2
√
cos π6 =
sin π6 =
Examples: We found on page 1 in section 3 that:
There are other trig ratios we can use but they are based on the standard ones. So
to find cosec, sec or cot we calculate sin, cos or tan respectively and then just ”turn
them upside down” because of the definitions:
1
1
cos θ
1
sec θ =
cot θ =
=
cosec θ =
sin θ
cos θ
tan θ
sin θ
(g) tan(− 17π
)
6
(f) sin(− 7π
)
4
)
(e) cos(− 7π
2
(d) sin(2π)
(c) tan( 7π
)
6
)
(b) cos( 5π
4
)
(a) sin( 5π
3
2. Now try the same process with these questions:
sin x + 1 = 0 =⇒ sin x = −1
√
) − 1 = 0 =⇒ cos(x +
2 cos(x + 5π
6
5π
)
6
=
√1
2




you may need to add or subtract 2π if you need bigger angles or negative angles for
the domain given in the question - that is, angles outside the standard 0 → 2π which
is only one revolution around the unit circle
5. Check the domain of the question
have you found all the solutions?
Note that θ is the basic angle found in step 2
So the unit circle on the right
might be useful here
4. Use knowledge of the unit circle to find
the basic angle in the right quadrants
3. Decide which quadrants the answers must be in - look at the sign of the trig ratio
e.g. if the sin ratio has a negative answer then angles must be in Q3 and Q4
if the cos ratio has a positive answer then angles must be in Q1 and Q4
so we are really looking at our CAST diagram ”backwards”
2. Find the basic angle that satisfies this ratio – this may involve looking at the angles
in the two special triangles or looking at the coordinates of points on the unit circle
where they intersect with the two axes.
e.g.
1. Rearrange the equation to make the trig ratio (usually sin, cos or tan) the subject
Solving Trig equations
...
6
x = π6 ,
=⇒
π
6
π
6
see special triangle 1
(mentalcheck : these values are in the domain 0 → 2π)
cos is + in Q. . . (θ) and Q. . . (2π − θ)
basic angle θ =
4
Note: if we added or subtracted 2π to either of our answers we would get values
outside the domain, so we don’t need to do anything else here
2π −
x = π6 ,
x ∈ (0, 2π)
=⇒
1. 2 cos x =
3
...
cos x =
...
√
Solve the following trig equations over the given domains (as in Exercise A you are guided
through the first few questions then you should use the same process to complete the rest):
PRACTICE EXERCISE B
B:
√
3=0
x = . . . . . . − 2π,
and
π
6
π
4
to RHS of equation to find x
use LCD to add fractions
add
π
6
to RHS of equation to find x
domain needs negative angles too
use LCD to add fractions
subtract
basic angle θ = . . . see special triangle 1
sin is – in Q3 (π + θ) and Q. . . (. . . . . . )
2 cos x + 1 = 0
3π
)
4
=0
5
x ∈ [0, 2π]
x ∈ (0, 2π)
7. 2 cos2 x + 3 cos x + 1 = 0
6. cos(x −
......
x ∈ (−2π, 2π)
......,
x ∈ (0, 4π]
......,
5. sin(x + π3 ) = 1
√
x = ......,
So the final solutions are:
Note: we need all four answers – the two negative and the original two positive ones
. . . . . . − 2π
......
x = ......,
...... −
......
=⇒
= ......,
π
6
......
x = . . . . . . − π6 ,
x+
=⇒
= ......,
π
6
x ∈ (−2π, 2π)
=⇒
x+
=⇒
sin(x + π6 ) = . . .
3. 8 sin(x + π6 ) + 4 = 0
4.
π
4
basic angle θ = . . . see special triangle 1
tan is . . . (+/–) in Q. . . (θ) and Q3 (. . . . . . )
(mental check: these values are inside the domain of 0 → 2π, and if we added or
subtracted 2π to either of them we would get answers outside the domain, so we
don’t need to do anything else here)
......
x = ......,
=⇒
...... +
......
x = . . . . . . + π4 ,
= ......,
π
4
x−
=⇒
......
x ∈ (0, 2π)
=⇒
= ......,
π
4
x−
=⇒
tan(x − π4 ) = . . .
2. tan(x − π4 ) −
Friday, 22 April 2011
3:05 PM
Trigonometry
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