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STUDY KNOWHOW PROGRAM
STUDY AND LEARNING CENTRE
Trigonometry
B Dec 2016
Contents
Pythagoras’ Theorem ............................................................................................................................................................... 3
Right Triangle Trigonometry................................................................................................................................................ 5
Sine Rule........................................................................................................................................................................................ 9
Cosine Rule ................................................................................................................................................................................ 13
Angular measurement - Radians...................................................................................................................................... 15
Circular Functions .................................................................................................................................................................. 17
Trigonometric Equations .................................................................................................................................................... 20
Graphs of Sine and Cosine Functions ............................................................................................................................. 21
B Dec 2016
2
PYTHAGORAS’ THEOREM
Right-angled triangles
A right-angled triangle has an angle of 90⁰. In a right-angled triangle the side opposite the right angle
is called the hypotenuse. It is also the longest side.
hypotenuse
Right angle
(90o)
Pythagoras’ Theorem
Pythagoras’ theorem states:
In a right angled triangle the square of the length of the hypotenuse (h) is equal to the sum of the
squares of the other two sides.
h
b
h2 = a2 + b2
a
Pythagoras’ Theorem can be used to find a side length of a right angled triangle when you know the
other two side lengths.
Examples:
Finding the hypotenuse
Find the length of the hypotenuse (h) in the following triangle.
Use Pythagoras theorem h2 = a2 + b2
with a = 6cm. and b = 8 cm:
h2 = a2 + b2
h2 = 62 + 82
h2 = 100
h = 10 cm
h
6
[Note:
The correct units can be put in at the end]
8
Finding a shorter side
Find the value of x in the triangle below.
x
2.7
4.2
B Dec 2016
h2 = a2 + b2
4.22 = 2.72 + x2
4.22 - 2.72 = x2
10.35 = x2
x = 3.22
3
Pythagorean triples
In some right-angled triangles all three sides have integer values. These three values form a
Pythagorean triple.
Some examples are triangles with sides: (3,4,5), (5,12,13), (7,24 25) and (8,15,17) Check!
Multiples of these, such as (6,8,10) and (9,12,15) are also Pythagorean triples.
Exercises
Find the missing sides in the following.
(1)
(2)
a
9mm
h
6.5
20.2
12mm
(3)
(4)
7cm
4.8 cm
a
c
Answers: (1) 15 mm.
B Dec 2016
14 cm
6.2 cm
(2) 19.13
(3) 7.84 cm.
(4) 12.12 cm.
4
RIGHT TRIANGLE TRIGONOMETRY
Trigonometry is a branch of mathematics involving the study of triangles, and has applications in fields
such as engineering, surveying, navigation, optics, and electronics. The ability to use and manipulate
trigonometric functions is necessary in other branches of mathematics, including calculus, vectors and
complex numbers.
Right-angled Triangles
In a right-angled triangle the three sides are given special names.
The side opposite the right angle is called the hypotenuse (h) – this is always the longest side of the
triangle.
The other two sides are named in relation to another known angle (or an unknown angle under
consideration).
If this angle is known or
under consideration
this side is called
the opposite side
because it is opposite
the angle
h

This side is called the adjacent side
because it is adjacent to or near the angle
Trigonometric Ratios
In a right-angled triangle the following ratios are defined for a given angle θ
sine θ =
θ
cosine θ =
θ
tangent θ =
θ
These ratios are abbreviated to sin θ, cos θ, and tan θ respectively.
A useful memory aid is SOH CAH TOA: Sin=Opp/Hyp Cos=Adj/Hyp Tan=Opp/Adj
These ratios can be used to find unknown sides and angles in right-angled triangles.
Examples
Evaluating ratios
In the right-angled triangle below evaluate sin θ, cos θ, and tan θ.
sin θ =
=
= 0.8
5
4

cosθ =
=
= 0.6
tan θ =
=
= 1.33
3
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5
Finding angles.
Find the value of the angle  in the triangle below
Method
1.
2.
3.
4.
19.7
13.4

Determine which ratio to use.
Write the relevant equation.
Substitute values from given information.
Solve the equation for the unknown
In this triangle we know two sides and need to find the angle .
The known sides are the opposite side and the hypotenuse.
The ratio that relates the opposite side and the hypotenuse is the sine ratio.
sin θ =
sin θ =
sin θ = 0 6082
θ = sin-1(0.6082)
[Using sin-1 function on the calculator]
θ = 42.9⁰
Finding side lengths
Find the value of the indicated unknown side length in each of the following right-angled triangles.
(a)
bc b
27o
42 cm
In this problem we know an angle, and the adjacent
side.
The side to be determined is the opposite side.
The ratio that relates these two sides is the tangent
ratio.
tan θ =
tan 27⁰ =
b = 42 × tan27⁰
b = 21.4cm
In this problem we know an angle, and
the adjacent side. The unknown side
is the hypotenuse.
The ratio that relates these two sides
is the cosine ratio.
x
(b)
35o
7
cos θ =
cos 35⁰ =
x = 7 × cos 35⁰
x = 8.55
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See Exercise 2
6
Special angles and exact values
There are some special angles for which the trigonometric functions have exact values rather than
decimal approximations.
Applying the rules for sine, cosine and tangent to the triangles below, exact values for the sine, cosine
and tangent of the angles 30⁰, 45⁰ and 60⁰ can be found.
Exercises
Exercise 1
Using the right-angled triangle below find:
(a) sin θ,
(b) tan θ, (c) cos α,
(d) tan α (Hint: Use Pythagoras theorem to find the hypotenuse)
Exercise 2
Find the value of the indicated unknown (side length or angle) in each of the following diagrams.
(a)
(b)
a
62o
47o
14cm
4.71 mm
a
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7
(c)
(d)
4.8 cm
x

6.2 cm
z
6.5
20.2

(e)
(f)
500
a
b
34
27o
42
(g) In a right-angled triangle sin = 0.55 and the hypotenuse is 21 mm. Find the length of each of the
other two sides.
Answers
Exercise 1
(a) 12/13 = 0.9231 (b) 12/5 = 2.4
Exercise 2
(a) = 6.6 cm
(f) b = 47.1
B Dec 2016
(c) 12/13 = 0.9231
(b) a = 3.4 mm
(c) z = 7.8,  = 37 7
(g) 11.6mm. and 17.5mm.
(d) 5/12 = 0.4167
(d) θ =18.8, x = 19.1
8
(e) a = 44.4
SINE RULE
The sine rule can be used to find angles and sides in any triangle (not just a right-angled triangle)
when given:
(i) One side and any two angles
OR (ii) Two sides and an angle opposite one of the given sides.
In the triangle ABC below:
angles A,B,C, are the angles at the vertices A,B,C respectively
a,b,c are the side lengths opposite the angles A,B,C respectively.
B
a
c
C
b
A
The sine rule states:
a
b
c
=
= or
sin A sin B sin C
sin A sin B sin C
=
=
a
b
c
or
Examples
1. In triangle PQR find:
Q
a) side length p
b) side length q
p
P
70°
30°
q
R
a) Side length p:
Use the sine rule in the form
p
q
r
=
=
sin P sin Q sin R
The relevant part of the formula is
p
r
=
sin P sin R
p
15
=
sin 70° sin 30°
p=
15 x sin 70°
sin 30°
p = 28 2 cm
b) Side length q:
Angle Q is found using the fact that the sum of the three interior angles of a triangle add to 180°.
 Q = 180°  (70° + 30°) = 80°
B Dec 2016
9
=
From the sine rule
28 2
q
=
sin 70° sin 80°
q=
28 2 x sin 80°
sin 70°
q = 29 6 cm
2. In triangle ABC find:
B
a) angle C
b) angle A
c) side length a.
126°
a
C
A
b = 20 m
a) Angle C:
sin A sin B sin C
=
=
Use the sine rule in the form
a
b
c
sin B sin C
=
b
c
The relevant part of the formula is
sin126° sin C
=
20
12
sin C =
12 𝑥 sin 126°
20
sin ⬚ C = 0 485
C = sin− ( 0 485)
C = 29°
b) Angle A:
A = 180°  (126° + 29°)
A = 25°
c) Side length a:
Use the sine rule in the form
a
b
c
=
=
sin A sin B sin C
The relevant part of the formula is
a
b
=
sin A sin B
a
20
=
sin 25° sin 126°
a=
20 x sin 25°
sin 126°
a = 10 4 m
B Dec 2016
10
3. Given a triangle ABC with angle A = 30°, adjacent side = 15 cm and opposite side = 8 cm as shown
below, find angle C.
In this case there are two possible solutions. This is called the ambiguous case of the sine rule.
B
30°
A
C1
C2
Use the sine rule in the form
sin A sin C
=
a
c
sin 30° sin C
=
8
12
12 x sin 30°
8
sin C = 0.75
sin C =
C = sin− ( 0 75)
C = 48.6°
Another solution to 3 is
ABC2
C = sin− (0 75)
This solution gives the triangle ABC1
= 131.4° = (180° 48.6°). This solution gives triangle
Note: In any non-right-angled triangle, where two sides and the non-included angle are given check
for the ambiguous case if:
 the angle is acute and
 the length of the side adjacent to the angle is greater than the length of the side
opposite the angle.
Exercises
Exercise 1
For the following triangles find the unknown sides.
B
a)
65°
c=?
Q
b)
10
r=?
C
A
30°
P
b=?
t=?
c)
S
130°
27
r=?
15°
T
11
38
45°
q=?
R
B Dec 2016
85°
R
Exercise 2
For the following triangles find all unknown angles and sides.
a)
b)
M
n= ?
21
80°
16
5
N
8.5
L
c)
B
A
35°
b=?
B
6.4
A
a=?
65°
4.7
C
Answers.
Exercise 1
a) b = 18.1, c =19.9.
b) q = 53.5, r = 41.2.
c) t = 9.1, r = 20.2.
Exercise 2
a) L = 35.4° , N = 64.6°, n = 7.8.
b) Ambiguous case: C = 131.2°, B = 13.8°, b = 6.7 or C = 48.8, B = 96.2, b = 27.7
c) B = 41.7°, A = 73.3°, a = 6.8
B Dec 2016
12
C
COSINE RULE
The Cosine Rule can be used to solve non-right triangles
Cosine Rule
a2 = b2 + c2 – 2bc cos α
b2 = a2 + c2 – 2ac cos β
c2 = a2 + b2 – 2ab cos Υ
The angles α, β and γ are respectively opposite the sides a, b, and c.
N.B. The side on the left hand side of the equation is opposite the angle listed at the end of the
Equation:
a2 = b2 + c2 – 2bc cos α
Use the Cosine Rule when you are given
 two sides and the angle between them
 three sides
Examples
1. Find the value of a in this triangle
a2 = b2 + c2 – 2bc cos α
a
a2 = 122 + 152 – 2×12×15 cos 83o
a2 = 144 + 225 – 360 × cos 83o
b =12
830
c = 15
a2 = 369 – 43.87
a2 = 325.13
a = 18.03
2. Find the size of angle β in this triangle:
b2 = a2 + c2 – 2ac cos β
112 = 52 + 72 – 2×5×7×cos β
121 = 25 + 49 – 70 cos β
b =11
a=5
°
47 = – 70 cos β
cos β =
β = 132⁰11’
B Dec 2016
13
c=7
Exercise
1. Use the sine OR cosine rule to find the pronumeral shown:
a)
b)
a
6.6
23.1
840
19.6
2.3
1610
b
c)
d)
0.93
13.6
780
350
8.2
1.25
a
c
2. Find θ
a)
b)
15.6
7.21
α0
9.2
β0
9.99
10.1
Answers
1 a) 28.7
2 a) 34..2o
B Dec 2016
b) 4.38
c) 8.33
4.93
d) 1.05
b) 42.9o
14
ANGULAR MEASUREMENT - RADIANS
Definition of a radian
Though angles have commonly been measured in degrees they may also be measured in units known
as radians. One radian is the angle created by bending the radius length around the arc of a circle.
Converting between radians and degrees
Because the circumference of a circle is given by the formula C = 2πr, we know 2π radians (2πc ) is a
complete rotation and the same as 360 degrees.
Similarly half a rotation or 180 degrees = radians (180⁰ = πc).
Angles that represent fractional parts of a circle can be expressed in terms of π:
Angle in Degrees
Angle in Radians
π
2
π
4
π
3
π
6
90
45
60
30
270
3 × 90 = 3 ×
=
For other angles rearranging πc = 180⁰ gives:
Examples
1. Convert 60⁰ to radians.
1c =
𝑂
𝜋
and 1𝑂 =
2. Convert 240⁰ to radians.
1 =
1 =
60 = 60 ×
240 = 240 ×
60 =
B Dec 2016
𝜋𝐶
= 1.05C
240 =
15
3. Convert
4. ‘Convert 6 5c to degrees
radians to degrees
°
1c =
=
4
×
1c =
°
°
6.5C = 6.5 ×
°
6.5C = 372.4⁰
= 45⁰
Note: The symbol for radian, c , is often omitted.
Exercise
1. Convert the following degrees to radians
a) 30⁰
d) 450⁰
b) 270⁰
e) 135⁰
c) 20⁰
f) 57.3⁰
2. Convert the following radians to degrees
a)
b)
d) 3 5π
e)
c)
π
f) 1 radian
Answers
1.
a)
b)
c)
d)
e)
f) 1 radian
2.
a) 90⁰
d) 630⁰
b) 225⁰
e) 180⁰
c) 330⁰
f) 57.3⁰
B Dec 2016
16
CIRCULAR FUNCTIONS
The trigonometric ratios that have been defined in
right-angled triangles can be extended to angles
greater than 90⁰ by considering angles as rotations
within a unit circle. The centre of the unit circle is
at the point (0,0) and it has a radius of one unit:
Angles are considered as rotations from the
positive x-axis.
Angles greater than 180⁰ and negative angles can also be defined in terms of the unit circle.
Anticlockwise rotations are considered positive and clockwise rotations are negative:
An angle of 135o or
An angle of -75⁰
c
If P(x,y) is any point on the circle, and
An angle of 210⁰ or – 150⁰
is the angle POQ in the triangle POQ as shown:
y
P(x, y)

y = sin ( )
1
O
x
Q
x = cos( )
B Dec 2016
17
Then from the trigonometric properties of a right triangle
cos( ) =
=
= x, sin( ) =
cos( ) = x,
=
= y,
tan( ) =
sin ( ) = y,
=
tan( ) =
Examples
P (0.45, 0.89)
63o
(i)
(ii)
(iii)
(iv)
sin 0 = 0
cos 180 = -1
sin (-90) = -1
sin 63o = 0.89
(v)
tan(-180) =
(vi)
tan 63o =
(vii)
cos (-297o) = 0.45
=0
−
= 1.98
Cos, sin and tan values can also be calculated using a scientific calculator.
[Hint: make sure your calculator is in degrees or radian mode accordingly].
Exact values
Using the table of exact values and the symmetry of the unit circle it is also possible to find the exact
values for multiples of 30⁰, 45⁰, 60⁰.
can be constructed:
30⁰ =
Sin
Cos
1
2
√
√3
2
√
1
Tan
45⁰ =
60⁰ =
=
√
√
=
√
1
2
1
√3
√3
Example
Evaluate sin 330⁰

Plotting 330⁰ on a unit circle shows that sin 330⁰
is closely related to sin 30⁰. The y-coordinates
differ only by sign because the distances from the
x-axis are the same.
1
sin 30⁰ = (from the table) and sin 330⁰ = 2 .

1) Evaluate cos 150⁰
B Dec 2016
18
Plotting 150⁰ on a unit circle shows that cos 150⁰
is closely related to cos 30⁰. The x-coordinates
differ only by sign because the distances from the
y-axis are the same.
√3
cos 30⁰ = √ (from the table) and cos 150⁰ =
3) Evaluate tan
Plotting
on a unit circle shows that tan
2
.
is
closely related to tan . Tan values of
diametrically opposite angles are the same.
tan = √3 (from the table) and tan
= √3 .
Note: By drawing a sketch diagram it is always possible to use the symmetry of the unit circle and the
values in the table to find the exact value of any multiple of 30⁰, 45⁰, 60⁰. The only difference will be a
change of sign and the following diagram may assist in determining whether a negative sign will be
required.
Rotating anticlockwise from the first quadrant:
All trigonometric functions are +ve in the first quadrant
Sin is +ve for angles in the second quadrant
Tan is +ve for angles in the third quadrant
Cos is +ve for angles in the fourth quadrant
It can be helpful to use a mnemonic such as
All Stations To Camberwell to remember these properties of
the trigonometric functions
Exercise
1) What are the coordinates for points on the unit circle that make the following angles with the
positive x-axis?
a) 30⁰
d) 270⁰
2) Find the exact value for:
b) 125⁰
e) -180⁰
c) -60⁰
f) 720⁰
a) sin 330
b) cos 210
c) sin( 30 )
d)cos 90
e) tan 300
f) cos 180
g) sin( 120 )
h) cos 315
Answers
1.
2.
a) (0.87,0.5)
a) -0.5
B Dec 2016
b) (-0.56,0.82)
b)
√
c) (0.5,- 0.87)
d) (0,-1)
e) (– 1,0)
f) (1,0)
c) -0.5
d) 0
e) √3
f) -1
19
TRIGONOMETRIC EQUATIONS
Often the value of the trigonometric function is given and the corresponding angle(s), within a given
domain, are required.
Examples
1) Given sin = 0.3 find all values of
in the domain 0
360⁰
The diagram shows that there are two angles in the given domain (one complete positive
rotation) that will have a sin value of 0.3.
Using the calculator to find the first value:
sin
= 0.3
= sin-1 (0.3)
= 17.46⁰
Using symmetry the angle in the 2nd. quadrant
is 180 – 17.46 = 162.54⁰
The solutions are
2) Solve
= 0.5 over the domain -2
= 17.46⁰, 162.54⁰
2
The diagram shows that there will be four angles in the given domain (one complete positive
rotation and one complete negative rotation) that will have a cos value of 0.5.
Using the table of exact values to find the first
value:
cos = 0.5
= cos-1 (0.5)
=
Use symmetry to find the other angles.
The solutions are
3) Solve tan 2x = -3 over the domain -90
=
,
,
180⁰
First, the domain must be adjusted to suit the angle 2x in the equation:
-90
x 180⁰ ⟹ -180
2x 360⁰
Using the calculator to find the first value:
tan 2 = -3
2 = tan-1 (-3)
2 = -71.57⁰
Tan values of diametrically opposite angles are the same and the
diagram shows how to use symmetry to find other angles in the
domain 180
2x 360⁰:
2x = -71.57⁰, 108.43⁰, 288.43⁰
Finally we divide by 2 to find solutions for x:
x = -35.78⁰, 54.22⁰, 144.22⁰
Exercises
a) If sin = 0.25 find
c) If cos = 0.4 find
e) If tan = -1.5 find
Answers
a) 14.5⁰ , 165.5⁰
d) 113.6⁰ , - 113.6⁰
B Dec 2016
for 0⁰
for -180⁰
for 0⁰
180⁰
b) If tan = 0.8 find
⁰
360 d) If cos = -0.4 find
360⁰
f) If cos = -0.3 find
b) 38.7⁰ , 218.7⁰
e) 123.7⁰ , 303.7⁰
for 0⁰
for -180⁰
for 0⁰
c) 66.4⁰ , -66.4⁰
f) 107.5⁰ , 252.5⁰
20
360⁰
360⁰
360⁰
GRAPHS OF SINE AND COSINE
FUNCTIONS
The functions y = sin
and y = cos have a domain of R and a range of [-1, 1]. The graphs of both
functions have an amplitude of 1 and a period of 2π radians (repeats every 2π units).
y = sin 𝑥
y = cos
y
1.5
1
0.5
-8
-6
-4
-2
2
4
x
6
-0.5
-1
-1.5
[Remember
3.142 so 2
6.284]
Change of amplitude and period
The graphs of both y = sin
and y = cos
Examples
1) y = 3 sin
have amplitude | | and period
2) y = 3cos 2
y
y
4
4
3
2
2
1
-6
-4
-2
2
4
6
-6
x
-1
-4
-2
2
4
6
x
-2
-2
-4
-3
y = 3 cos 2 𝑥 has:
amplitude 3
(|𝑎| = 3)
2π
period
= π (n = 2)
y = 3 sin 𝑥 has:
amplitude 3 (|𝑎| = 3)
period 2π
(n = 1)
Vertical translation
The graphs of y = sin
+ k, y = cos
up k units for k > 0 and down k units for k < 0.
are the graphs of y = sin
y = cos
The graphs of y = sin 𝑥 and y = 2 + sin 𝑥 are shown
B Dec 2016
21
translated
Horizontal translation
Replacing x with (x ) shifts the graphs of y = sin
Replacing x with (x+ ) shifts the graphs of y = sin
Examples
1) y =
(
and y = cos
and y = cos
2) y =
)
(
y
-6
-4
y
1.5
1
1
0.5
0.5
2
4
6
8
x
-6
-4
-2
2
-0.5
-0.5
-1
-1
-1.5
-1.5
(4
First change y = 3
the graph is clear.
y
4
4
6
8
x
)
) to the form y = 3
(4
y = 3 𝑠𝑖𝑛 4( 𝑥
𝜋
4(
) so that the horizontal translation of
)
3
𝜋
1
1
2
3
𝜋
The period is
=
The amplitude is 3.
There is a horizontal shift right of
[NB: Dotted graph is y = 3 𝑠𝑖𝑛 𝑥]
2
-1
units to the right.
units to the left.
)
1.5
-2
3) y = 3
-2
horizontally
horizontally
x
𝜋
.
-1
-2
-3
Reflection
Changing the sign of
x – axis.
B Dec 2016
in the equations y = sin
or y = cos
22
results in a reflection about the
Example
y= 3
2
y
y = 3cos 2𝑥
4
3
The graph of y = 3cos 2𝑥 is a reflection of
y = 3cos 2𝑥 (dotted) in the x-axis
2
1
-3
-2
The period is
-1
1
2
3
x
4
𝜋
= 𝜋
The amplitude is 3.
-1
-2
-3
Exercises
1.Sketch the graph of the following functions for one complete cycle stating the amplitude and period:
(a) y = 2cos
(b) y = 2sin 3
(c) y = sin 2
(d) y = 3cos
(e) y = -2sin 3
2. Sketch the graph of the following functions for one complete cycle stating the amplitude and
period.
(a) y = 2sin(
)
(b) 3cos(x
)
3. Sketch the graph of the following functions for one complete cycle stating the amplitude and
period.
(a) y = 2sin(3
)
(b) 3cos(4x 2 )
(c) y = 2sin(2x + )
Answers
1(a)
Amplitude = 2. Period = 2
1(b)
Amplitude = 2. Period = 2/3
y
y
2
2
1
1
1(c)
Amplitude = 0.5. Period = 
y
0.5
1
2
3
4
5
x
6
0.5
-1
-1
-2
-2
1(d)
1
1.5
2
x
0.5
-0.5
1(e)
Amplitude = 3. Period = 4
y
Amplitude = 2. Period = 2/3
y
3
2
2
1
1
2
4
6
8
10
12
x
-1
0.5
1
1.5
2
x
-2
-1
-3
-2
B Dec 2016
23
1
1.5
2
2.5
3
x
2(a)
Amplitude = 2. Period = 2
2(b)
Amplitude = 1. Period = 2
y
1.5
y
1
2
0.5
1
1
1
2
3
4
5
6
x
2
3
4
5
x
-0.5
-1
-1
-1.5
-2
3(a)
Amplitude = 2. Period = 2/3
3(b)
3(c)
Amplitude = 3. Period = /2
Amplitude = 2. Period = 
y
y
y
2
3
2
2
1
1
1
0.5
1
1.5
2
x
0.5
1
1.5
x
0.5
-1
-1
-1
-2
-2
B Dec 2016
-2
-3
24
1
1.5
2
x