Download trigonometry - Hinchingbrooke

TRIGONOMETRY
A
1. The Sine Rule
a
b
c


sin A sin B sin C
b
c
B
C
a
We can prove the sine rule like this. Let x and y be the perpendiculars from vertices A and B to
the opposite sides a and b.
Considering the two right-angled triangles with x as one of the
A
sides,
x
c
x
sin C 
b
sin B 
b
c
x
y
B
a
C
Comparing expressions for x,
b sin C  c sin B
b
c

sin B sin C
Similarly, considering the two right-angled triangles with y as one of the sides,
y
c
y
sin C 
a
sin A 
Comparing expressions for y,
a sin C  c sin A
a
c

sin A sin C
…and so we have the sine rule.
We now move onto using the sine rule. In order to use it we must have a combination of two
sides and one angle, or one side and two angles.
Example 1 : Find the missing angles and sides in the triangle below.
A
31
B
40
12cm
C
The angles of the triangle add up to 180°.
C  180  31  40
 109
To find b,
a
b

sin A sin B
12
b

sin 31 sin 40
12sin 40
b
sin 31
 14.98 cm
To find c, we use a rather than b since a is an integer and therefore more convenient.
a
c

sin A sin C
12
c

sin 31 sin109
12sin109
c
sin 31
 22.03 cm
Example 2 : solve the triangle with A  30 , b = 5cm, a = 6cm.
Construction : draw the base b and a line leaving A at an angle of 30°. Set compasses to 6cm and
draw an arc from C. The point of intersection of the arc and the line leaving A is the vertex B.
B
6cm
A
30
5cm
C
Solving using the sine rule (we put the sines in the numerators to make the algebra easier),
sin A sin B

b
1a–
0.5
– 0.5
0.5
0.5
1
90
180
sin 30 sin B

6
5
5sin 30
sin B 
6
 0.416
A look at the graph of the sine function tells us that there
are two possible solutions to this equation in between 0°
and 180°. Solving the equation we have B  24.6 and
B  180  24.6  155.4
However B cannot be 155.4°, since A is already 30°, so
B = 24·6° and C  180  30  24.6  125.4 .
1
0.5
90
– 0.5
24.6
180
155.4
Finally we calculate c.
a
c

sin A sin C
6
c

sin 30 sin125.4
6sin125.4
c
sin 30
 9.78 cm
C2 p20 Ex 2A, p22 Ex 2B
Example 3 : Now for an ambiguous case - solve the triangle with A  30 , b = 6cm, a = 4cm.
Construction : very similar to the previous example, although note that now the arc crosses the
line leaving A at two different points. This means that there are two solutions to the triangle.
1–
0.5
– 0.5
0.5
1
0.5
90
180
B
4cm
B
A
4cm
30
6cm
C
sin A sin B

a
b
sin 30 sin B

4
6
6sin 30
sin B 
4
 0.75
B  48.6, B  131.4
1
0.5
90
– 0.5
48.6
180
131.4
Both of these angles are possible, as they are less than 180  30  150 , so we consider each case.
• If B = 48.6°, then C = 180° − 30° − 48.6° = 100.4°. To find c,
a
c

sin A sin C
4
c

sin 30 sin100.4
4sin100.4
c
sin 30
 7.84 cm
• If B = 131.4°, then C = 180° − 30° − 131.4° = 18.6°. To find c,
a
c

sin A sin C
4
c

sin 30 sin18.6
4sin18.6
c
sin 30
 2.55 cm
C2 p23 Ex 2C
2. The Cosine Rule
In a right angled triangle, c 2  a 2  b 2 . However, for the acute-angled triangle shown below, it is
clear that c 2  a 2  b 2 . And for the obtuse-angled triangle, c 2  a 2  b 2 . Therefore, we need a
modified version of Pythagoras’ theorem for non-right angled triangles.
c
c
b
c
b
a
a
a
acute-angled
right-angled
b
C
C
obtuse-angled
In fact the formula is the cosine rule
c 2  a 2  b 2  2bc cos C
Notice how if C is acute, then cosC is positive, and c 2  a 2  b 2 . Similarly, if C is obtuse, then
cosC is negative, and c 2  a 2  b 2 . In fact the formula appears in three forms…
a 2  b 2  c 2  2bc cos A
b 2  a 2  c 2  2ac cos B
c 2  a 2  b 2  2ab cos C
To prove the cosine rule, drop a perpendicular of length h from C to the line
AB. Let the base of the unshaded triangle be x. Notice that x  b cos A .
Using Pythagoras’ rule on the unshaded triangle,
b
x h b
2
2
a
h
2
h2  b2  x 2
Using Pythagoras’ rule on the shaded triangle,
x
A
(c  x)  h  a
2
C
2
2
Substituting h 2 from the first equation,
(c  x ) 2  b 2  x 2  a 2
c 2  2cx  x 2  b 2  x 2  a 2
a 2  b 2  c 2  2cx
a 2  b 2  c 2  2bc cos A
c
B
Because the formula contains three sides and one angle, we can use it if we know three sides, or
two sides and the angle they enclose.
Example 1 : Solve the triangle with a = 8cm, b = 6cm and c = 4cm.
To find A,
A
a 2  b 2  c 2  2bc cos A
1– 1
0.5
0.5
–1
0.5
82  62  42  2  6 10.5
904 cos A
180
64  52  48cos A
6cm
4cm
12
48
A  104.5
cos A  
B
We see from the cosine function that there is only one
solution between 0° and 180°, and so we do not have to
worry about any ambiguous cases. This is consistent
with our experience that three sides uniquely define a
triangle. It also means that the cosine rule may be less
prone to human error than the sine rule!
C
8cm
1
104.5
0.5
90
180
– 0.5
–1
To find B, we can use either the cosine rule or the sine rule. The cosine rule is marginally
preferable as with the sine rule we would have to use the calculated value of A, rounding it off in
the process. Choosing the cosine rule,
b 2  a 2  c 2  2ac cos B
62  82  42  2  8  4 cos B
36  80  64 cos B
44
64
B  46.6
cos B 
Finally we calculate C,
C  180  104.5  46.6
 29.0
If this answer looks like it should be 28.9, then remember we must use the full calculator values
for A and B, not the rounded off ones.
Example 2 : Solve the triangle with a = 7cm, b = 5cm and C = 119°.
To find c,
C
c 2  a 2  b 2  2bc cos C
 7 2  52  2  7  5cos119
 74  70 cos119
c  10.39 cm
5cm
A
119
7cm
B
We can now use the sine or the cosine rule to calculate A or B. Choosing the sine rule for A,
sin A sin C

a
c
sin A sin119

7
10.39
7 sin119
sin A 
10.39
A  36.1, A  143.9
We reject the solution A = 143.9° as there are not enough degrees in the triangle. Finally,
B  180  119  36.1
 24.9
C2 p31 Ex 2D, p29 Ex 2E
3. Applications of the Sine and Cosine Rules
Example 1 : Find the angle between the hour hand and the minute hand of a clock at 12.15. If the
hour hand is 8cm long and the minute hand is 10cm long, how far apart are the tips
of the hands at this time?
The hour hand covers 360° ÷ 12 = 30° in one hour, and therefore 30° ÷ 4 = 7.5° in quarter of
an hour. So the angle between the hands is 90° – 7.5° = 82.5°.
C
By the cosine rule,
a 2  b 2  c 2  2bc cos A
 10  8  2 10  8cos82.5
 164  160 cos82.5
a  11.96 cm
2
8cm
2
A
82.5
B
10cm
Example 2 : A boat is sailing directly towards the foot of a cliff. The angle of elevation of a point
on top of the cliff, and directly ahead of the boat, increases from 8° to 12° as the
boat sails 100m. Find
a) the original distance from the boat to the point on top of the cliff
b) the height of the cliff.
C
a) We have A = 168°, B = 8°, c = 100m.
C  180  168  8
 4
cliff
12
168
A
8
100m
B
To find a,
a
c

sin A sin C
a
100

sin 68 sin 4
100sin168
a
sin 4
 298.05 m
The original distance from the boat to the top of the cliff is 298.05m.
b) To find the height of the cliff,
height of cliff  a sin B
 298.05sin 8
 41.48 m
Example 3 : A ship leaves a port P and sails 5 nautical miles on a bearing of 052°, followed by 6
nautical miles on a bearing of 205° to arrive at the next port Q.
Find the distance and bearing of P from Q.
To find the angle X of the triangle, we use the property of
‘interior angles’.
X
X  380  205  (180  52)
 27
Using the cosine rule,
x  p  q  2 pq cos X
2
2
5 miles
52
P
2
 62  52  2  6  5cos 27
 61  60 cos 27
x  2.75 miles
205
6 miles
Q
Using the sine rule,
sin Q sin X

q
x
sin Q sin 27

5
2.75
5sin 27
sin Q 
2.75
Q  55.8, Q  124.2
We reject the solution Q  124.2 as it does not fit with our diagram. The bearing of P from Q is
given by 360  (55.8  25)  329.2 .
C2 p31 Ex 2F
4. The Area of a Triangle
Suppose we label the angles of a triangle A, B and C. Then we label the side opposite angle A as
a, the side opposite angle B as b, and the side opposite angle C as c. We can find a formula for
the area of a triangle using this notation.
A
area of triangle  12  base  height
b
c
 12  a  b sin C
 12 ab sin C
h
B
C
a
Clearly we could also have these alternative expressions for the area : 12 bc sin A or 12 ac sin B .
Example 1 : Find the area of this triangle.
C
7cm
100
A
B
5cm
area of triangle  12 ac sin B
 12  7  5  sin100
 17.23 cm2
Example 2 : The area of a triangle is 36 cm². Two sides of the triangle are 8cm and 10cm. Find
the two possible angles between these two sides.
area of triangle  12 bc sin A
1–
0.5
– 0.5
0.5
1
0.5
90
180
C
36   8 10  sin 
1
2
10cm
36
40
  64.2,   115.8,
sin  
A

8cm
B
A small sketch of the sine curve between 0° and 180° illustrates the last step in the calculation.
1
0.5
90
– 0.5
64.2
180
115.8
C2 p34 Ex 2G
5. Radian Measure
One radian is defined as the angle subtended at the centre of a
circle, by an arc equal in length to its radius. In general,
arc length
angle in radians 
radius
Therefore, for one revolution,
2 r
angle in radians 
 2 .
r
So 2π radians = 360°, and one radian is approximately 57°.
r
1 rad
r
r
It is important to be able to convert between degrees and radians fluently. The table below shows
some common angles in both units.
degrees 0 30 45 60 90 120 135 150 180 240 270 300 360
2
3
5
4
3
5





2
radians 0
3
4
6
3
2
3
6
4
3
2
The symbol for radians is a small superscript c, and so 2 radians is written 2c or 2 rad. If units are
omitted, then an angle is in radians, not degrees!
Example 1 : Convert the following angles into degrees.
a) 79  rad
b) 2 rad
a)
7
9
180  140
b) 2 
180

 114.6 (1 d.p.)
Example 2 : Convert the following angles into radians.
a) 24°
b) 7°


7
 152 
 180
 or 0.122
a) 24 
b) 7 
180
180
When angles are measured in radians, we have several convenient formulas concerning circles…
C2 p81 Ex 6A
6. Arc Length
Consider the arc PQ, of length s, subtending angle θ radians at the
P
centre of the circle. Now
r
arc length
angle in radians 
radius
O 
s
s

r
r
Q
And so
s  r
So a length of arc is found by multiplying the radius by the angle (measured in radians)
subtended by the arc.
In fact, two points P and Q on a circle define two arcs, a shorter one (known as the minor arc)
and a longer one (known as the major arc).
Example 1 : Find the length of the minor arc PQ,
We first write 135° as 34  radians.
P
s  r
 8  34 
O 135
 6 cm
8cm
Q
Example 2 : Find the perimeter of the shape opposite.
The perimeter consists of two arcs and two straight lines.
We first write 60° as 13  radians.
2cm
perimeter  5    7    2  2
1
3
1
3
 4  4 cm
60
Example 3 : A road tunnel 60m long has the cross-section
shown opposite, consisting of an arc centre O and
a straight line. It is required to paint the inside of
the tunnel (not including the road itself). What
area is to be painted?
We first find the angle θ. Splitting the isosceles triangle into
two right-angled triangles gives us
sin
    1.25
1.5
1
2
1
2
  0.985
  1.970 radians
The arc subtends an angle of 2 1.970 radians.
arc length  1.5  (2  1.970)  2.5
 6.469 m
The internal area is therefore
area  6.469  60
 388.2 m2
C2 p83 Ex 6B
5cm
O
1.5m

2.5m
1.5m
7. Area of a Sector
Similarly, we also have minor sectors and major sectors. Consider the sector OPQ, with area A.
Now
area of sector
fraction of circle 
P
area of circle
r

A
 2
O  A
2  r
And so
r
A  12 r 2
Q
P
Example 1 : Find the area of the major sector OPQ,
The major sector subtends 2  34   54  .
A  12 r 2
 8  
1
2
2
5
4
O 135
8cm
 40 cm 2
Q
Example 2 : A shelf has the shape of a sector of a circle of
radius 8cm, with a concentric sector of radius
5cm removed, as shown in the diagram below. If
the shelf has an area of 50cm2, find the angle
subtended by the sectors.
1
2
 82    12  52    50
19.5  50
3cm
5cm

  2.56 radians
Example 3 : Find a formula for the area A of the segment below. Hence find the area of the
segment if r = 6cm and θ = 1.5 rad.
P
area of segment  area of sector  area of triangle
r
 12 r 2  12 ab sin C
 12 r 2  12 r 2 sin 
 12 r 2 (  sin  )
Substituting,
area of segment 
12
2
62 (1.5  sin1.5)
 9.045 cm 2
O

A
r
Q
Example 4 : This shape is made up of arcs centered on the corners of an
equilateral triangle of side length r. Show that the area of the
shape is given by 12 r 2   3 .


area  3  sector  2  triangle
 3  12 r 2
 12 r 2 

   2
1

3
1
2
r 2 sin
 
r
1

3
3 2
r
2
 12 r 2   3

Example 5 : The diagram shows a triangle, and an arc whose centre is the point P.
a) Show that   1.330 rad
5cm
b) Find the perimeter of the region R.
4cm
c) Find the area of the region R.
P 
R
10cm
4cm
3cm
a) Using the cosine rule,
a 2  b 2  c 2  2bc cos A
102  92  7 2  2  9  7 cos 
100  81  49  126 cos 
30
126
  1.330... rad
We only get the answer in radians if our calculator is in radian mode!
cos  
b) The perimeter is an arc length added to three straight edges.
perimeter  4 1.330...  3  10  5
 23.32 cm (2 d.p.)
c) The area is the triangle subtract the sector.
area  12  9  7  sin1.330...  12  42  1.330...
 30.594...  10.643...
 19.95 cm 2 (2 d.p.)
Here it is essential to have our calculator is in radian mode.
C2 p87 Ex 6C
8. Sine, Cosine and Tangent of Common Angles
Sines, cosines and tangents are commonly used with right-angled triangles.
opp
adj
opp
sin  
cos  
tan  
hyp
hyp
adj
hyp
opp

adj
Activity 1 : Do the Sine, Cosine and Tangent of Common Angles worksheet. Encourage students
to learn these values.
sin 
cos
tan 

0°
0
1
0
1
1
3
30°
2
3
2
45°
60°
90°
1
2
1
2
3
2
1
1
2
0
1
3
–
y
9. The Unit Circle
However sines, cosines and tangents also have a close
relationship with the unit circle.
The diagram shows a circle whose centre O is at (0, 0), and
whose radius is 1 unit. Let P be a point on the unit circle.
If we want to tell someone exactly where P is, we can give – 1
the angle between the positive x-axis and the line OP. We
shall call this angle θ, and we shall always measure it anticlockwise, starting from the positive x-axis.
1
P

1
O
x
–1
Another way to specify the position of P is to give its coordinates. We can find them by
considering the right-angled triangle in the diagram below.
The x-coordinate is the distance OA. Now
adj
cos  
hyp
OA

1
 OA
The y-coordinate is the distance AP. Now
opp
hyp
AP

1
 AP
y
1
P

–1
O
sin  
–1
1
A
x
So the coordinates of P are (cos  , sin  ) . So we define sine and cosine as the y and x
opp
adj
coordinates of the point P, rather than sin  
and cos  
.
hyp
hyp
The unit circle is divided by the x and y axes into four
quadrants, which are numbered anticlockwise as shown
here.
If the angle θ is greater than 90°, then P will be in the 2nd,
3rd or 4th quadrants. We can now give a meaning to sines
and cosines of angles greater than 90°, by simply saying
that
–1
cos   x  coordinate of P
sin   y  coordinate of P
y
1
1st
2nd
1
O
4th
3rd
For example, when θ = 130°, we find from a calculator that
cos θ = –0.643, and sin θ = 0.766. We can see why by
looking at the unit circle : when θ = 130°, the coordinates
of P are (–0.643, 0.766).
x
–1
Activity 2 : Use the Unit Circle worksheet to find the coordinates of points on the unit circle at
angles 50°, 120°, 255° and 305°. Use calculators to check that the coordinates are
indeed (cos θ, sin θ). Alternatively, let students choose their own points within each
quadrant.
As an extension, investigate the sine and cosine of 430° and –65°.
0.5
1
–
0.5
0.5
1
–1
0.5
1
90
180
270
360
10. The Sine Function
The sine of θ is the y-coordinate of the point P on the unit circle. Because
of this, it is clear that...
y
x
sin 0° = 0, sin 90° = 1, sin 180° = 0, sin 270° = –1, sin 360° = 0.
If we plot a graph of sin θ against θ, we find it looks like this.
sin
y
1
0.5

90
x
– 0.5
– 1
180
270
360 
180
360
540
720
– 360
180
But after 360°, P is back to the start of the circle (1, 0) again. So the graph repeats itself every
360°. We say it is periodic, with period 360°.
sin
1
0.5
– 360
– 180
180
360
540
720 
– 0.5
–1
This graph is called the sine function.
0.5
1
–
0.5
0.5
1
–1
0.5
1
90
180
270
360
y
11. The Cosine Function
The cosine of θ is the x-coordinate of the point P on the unit circle.
Because of this, it is clear that...
x
cos 0° = 1, cos 90° = 0, cos 180° = –1, cos 270° = 0, cos 360° = 1.
If we plot a graph of cos θ against θ, we find it looks like this.
cos
x
1
0.5

90
y
1– 1
0.5
0.5
180
360
540
720
– 360
180
180
270
360 
– 0.5
–1
Clearly the cosine function is periodic, with period 360°, in exactly the same way as the sine
function.
cos
1
0.5
– 360
– 180
180
– 0.5
–1
This graph is called the cosine function.
360
540
720 
0.5
2
1.5
1
–
0.5
1
1.5
0.5
1
1.5
2
–2
0.5
1
1.5
2
90
180
270
360
12. The Tangent Function
There are two ways of thinking about the tangent function.
Firstly, we can draw the tangent with equation x  1 to the unit
circle.
Now consider the line OP extrapolated until it cuts the tangent
at B. Let the point A be (1, 0).
y
1
P

opp
–1
adj
AB

1
 AB
So the tangent of  is the distance up the tangent between
(1, 0) and the place where the extrapolated radius cuts the tangent.
O
tan  
–1
Activity 3 : Use the Tangent Function worksheet. This has extrapolated radii every 15° up to
360°. This will make the fact that we have asymptotes at 90° and 270°, and that the
period of the tangent graph is 180°, very obvious!
tan
y
4– 4
2
2
180
360
– 360
180
2
1.5
1
0.5

x
90
– 0.5
–1
– 1.5
–2
180
Notice that the period of the tangent function is 180°.
tan
4
2
– 360
– 180
180
–2
–4
360 
270
360 
B
A
1 x
Alternatively, we can define tangent in terms of sine and cosine.
For a right-angled triangle, we have
 opp 


opp  hyp  sin 
tan  


adj  adj  cos 
 hyp 


hyp
opp

adj
So the tangent of θ is found by dividing the sine of θ by the cosine of θ.
1– 1
0.5
0.5
90
180
– 180
90
Activity 4 : Now distribute the Handy Graphs of Trigonometric Functions worksheet for future
reference. Look carefully at the sine and cosine graphs and then the tangent graph.
sin 
Discuss how the graphs confirm that tan  
.
cos 
13. Sine, Cosine and Tangent of any Angle
From the unit circle we can see that cos(–50°) is the same as cos 50°. This is also clear from the
graph.
cos
y
1
0.5
50
50
– 180
x
– 90
90
180

– 0.5
–1
1– 1
0.5
0.5
90
180
– 180
90
In general, cos( )  cos  . This means the cosine function is symmetrical about the y-axis.
(Functions with this property are known as even functions).
From the unit circle, we can see that sin(–50°) is the exact negative of sin50°. This is also clear
from the graph.
sin
y
1
0.5
50
50
x
– 180
– 90
90
180

– 0.5
–1
In general, sin( )   sin  . This means the sine function has rotational symmetry about the
origin. (Functions with this property are known as odd functions).
1– 1
0.5
0.5
90
180
– 180
90
We also have
sin( )
cos( )
 sin 

cos 
  tan 
So the tangent function is also an odd function, with rotational symmetry about the origin. This is
confirmed by the graph of the tangent function.
tan( ) 
Example 1 : Write the following as trigonometric functions of acute angles
a) cos(−110°)
b) sin 220°
c) tan(−110°)
a) cos(−110°) = cos 110° = −cos 70°
cos
y
1
0.5
1–
0.5
–1
0.5
1
1
0.5
0.5
90
180
270
360
70
x
– 180
– 90
90
180

– 0.5
–1
b) sin 220° = −sin 140° = −sin 40°
sin
y
1
0.5
40
90
x
180
270
360 
– 0.5
–1
c) tan(−110°) = tan 70°, since tan has a period of 180°.
Example 2 : Write down the values of these in surd form.
a) sin 120°
b) cos 300°
c) tan 300°
sin120  sin 60
tan 300  tan(60)
a)
c)

b)
  tan 60
3
2
d)
d) sin 330°
e) cos 225°
cos 225  cos135
e)
  cos 45
 3
sin 330  sin( 30)
  sin 30
  12
C2 p113 Ex 8B, p117 Ex 8C, p118 Ex 8D, p120 Ex 8E

1
2
14. The Pythagorean Relationship
Consider the right-angled triangle with hypotenuse 1. The opposite
has length sin  , and the adjacent has length cos . Using
Pythagoras’ rule we have
sin 2   cos 2   1
1
sin

cos
With this Pythagorean relationship, we can prove other identities (and solve certain trigonometric
equations - see next section).
Example 1 : Prove the following identities.
a) (sin   cos  )2  (sin   cos  )2  2 .
1
.
cos 2 x
c) cos 4   sin 4   1  2 cos 2  .
It is usual to take the more ‘complicated’ side (often the LHS), and reduce it to the form of the
‘simpler’ side.
a) (sin   cos  ) 2  (sin   cos  ) 2  sin 2   2sin  cos   cos 2   sin 2   2sin  cos   cos 2 
b) tan 2 x  1 
 2  sin 2   cos 2  
 2 1
2
b)
c)
2
sin x
1
cos 2 x
sin 2 x  cos 2 x

cos 2 x
1

cos 2 x
tan 2 x  1 
cos 4   sin 4   1   cos 2   sin 2   cos 2   sin 2    1
 cos 2   sin 2   1
 cos 2   cos 2 
 2 cos 2 
Example 2 : Given that x  cos and y  tan  , express y in terms of x.
y  tan 
sin 

cos 

 1  cos 2 
cos 

1  x2
x
Example 3 : Given that cos  23 , and that θ is an acute angle, find sin  and tan  .
We can use this right-angled triangle. A simple application of Pythagoras
shows that the opposite is 5 . We now have
3

5
3
5
tan  
2
sin  
5
2
Example 4 : Given that cos  23 , and that θ is a reflex angle, find sin  and tan  .
Since θ is reflex, we cannot use a right-angled triangle, and so we use the identity…
sin 2   cos 2   1
4
1
9
5
sin 2  
9
sin 2  
5
3
Notice we take the negative root here. This is because the sine of a reflex angle is negative.
sin 
tan  
cos 
sin   



5
3
2
3
5
2
Example 5 : Given that tan   247 , and that θ is an acute angle, find sin  and cos . How would
your answers change if θ is reflex?
We can use this right-angled triangle.
sin  
7
25
cos  
24
25
25

24
If θ is reflex, then because tan  is positive, θ must be in the third quadrant, and so both sin 
and cos are negative. We therefore have
sin    257
cos    24
25
C2 p145 Ex 10A
7
1– 1
0.5
0.5
90
180
270
360
15. Solving Simple Trigonometric Equations
Always do a sketch graph, unless the equation involves tan.
Example 1 : Solve the equation sin x  0.2, 0  x  360 .
sinx
1
From the calculator we have x  11.5 , and by the
symmetry of the sine function we also have
1
0.5
–
0.5
0.5
1
–1
0.5
1
x  180 11.5  168.5
90
180
270
360
0.5
90
180
270
360 x
90
180
270
360
– 0.5
–1
Example 2 : Solve the equation cos x  0.6, 0  x  360 .
From the calculator we have x  126.9 , and by the
symmetry of the cosine function we also have
x  360 126.9  233.1
cosx
1
0.5
x
– 0.5
–1
Example 3 : Solve the equation tan x  2.6, 0  x  360
1–
0.5
–but
0.5 this solution is not in the required range. By the
1
0.5
1
From the calculator we have x  69.0 , 10.5
90
180
270
360
periodicity of the tangent function we have the solutions x  69.0 180  111.0 and
x  111.0 180  291.0 . Notice that we do not need to sketch the tangent function, as it is very
simple to ‘hop’ from one solution to another (simply add or subtract 180°).
Note that simple sine, cosine and tangent equations like the ones above always have two
solutions in a 360° range (unless we are solving something like sin x  1 ).
Example 4 : Solve sin  x  90  34 , 0  x  360 .
Although the equation involves the sine of x  90 , we still
sketch the ordinary sine curve. The values on the x-axis are
0.5
1
–1
0.5
the relevant values of x  90 .
90
180
– 180
90
x  90  48.6, 131.4
x  138.6, 221.4
sinx
1
0.5
90
180
270
360 x
90
180 x
– 0.5
–1
cosx
Example 5 : Solve cos  x  30   , 180  x  180 .
1
2
1
x  30  120, 120
x  150, 90
0.5
– 180
– 90
– 0.5
–1
1– 1
0.5
0.5
90
180
– 180
90
Example 6 : Solve sin  x  60  12 , 180  x  180 .
sinx
1
x  60  30, 150 ,  210
1– 1
0.5
0.5
0.5
1
–1
0.5
x  90, 210 ,  150
90
180
270
360
0.5
We notice that adding 60° will take one of the solutions out
of range, so we go back a period from 150°.
– 180
– 90
90
180 x
– 0.5
– 1
cosx
Example 7 : Solve cos  x  50  0.8, 0  x  360 .
0.5
1
–
0.5
0.5
1
–1
0.5
1
90
180
270
360
x  50  36.9 , 323.1,396.9
1
0.5
x  13.1 , 273.1,346.9
We notice that subtracting 50° will take one of the solutions
out of range, so we go forward a period from 36·9°.
Example 8 : Solve sin 2 x 
3
, 0  x  360 .
2
2 x  60, 120, 420, 480
180
270
360
90
180
270
360 x
90
180 x
– 0.5
–1
sinx
1
0.5
x  30, 60, 210, 240
Notice that since we divide by two, we have twice as many
solutions as normal. This is because the graph of
y  sin 2 x is a horizontally ‘squashed’ version of the
y  sin x graph.
x
90
– 0.5
–1
Example 9 : Solve tan  3x 15  1, 0  x  360
This time, we need to find six solutions.
3x  15  45, 225, 405, 585, 765, 945
0.5
1
–1
0.5
90
90 , 600, 780, 960
180
3x  60, 240180
,– 420
x  20, 80, 140, 200, 260, 320
C2 p148 Ex 10B, p150 Ex 10C
Example 10 : Solve 3cos2 x  cos x  0,  180  x  180 .
cosx
1
0.5
– 180
– 90
– 0.5
–1
3cos 2 x  cos x  0
cos x(3cos x  1)  0
cos x  0  x  90
cos x   13  x  109.5
Notice that we factorise in the working above. Never divide by cos x , as this loses the solutions
of cos x  0 .
Example 11 : Solve 2 tan 2 x  tan x  3,  180  x  180
2 tan 2 x  tan x  3
2 tan 2 x  tan x  3  0
(2 tan x  3)(tan x  1)  0
1–
0.5
–1
0.5
1
1
0.5
0.5
90
180
270
360
tan x
  32  56.3, 123.7
tan x  1  x  135, 45
Example 12 : Solve tan x  2sin x, 0  x  360
tan x  2sin x
cosx
sin x
 2sin x
cos x
sin x  2sin x cos x
2sin x cos x  sin x  0
1
0.5
sin x(2 cos x  1)  0
sin x  0  x  0, x  180, x  360
90
180
270
360
x
– 0.5
cos x  12  x  60, x  300
–1
Example 13 : Solve tan 2 x  3, 0  x  2 .
Notice that as there is no ° symbol, the angles are measured in radians. The 2 is another big
clue.
tan 2 x  3
tan x   3
1–
0.5
–1
0.5
1
1
0.5
0.5
tan x 23 3  x  13  ,
22
tan x   3  x  23  ,
4
3

5
3

To get from one solution to the next we add π radians, as this is the radian equivalent of 180°, the
period of the tangent function.


Example 14 : Solve sin 2 x  16   12 , 0  x  2
sinx
1
0.5

– 0.5
–1
2

3
2
2 x


sin  x     
sin 2 x  16  
1
2
1
2
1
1
x  6   4  , 43  ,
1
6
x  121  ,
7
12
,
5
4
,
13
12
7
4
,

19
12

22
Example 15 : Solve 4  sin   4cos2  , 0    2
Using the Pythagorean identity,
sin
4  sin   4 1  sin 2  
1
4  sin   4  4sin 2 
0.5
4sin 2   sin   0

sin  (4sin   1)  0
sin   0  
sin   14  
1–
0.5
–1
0.5
1
1
0.5
 0,    ,  0.5
3
2
 2
22
 0.253,   2.889
– 0.5
(3 d.p.)

2
3
2
2 
–1
The last two solutions are best obtained by having your calculator in radian mode.
Example 16 : Solve 8 tan   3cos  , 0    2 .
8 tan   3cos 
sin
8sin 
 3cos 
cos 
8sin   3cos 2 
1
0.5
8sin   3 1  sin 2  
3sin   8sin   3  0

– 0.5

3
2

3
2
2
2 
2
(3sin   1)(sin   3)  0
1–
0.5
–1
0.5
1
1
0.5
0.5
3
2

22
–1
sin   13    0.340,   2.802
sin   3  no solutions
Example 17 : Solve sin  tan   3, 0    2
sin  tan   3
sin 
cos
sin 
3
cos 
sin 2   3cos 
1
0.5
1  cos 2   3cos 

– 0.5
cos 2   3cos   1  0
3  32  4 1 1
2 1
3  13

2
3  13
cos  
   1.263,   5.020
2
3  13
cos  
 no solutions
2
cos  
C2 p153 Ex 10D
–1
2
2 
16. Transformations of Trigonometric Graphs
recommended that Autograph and a projector are used to deliver this section of the notes.
Students can discuss the effect of a change in the function before it is plotted. The animation
feature is also very useful.
a) Translation
• If we translate the graph of y  sin x by a units vertically upwards, we get the graph of
y  sin x  a .
3– 2
1
2
1
90
180
270
360
–It360
90is
180
270
y
3
y = sinx + 2
2
1
y = sinx
– 360 – 270 – 180 – 90
90
180
270
360
x
y = sinx – 1
–1
–2
1– 1
0.5
0.5
90
180
270
360
– 360
90
180
270
• If we translate the graph of y  sin x by a units to the right, we get the graph of
y  sin( x  a) . To understand this, consider the graph of y  sin( x  30) . We can think of
this as clock running 30 minutes slow. When x is 30°, we are taking the sine of only
30  30  0 . So our graph is doing at 30° what the ordinary sine graph does at 0°, and so
the whole graph is shifted 30° to the right.
Similar reasoning means that if we translate the graph of y  sin x by a units to the left,
we get the graph of y  sin( x  a) .
y
1
y = sin(x + 30)
y = sinx
0.5
y = sin(x – 30)
– 360 – 270 – 180 – 90
– 0.5
2– 2
1
1
90
180
270
360
– 360
90
180
270
90
180
270
360
x
– 1
b) Stretches
• If we stretch the graph of y  sin x vertically by scale factor a, we get the graph of
y  a sin x .
y
2
y = 2sinx
1
y = sinx
– 360 – 270 – 180 – 90
90
180
–1
–2
1
y = sinx
2
270
360
x
1– 1
0.5
0.5
90
180
270
360
– 360
90•
180
270
If we stretch the graph of y  sin x horizontally by scale factor a, we get the graph of
x
y  sin   .Consider the situation when a  2 . We can think of this as a clock running at
a
half its normal speed. When x is 180°, we are taking the sine of only 90°. So our graph is
doing at 180° what the ordinary sine graph does at 90°, and so the whole graph is twice as
wide.
y
y = sin
1
x
2
y = sinx
0.5
– 360 – 270 – 180 – 90
– 0.5
1– 1
0.5
0.5
90
180
270
360
– 360
90
180
270
90
180
270
360
x
–1
• In a similar way, the graph of y  sin 2 x is a horizontally ‘squashed’ version of y  sin x .
We can think of this as a clock running at twice its normal speed. When x is 90°, we are
taking the sine of 180°. So our graph is doing at 90° what the ordinary sine graph does at
180°, and so the whole graph is half as wide.
y
y = sin 2x
1
y = sinx
0.5
– 360 – 270 – 180 – 90
– 0.5
1– 1
0.5
0.5
90
180
270
360
– 360
90
180
270
90
180
270
360
x
– 1
c) Reflection
• Reflection in the x-axis changes the sign of the y-coordinate. So y  sin x becomes
y   sin x on reflection in the x-axis.
y = – sinx
y
1
y = sinx
0.5
– 360 – 270 – 180 – 90
– 0.5
–1
90
180
270
360
x
1
3
2
1
2
1–
2
3
–3
1
2
3
90
180
270
360
When sketching transformed trig curves, always indicate maxima, minima, and points of
intersection with the axes.
Example 1 : Sketch the graph of y  3sin 2 x for 0  x  360 .
1
vertical stretch S.F. 3
2
y  sin x 
 y  sin 2 x 
 y  3sin 2 x
horizontal stretch S.F.
y
(225, 3)
(45, 3)
3
2
1
90
180
270
360 x
–1
–2 2
1
3
1
1–
2
3
–1
90
180
270
360
–3
(315, -3)
(135, -3)
Example 2 : Sketch the graph of y  1  2 cos x for 0  x  360 .
vertical stretch S.F. 2
translation 1 unit up
y  cos x 
 y  2cos x 
 y  1  2cos x
The graph crosses the x-axis when
y
3
1  2 cos x  0
cos x   12
(360, 3)
(0, 3)
2
x  120, x  240
(120, 0)
1
90
1
2
1
2
3
1–
2
–4
1
2
3
4
90
180
270
360
180
360 x
270
–1
(240, 0)
(180, -1)
Example 3 : Sketch the graph of y  3sin x  1 for 0  x  360 .
vertical stretch S.F. 3
translation 1 unit down
y  sin x 
 y  3sin x 
 y  3sin x  1
The graph crosses the x-axis when
3sin x  1  0
sin x 
1
3
x  19.5, x  160.5
y
(90, 2)
2
(160.5, 0)
1
(0, -1)
x
90
180
270
–1
–2
(19.5, 0)
–3
–4
(270, -4)
360
4–
2
–4
2
4
4
2
2
90
180
270
360
Example 4 : Sketch the graph of y  tan x  1 for 0  x  360 .
translation 1 unit down
y  tan x 
 y  tan x  1
The graph crosses the x-axis when
y
4
tan x  1  0
tan x  1
2
x  45, x  225
(225, 0)
90
(0, -1)
–2
1
3
2
1–
2
3
–1
1
90
180
270
360
180
270
360
x
(45, 0)
–4
Example 5 : Sketch the graph of y  1  2 cos( x  90) for 0  x  360 .
translation 90 left
vertical stretch S.F. 2
y  cos x 
 y  cos( x  90) 
 y  2cos( x  90)
translation 1 unit up

 y  1  2cos( x  90)
The graph crosses the x-axis when
1  2 cos( x  90)  0
cos( x  90)   12
x  90  120, 240
x  30, x  150
y
(270, 3)
3
2
(0, 1)
1
(30, 0)
90
–1
(90, -1)
180
270
360 x
(150, 0)
Example 6 : Sketch the graph of y  tan(45  x) for 0  x  360 .
Whenever the function has a negative x, we must make it positive by using one of the following
relationships. Remember that cos is different!
4–
2
–4
2
4
4
2
2
90
180
270
360
sin( x)   sin x
cos( x)  cos x
tan( x)   tan x
So we rewrite our function as y   tan( x  45) .
translation 45 right
reflection in x -axis
y  tan x 
 y  tan( x  45) 
 y   tan( x  45)
The graph crosses the x-axis when
 tan( x  45)  0
x  45  90, 270
x  45, x  225
y
4
2
(0, 1)
(225, 0)
(45, 0)
–2
–4
90
180
270
360
x
1–
–3
1
2
3
1
1
2
90
180
270
360
Example 7 : Sketch the graph of y  2 cos(30  x)  1 for 0  x  360 .
We first rewrite the function as y  2 cos( x  30)  1 .
translation 30 right
vertical stretch S.F. 2
y  cos x 
 y  cos( x  30) 
 y  2cos( x  30)
translation 1 unit down

 y  2cos( x  30)  1
The graph crosses the x-axis when
2 cos( x  30)  1  0
cos( x  30) 
1
2
x  30  60, 300
x  90, x  330
The graph crosses the y-axis at
y
 0,
(30, 1)
1

3 1
(90, 0)
90
–1
(330, 0)
180
(0, 1)
–2
2cos30  1  3  1
–3
(210, -3)
C2 p124 Ex 8F
Topic Review : Trigonometry
270
360 x