Trigonometry-1.ppt

© Boardworks Ltd 2005
1 of 47
These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page.
This icon indicates the slide contains activities created in Flash. These activities are not editable.
For more detailed instructions, see the Getting Started presentation.
1 of 47
AS-Level Maths:
Core 2
for Edexcel
C2.4 Trigonometry 1

2 of 47
Contents
2 of 47
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θ and tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions

3 of 47
θ
O
P
P
O
S
I
T
E
H
Y
P
O
T
E
N
U
S
E
A D J A C E N T
The three trigonometric ratios
Sin θ =
Opposite
Hypotenuse S O H
Cos θ =
Adjacent
Hypotenuse C A H
Tan θ =
Opposite
Adjacent T O A
Remember: S O H C A H T O A
The three trigonometric ratios, sine, cosine and tangent, can
be defined using the ratios of the sides of a right-angled
triangle as follows:

4 of 47
x
y
O
P(x, y)
r
These definitions are limited because a right-angled triangle
cannot contain any angles greater than 90°.
To extend the three trigonometric ratios to include angles
greater than 90° and less than 0° we consider the rotation of a
straight line OP of fixed length r about the origin O of a
coordinate grid.
Angles are then measured
anticlockwise from the
positive x-axis.
For any angle θ there is an
associated acute angle α
between the line OP and
the x-axis.
α
θ

5 of 47
The three trigonometric ratios are then given by:
sin =
y
r

cos =
x
r

tan =
y
x

The x and y coordinates can be positive or negative, while r is
always positive.
This means that the sign of the required ratio will depend on
the sign of the x-coordinate and the y-coordinate of the point P.

6 of 47
The relationship between θ measured from the positive x-axis
and the associated acute angle α depends on the quadrant
that θ falls into.
If we take r to be 1 unit long then these ratios can be written
as:
sin = =
1
y
y

cos = =
1
x
x

tan =
y
x
 
sin
tan =
cos



For example, if θ is between 90° and 180° it will fall into the
second quadrant and α will be equal to (180 – θ)°.

7 of 47
The sine of any angle
If the point P is taken to revolve about a unit circle then sin θ is
given by the y-coordinate of the point P.

8 of 47
The cosine of any angle
If the point P is taken to revolve about a unit circle then cos θ
is given by the x-coordinate of the point P.

9 of 47
The tangent of any angle
Tan θ is given by the y-coordinate of the point P divided by the
x-coordinate.

10 of 47
The tangent of any angle
Tan θ can also be given by the length of tangent from the point
P to the x-axis.

11 of 47
3rd quadrant
2nd quadrant 1st quadrant
4th quadrant
Tangent is positive
T
Sine is positive
S All are positive
A
Remember CAST
We can use CAST to remember in which quadrant each of
the three ratios are positive.
Cosine is positive
C

12 of 47
The sin, cos and tan of angles in the first quadrant are positive.
In the second quadrant: sin θ = sin α
cos θ = –cos α
tan θ = –tan α
In the third quadrant: sin θ = –sin α
cos θ = –cos α
tan θ = tan α
In the fourth quadrant: sin θ = –sin α
cos θ = cos α
tan θ = –tan α
where α is the
associated
acute angle.

13 of 47
The value of the associated acute angle α can be found using
a sketch of the four quadrants.
For angles between 0° and 360° it is worth remembering that:
when 0° < θ < 90°, α = θ
when 90° < θ < 180°, α = 180° – θ
when 180° < θ < 270°, α = θ – 180°
when 270° < θ < 360°, α = 360° – θ
For example, if θ = 230° we have:
230° is in the third quadrant where only tan is positive and so:
α = 230° – 180° = 50°
sin 230° = –sin 50°
cos 230° = –cos 50°
tan 230° = tan 50°

14 of 47
Contents
14 of 47

15 of 47
The graph of y = sin θ

16 of 47
The graph of y = sin θ
The graph of sine is said to be periodic since it repeats itself
every 360°.
We can say that the period of the graph y = sin θ is 360°.
Other important features of the graph y = sin θ include the
fact that:
It passes through the origin, since sin 0° = 0.
The maximum value is 1 and the minimum value is –1.
Therefore, the amplitude of y = sin θ is 1.
It has rotational symmetry about the origin. In other words, it
is an odd function and so sin (–θ) = –sin θ.

17 of 47
The graph of y = cos θ

18 of 47
The graph of y = cos θ
Like the graph of y = sin θ the graph of y = cos θ is periodic
since it repeats itself every 360°.
We can say that the period of the graph y = cos θ is 360°.
Other important features of the graph y = cos θ include the
fact that:
It passes through the point (0, 1), since cos 0° = 1.
The maximum value is 1 and the minimum value is –1.
Therefore, the amplitude of y = cos θ is 1.
It is symmetrical about the y-axis. In other words, it is an
even function and so cos (–θ) = cos θ.
It the same as the graph of y = sin θ translated left 90°. In
other words, cos θ = sin (90° – θ).

19 of 47
The graph of y = tan θ

20 of 47
The graph of y = tan θ
You have seen that the graph of y = tan θ has a different
shape to the graphs of y = sin θ and y = cos θ.
Important features of the graph y = tan θ include the fact that:
It passes through the point (0, 0), since tan 0° = 0.
It is symmetrical about the origin. In other words, it is an odd
function and so tan (–θ) = –tan θ.
It is periodic with a period of 180°.
tan θ is not defined for θ = ±90°, ±270°, ±450°, … that is, for
odd multiples of 90°. The graph y = tan θ therefore contains
asymptotes at these points.
Its amplitude is not defined, since it ranges from +∞ to –∞.

21 of 47
Contents
21 of 47

22 of 47
Sin, cos and tan of 45°
A right-angled isosceles triangle has two acute angels of 45°.
45°
45°
Suppose the equal sides are of
unit length.
1
1
Using Pythagoras’ theorem:
We can use this triangle to write exact values for sin, cos and
tan 45°:
cos 45° = tan 45° = 1
2
 2
The hypotenuse  
2 2
1 1
1
2
sin 45° =
1
2

23 of 47
2 2
2
60° 60°
60°
2
60°
30°
1
3
Suppose we have an equilateral triangle of side length 2.
We can use this triangle to write exact values for sin, cos and
tan 30°:
If we cut the triangle in half then we have
a right-angled triangle with acute angles
of 30° and 60°.
The height of the triangle  
2 2
2 1
 3
sin 30° =
1
2
cos 30° =
3
2
tan 30° =
1
3

24 of 47
Suppose we have an equilateral triangle of side length 2.
We can also use this triangle to write exact values for sin, cos
and tan 60°:
 3
sin 60° =
3
2
cos 60° =
1
2
tan 60° = 3
2 2
2
60° 60°
60°
2
60°
30°
1
3
If we cut the triangle in half then we have
a right-angled triangle with acute angles
of 30° and 60°.
The height of the triangle  
2 2
2 1

25 of 47
Sin, cos and tan of 30°, 45° and 60°
The exact values of the sine, cosine and tangent of 30°, 45°
and 60° can be summarized as follows:
30°
sin
cos
tan
45° 60°
Use this table to write the exact value of cos 135°
cos 135° = –cos 45° =
1
2
1
2
1
2
1
3
1
2
3
2
3
2
3
1
1
2

26 of 47
Sin, cos and tan of 30°, 45° and 60°
Write the following ratios exactly:
1) cos 300° =
3) tan 240° =
5) cos –30° =
7) sin 210° =
2) tan 315° =
4) sin –330° =
6) tan –135° =
8) cos 315° =
1
2
1
2
1
2
3
2
1
2
3
-1
1

27 of 47
Contents
27 of 47

28 of 47
Equations of the form sin θ = k
Equations of the form sin θ = k, where –1 ≤ k ≤ 1, have an
infinite number of solutions.
If we use a calculator to find arcsin k (or sin–1 k) the calculator
will give a value for θ between –90° and 90°.
This is called the principal solution of sin θ = k.
Other solutions in a given range can be found using the
graph of y = sin θ or by considering the unit circle.
For example:
There is one and only one solution is this range.
Solve sin θ = 0.7 for –360° < θ < 360°
arcsin 0.7 = 44.4° (to 1 d.p.)

29 of 47
44.4°
y = 0.7
y = sin θ
Using the graph of y = sin θ between –360° and 360° and the
line y = 0.7 we can locate the other solutions in this range.
So the solutions to sin θ = 0.7 for –360° < θ < 360° are:
θ = –315.6°, –224.4°, 44.4°, 135.6° (to 1 d.p)
135.6°
–224.4°
–315.6°

30 of 47

31 of 47
44.4°
We could also solve sin θ = 0.7 for –360° < θ < 360° by
considering angles in the first and second quadrants of a unit
circle where the sine ratio is positive.
Start by sketching the principal solution 44.4° in the first
quadrant.
Next, sketch the associated acute
angle in the second quadrant.
135.6°
–224.4° –315.6°
Moving anticlockwise from the
x-axis gives the second solution:
Moving clockwise from the
x-axis gives the third and fourth
solutions:
180° – 44.4° = 135.6°
–(180° + 44.4°) = –224.4°
–(360° – 44.4°) = –315.6°

32 of 47
Equations of the form cos θ = k and tan θ = k
Equations of the form cos θ = k, where –1 ≤ k ≤ 1, and
tan θ = k, where k is any real number, also have infinitely
many solutions. For example:
Solve tan θ = –1.5 for –360° < θ < 360°
Using a calculator, the principal solution is θ = –56.3° (to 1 d.p.)
–56.3°
–236.3°
123.7°
303.7°
Now look at angles in the second and
fourth quadrants of a unit circle where
the tangent ratio is negative.
This gives us four solutions in the
range –360° < θ < 360°:
θ = –236.3°, –56.3°, 123.7°, 303.7°

33 of 47
Equations of the form cos θ = k

34 of 47
Equations of the form tan θ = k

35 of 47
Equations involving multiple or compound angles
Multiples angles are angles of the form aθ where a is a given
constant.
Compound angles are angles of the form (θ + b) where b is a
given constant.
When solving trigonometric equations involving these types of
angles, care should be taken to avoid ‘losing’ solutions.
Solve cos 2θ = 0.4 for –180° < θ < 180°
Start by changing the range to match the multiple angle:
Next, let x = 2θ and solve the equation cos x = 0.4 in the range
–360° ≤ x ≤ 360°.
–180° < θ < 180°
–360° < 2θ < 360°

36 of 47
Now, using a calculator: x = 66.4° (to 1 d.p.)
Using the unit circle to find the values
of x in the range –360° ≤ x ≤ 360°
gives:
66.4°
293.6°
–66.4°
–293.6°
x = 66.4°,
But x = 2θ so:
θ = 33.2°, 146.8°, –33.2°, –146.8°
This is the complete solution set in the range –180° < θ < 180°.
293.6°, –66.4°, –293.6°

37 of 47
Solve tan(θ + 25°)= 0.8 for 0° < θ < 360°
Start by changing the range to match the compound angle:
Next, let x = θ + 25° and solve the equation tan x = 0.8 in the
range 25° < x < 385°.
0° < θ < 360°
25° < θ + 25° < 385°
Using the unit circle to find the values
of x in the range 25° ≤ x ≤ 385° gives:
x = 38.7°, 218.7° (to 1 d.p.)
But x = θ + 25° so:
θ = 13.7°, 193.7.7° (to 1 d.p.)
Using a calculator: x = 38.7° (to 1 d.p.)
38.7°
218.7°

38 of 47
Trigonometric equations involving powers
Sometimes trigonometric equations involve powers of sin θ,
cos θ and tan θ. For example:
Notice that (sin θ)2 is usually written as sin2θ.
Solve 4sin2θ – 1= 0 for –180° ≤ θ ≤ 180°
2
4sin 1= 0
 
2
4sin =1

2 1
4
sin =

1
2
sin =
 
1
2
When sin θ = ,
1
2
When sin θ = – ,
θ = 30°, 150°
θ = –30°, –150°
So the full solution set is θ = –150°, –30°, 30°, 150°

39 of 47
Trigonometric equations involving powers
Treat this as a quadratic equation in cos θ.
Solve 3cos2θ – cos θ = 2 for 0° ≤ θ ≤ 360°
When cos θ = 1, θ = 0°, 360°
θ = 131.8°, 22.8.2°
So the full solution set is θ = 0°, 131.8°, 228.2°, 360°
3cos2θ – cos θ = 2
3cos2θ – cos θ – 2 = 0
Factorizing:
(cos θ – 1)(3cos θ + 2) = 0
cos θ = 1 or cos θ = – 2
3
When cos θ = – ,
2
3

40 of 47
Contents
40 of 47

41 of 47
Two important identities that must be learnt are:
sin
tan (cos 0)
cos

 

 
sin2θ + cos2θ ≡ 1
An identity, unlike an equation, is true for every value of the
given variable so, for example:
The symbol ≡ means “is identically equal to” although an
equals sign can also be used.
sin24° + cos24° ≡ 1, sin267° + cos267° ≡ 1, sin2π + cos2π ≡ 1, etc.

42 of 47
We can prove these identities by considering a right-angled
triangle:
x
y
r
θ
y
r

sin =
x
r

and cos =
sin
=
cos



y
r
x
r
=
y
x
= tan as required.

Also:
y x
r r
 
   
   
   
2 2
2 2
sin +cos = +
2 2
2
+
=
x y
r
But by Pythagoras’ theorem x2 + y2 = r2 so:
2
2 2
2
sin +cos = =1 as required.
r
r
 

43 of 47
71.6°
One use of these identities is to simplify trigonometric
equations. For example:
Solve sin θ = 3 cos θ for 0° ≤ θ ≤ 360°
Dividing through by cos θ:
sin 3cos
=
cos cos
 
 
tan = 3

Using a calculator, the principal solution is θ = 71.6° (to 1 d.p.)
251.6°
So the solutions in the given range are:
θ = 71.6°, 251.6° (to 1 d.p.)

44 of 47
Solve 2cos2θ – sin θ = 1 for 0 ≤ θ ≤ 360°
We can use the identity cos2θ + sin2 θ = 1 to rewrite this
equation in terms of sin θ.
2(1 – sin2 θ) – sin θ = 1
2 – 2sin2θ – sin θ = 1
2sin2θ + sin θ – 1 = 0
(2sin θ – 1)(sin θ + 1) = 0
So: sin θ = 0.5 or sin θ = –1
If sin θ = 0.5, θ = 30°, 150°
If sin θ = –1, θ = 270°

45 of 47
Contents
45 of 47

46 of 47
Examination-style question
Solve the equation
3 sin θ + tan θ = 0
for θ for 0° ≤ θ ≤ 360°
Rewriting the equation using
sin
tan :
cos




sin
3sin + = 0
cos



3sin cos +sin = 0
  
sin (3cos +1) = 0
 
So sin θ = 0 or 3 cos θ + 1 = 0
3 cos θ = –1
cos θ = – 1
3

47 of 47
Examination-style question
In the range 0° ≤ θ ≤ 360°,
when sin θ = 0, θ = 0°, 180°, 360°.
1
3
when cos θ = – :
cos–1 – = 109.5° (to 1 d.p.)
1
3
cos θ is negative in the 2nd and 3rd quadrants so the second
solution in the range is:
250.5°
109.5°
θ = 250 .5° (to 1 d.p.)
So the complete solution set is:
θ = 0°, 180°, 360°, 109.5°, 250 .5°

Trigonometry-1.ppt

Recommended

Recommended

More Related Content

Similar to Trigonometry-1.ppt

Similar to Trigonometry-1.ppt (20)

Recently uploaded

Recently uploaded (20)

Trigonometry-1.ppt