Trigonometry-1.ppt

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AS-Level Maths:
Core 2
for Edexcel
C2.4 Trigonometry 1

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Contents
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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

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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:

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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

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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.

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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)°.

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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.

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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.

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

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

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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

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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.

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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°

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Contents
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The graph of y = sin θ

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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.

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The graph of y = cos θ

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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).

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The graph of y = tan θ

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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 –∞.

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Contents
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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

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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

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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

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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

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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

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Contents
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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.)

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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°

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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°

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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°

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Equations of the form cos θ = k

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Equations of the form tan θ = k

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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°

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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°

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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°

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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°

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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

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Contents
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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.

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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
 

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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.)

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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°

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Contents
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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

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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

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