Trigonometric
functions
The course of many phenomena, sometimes from completely different fields,
can be described by a sinusoidal graph. This is the case with phenomena related
to rotational motion. For example, using trigonometric functions, it is possible
to describe changes in the length of the day in consecutive days of the year
or changes in the alternating current voltage over time.
Trigonometric functions of an acute angle
and negative measures
Angles with positive
Trigonometric functions of any angle
Basic relationships between trigonometric functions
Function graph y = sin 𝛼
graph y = tan 𝛼
Function graph y = cos 𝛼
Reduction formulas
Arc measure of an angle
Trigonometric functions of a real variable
y = sin a x ...
Function
Functions y = a sin x,
Trigonometric equations and inequalities
Sine, cosine and tangent of sum and difference of angles
MLR3-1 [R] str. 5
TRIGONOMETRIC FUNCTIONS
OF AN ACUTE ANGLE
I In this chapter, we review information about the trigonometric functions
of an acute angle in a right-angled triangle. Below we remind you how we
define the trigonometric functions of the acute angle.
sin α = a
c
cos α = b
c
tan α = a
b
EXERCISE A Calculate the sine, cosine and tangent of the angles α and β
marked in the figures below.
If we know the measure of an angle, the values of the trigonometric
functions of this angle can be calculated using a calculator or read from
trigonometric tables. Conversely, if we know the value of a trigonometric
function of any acute angle, we can find its measure.
EXERCISE B a) Using the calculator or the table, enter the values (in rounds to
hundredths): sin 3◦, tan 27◦, cos 65◦.
b) Find measures of acute angles α, β and γ, if sin α = 0,6,
tan γ = 10. Give the results rounded to whole degrees.
For most angles, the values of the trigonometric functions, which can be read from
trigonometric tables or calculated with a
calculator, are approximations. The table
opposite gives the exact values of the
trigonometric functions of the angles 30◦,
45◦ and 60◦.
α
sin α
cos α
tan α
30◦
1
2
√
3
2
√
3
3
cos β = 1 ,
3
45◦
60◦
√
2
2
√
2
2
√
1
3
2
1
2
√
3
EXERCISE C Using the fact that a triangle with angles of 90◦, 60◦, 30◦ is half
an equilateral triangle and a triangle with angles 90◦, 45◦, 45◦ is half a square,
justify the values of the trigonometric functions given in the table.
6
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 6
The shorter leg (cathetus) of a right-angled triangle is 4 long and
one of the angles is 75◦. Calculate the lengths of the remaining sides of the
triangle.
EXAMPLE 1
180◦ − 90◦ − 75◦ = 15◦
The shortest side in a triangle is opposite the
smallest angle.
x
= tan 75◦
4
4
= sin 15◦
y
x = 4 × tan 75◦
y =
x ≈ 4 × 3,7321 ≈ 14,9
y ≈ 0,2588 ≈ 15,5
4
sin 15◦
4
The values of tan 75◦ and
sin 15◦ we read from the table or calculator.
Ans. The lengths of other sides of the triangle are about 14,9 and about 15,5.
PROBLEM
The hypotenuse of a right-angled triangle is 8 long and one the angles
has the measure of 22◦. Calculate the lengths of the legs.
In a right-angled triangle, one of the legs has a length of 4, and
the hypotenuse has a length of 6. Calculate the measures of acute angles of this
triangle.
EXAMPLE 2
cos α =
4
≈ 0,6667
6
α ≈ 48◦
We calculate the value of the adequate trigonometric function.
β = 90◦ − α ≈ 42◦
The angle’s measure we read from
the table or calculator.
Ans. The acute angles of the triangle have measures about 48◦ and about 42◦.
PROBLEM
In a right-angled triangle, the legs are 7 and 9 long. Calculate the measures of acute angles of this triangle.
PROBLEMS
1. Calculate the values of the given trigonometric functions of angles α and β.
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE
MLR3-1 [R] str. 7
7
2. Calculate the length of the side marked with a letter.
3. Calculate the length of the sides marked with a letter.
4. a) Order the numbers sin α, cos α and tan α in the order from the smallest to
the largest, knowing that 0◦ < α < 1◦.
b) Order the acute angles α, β and γ in the order from the angle with the smallest
measure to the angle with the greatest measure, knowing that:
sin α = cos β = tan γ = 1
10
5. a) Find the area of an isosceles triangle where the base length is a and the angle
at the base is α.
b) Calculate the circumference of the rhombus with the acute angle β and the
shorter diagonal of length d.
c) Calculate the length of the chord of the circle with radius r on which the central
angle with the measure γ is subtended.
ANGLES WITH POSITIVE
AND NEGATIVE MEASURES
The photo shows the dial of an old alarm
clock. The yellow hand that is used to
set the alarm time shows 700 . Imagine
that this hand was turned 30◦. What time
would the alarm clock be set then? The
answer depends which direction you turn
the pointer to.
If we had turned the hand clockwise, the alarm would be set to 800 , and if
we turned it counterclockwise, it would be set to 600 .
8
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 8
The ray that revolves around its origin delineates a certain angle. We
assume that the measure of this angle is positive if the ray turns counterclockwise. When the ray is rotated clockwise, the plotted angle has a
negative measure.
Each picture shows the
angle formed by the rotating ray. The arrow in
the drawing shows the
direction of the ray’s rotation and allows you to
define which arm is the
starting and which the
end one.
EXERCISE A Below the figures there are measures of the angles marked with
black arrows. Give measures of the other angles marked with arrows in these
drawings.
Another photo was taken after the yellow
hand was moved from 700 o’clock to 500
o’clock. Consider how this hand was rotated. This could be done in many ways. For
example, make the pointer turn 360◦ and
then rotate it 60◦.
To describe a situation in
which a ray makes more
than one complete revolution around its origin, we
assume that the angle measure can be greater than
360◦ or less than −360◦.
ANGLES WITH POSITIVE AND NEGATIVE MEASURES
MLR3-1 [R] str. 9
9
EXERCISE B In each of the pictures below, the acute angle made up of the two
rays is 32◦. Determine measures of the marked angles.
When we talk about an angle located in the coordinate system, we always
mean the angle whose vertex is at the origin of the
coordinate system, and the
starting arm is on the nonnegative part of the x-axis.
If the end arm of an angle in the coordinate system lies on the y-axis or
on the x-axis, its measure is a multiple of 90◦.
EXERCISE C Give two examples of positive measures of angles and two examples of negative measures of angles placed in the coordinate system so that
they have a common end arm that lies:
a) on the positive part of the y-axis,
b) in the first quadrant of the system,
c) in the third quadrant of the system,
d) in the fourth quadrant of the system.
Notice that the end arms of the angles α,
α+360◦, α−360◦, α+2×360◦, α−2×360◦,
α + 3 × 360◦ etc. coincide. In other words,
for any integer k the end arm of the angle
α coincides with the end arm of the angle
α + k × 360◦.
10
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 10
PROBLEMS
1. The
knob on the old combination lock is set on the
digit 0 (see figure).
a) To open the lock, turn the knob successively through
angles of 36◦, −108◦ and −396◦. Determine which three
digits the combination of this lock consists of.
b) At what angles should you turn the knob one by one
to open the lock, which is locked with the code 375?
2. The measure of the angle α is given below the figure.
Determine measures of the other angles marked with
arrows.
two lines in the picture intersect at an angle of 37◦. Give the measure for
the marked angle.
3. The
4. Sketch in the coordinate system the angle with the given measure.
a) 45◦
b) 135◦
c) −135◦
d) −315◦
e) 405◦
5. Angles α and β are placed in the coordinate system. Determine which quadrants
their end arms lie in if:
a) α = 370◦, β = −370◦
d) α = 545◦, β = −545◦
b) α = 480◦, β = −480◦
e) α = 1200◦, β = −1200◦
c) α = 800◦, β = −800◦
f) α = 4580◦, β = −4580◦
6. Give examples of a few obtuse angles with the end arm lying in the II quadrant
of the coordinate system, the measures of which are:
a) less than −1000◦,
ANGLES WITH POSITIVE AND NEGATIVE MEASURES
MLR3-1 [R] str. 11
b) greater than 1000◦.
11
TRIGONOMETRIC FUNCTIONS
OF ANY ANGLE
EXERCISE A Mark the points A = (4, 0) and B = (4, 3) in the coordinate system.
These points and the point O = (0, 0) are the vertices of a right-angled triangle.
Calculate the values of the trigonometric functions of the angle AOB in this
triangle.
Trigonometric functions can be defined not only for acute angles of a
right-angled triangle. They can also be defined for any angle.
Let α be any angle in the coordinate system
(the vertex of the angle α lies at the point
(0, 0), and the starting arm is on the nonnegative part of the x axis).
On the end arm of angle α, we choose a
point P different from the angle’s vertex.
(Recall that the first coordinate of a point
in the coordinate system is called abscissa,
and the second — ordinate).
r = x2 + y 2
Sine of angle α is the ratio of the ordinate
of point P to the distance r of this point
from the origin of the coordinate system.
y
sin α = r
Cosine of angle α is the ratio of the abscissa of point P to the distance r of this point
from the origin of the coordinate system.
cos α = x
r
If the abscissa of point P is different from
zero, then the tangent of angle α is the
ratio of the ordinate of point P to the abscissa of this point.
y
tan α = x
Note that if the points P1 = (x1 , y1 ) and
P2 = (x2 , y2 ) lie on the end arm of angle
α, then the similarity of the triangles
OP1 A1 and OP2 A2 implies the equali−x
y
r
ties −x1 = y1 = r1 . There from we get:
2
r1 = |OP1 |
r2 = |OP2 |
x1
x
= r2
r1
2
2
2
y1
y
= r2
r1
2
y1
y
= x2
x1
2
Thus, the values of the trigonometric functions of angle α do not depend
on which of point on the end arm of the angle we choose.
Note. It is easy to justify that also when the end arm of an angle lies on one of
the axes of the coordinate system, the values of the trigonometric functions do
not depend on what point on the end arm of the angle we choose.
12
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 12
The figure shows the angle α whose end
arm passes through the point (−6, −2). It
follows that:
√
y
−2 = − 10
= √
r
10
2 10
√
x
−6
−3
10
cos α = = √ =
r
10
2 10
sin α =
tan α =
EXERCISE B
y
= −2 = 1
x
−6
3
Calculate the trigonometric functions values of angles α, β and γ.
In the coordinate system, sketch the
graph of the function y = − 25 x. Select any angle
whose end arm is a part of the graph lying in the
4th quadrant of the coordinate system. Calculate the
values of trigonometric functions of this angle.
EXAMPLE
If x = 5, then y = −
2
× 5 = −2.
5
We chose any point on the end arm, e.g.,
with abscissa 5.
P = (5, −2)
√
|OP| = 52 + (−2)2 = 29
√
−2
−2 29
=
sin α = √
29
29
5
cos α = √
29
=
√
5 29
29
tan α = −
2
5
PROBLEM
The end arm of angle α is a part of the line with the equation y = 0,7x,
which lies in the third quadrant of the coordinate system. Calculate sin α, cos α and
tan α.
EXERCISE C
a) Calculate sin 90◦, cos 90◦, sin 180◦, cos 180◦, tan 180◦.
b) The tangent of 90◦ is undefined. Give examples of other angles that the
tangent is undefined for.
c) The end arm of a certain angle α lies in the III quadrant of the coordinate
system. Determine which numbers among sin α, cos α, tan α are negative.
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE
MLR3-1 [R] str. 13
13
From the definition of trigonometric functions, it follows that tan α can be
any number, and sin α and cos α are numbers in the interval −1 ; 1 , be
cause for any point P = (x, y), the inequalities |x| ≤ x2 + y 2 , |y| ≤ x2 + y 2
are met.
Information about the values of the trigonometric functions of angles from
0◦ to 360◦ are summarized in the table below.
I
quadrant
II
α
0◦
(0◦ ; 90◦)
90◦
(90◦ ; 180◦)
sin α
cos α
tan α
0
1
0
+
+
+
1
0
×
+
−
−
III
IV
180◦ (180◦ ; 270◦) 270◦ (270◦ ; 360◦) 360◦
0
−1
0
−
−
+
−1
0
×
−
+
−
0
1
0
The + sign means a positive value, the − sign means a negative value, × means
that the function is undefined.
For example, we can read from the table that if the end arm of an angle
lies in the I quadrant of the coordinate system, then the values of all
trigonometric functions of this angle are positive. On the other hand, when
the end arm lies in the II quadrant of the coordinate system, only sine of
the angle has a positive value. The rules for the signs of trigonometric
functions help to remember the following rhyme:
In the first — all positive,
in the second — only sine,
in the third — only the tangent,
and in the fourth — cosine.
We already know that for any integer k the end arm of the angle α coincides with the end arm of the angle α + k × 360◦. Hence, the following
equations hold.
sin (α + k ×360◦) = sin α
cos (α + k ×360◦) = cos α
tan (α + k ×360◦) = tan α
In one of the next topics, we will show that there is also equality
tan (α + k×180◦) = tan α, where k ∈ .
14
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 14
PROBLEMS
1. Find a point with both integer coordinates on the end arm of the marked angle.
Calculate the values of the trigonometric functions of this angle.
2. Calculate the values of the trigonometric functions of the marked angle.
3. The
figure next to shows the angle of
inclination of the graph of the function
y = ax + b to the x axis. Show that the
tangent of this angle is equal to the slope
coefficient a.
4. Determine what is the lowest and the highest value that can have the expression.
a) 2 sin α + 5
b) 4 − 1 cos α
2
c) sin2 α − 1
d) 3|cos α| − 0,6
5. Using
the definition of trigonometric functions, find all the angles that satisfy
the condition 0◦ ≤ α ≤ 360◦ and the equality:
a) sin α × cos α = 0
b) sin α = cos α
c) tan α = −1
6. Sketch an angle α in the coordinate system that satisfies the given conditions.
a) sin α = −0,8, cos α > 0
b) cos α = 0,2, sin α < 0
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE
MLR3-1 [R] str. 15
c) tan α = 1 , sin α < 0
3
15
7. Present in the coordinate system two different angles α and β (those whose end
arms do not coincide) satisfying the given condition.
a) sin α = sin β = − 1
3
b) cos α = cos β = 0,4
c) tan α = tan β = 2
8. Point
P = (5, 2) lies on the end arm of angle α (see figure). Calculate sin α
and cos α and the sines and cosines of the angles on the end arms of which the
points P1 , P2 , P3 and P4 lie.
Worth knowing!
A triangle with angles of 90◦,
30◦ and 60◦ is half an equilateral triangle. Therefore, if the
shorter leg of this triangle has
length 1, the other sides have
√
lengths 2 and 3.
We place the angle of 150◦ in
the coordinate system (see figure below).
Since 150◦ = 180◦ − 30◦, in the drawing we can
√
mark a right-angled triangle with sides 1, 2, 3
long.
The vertex P of the marked triangle has the coor√
dinates (− 3, 1), so:
√
√
1
=− 3
sin 150◦ = 1 , cos 150◦ = − 3 , tan 150◦ = √
2
2
− 3
3
9. Using the method in the box above, calculate:
a) sin 120◦
b) cos 210◦
c) cos 240◦
d) tan 300◦
e) sin (−510◦)
10. Find the value of the expression:
a) cos α + cos 2α + cos 3α + cos 4α, when α = −30◦
b) tan α + tan 2α + tan 3α + tan 4α, when α = 60◦
16
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 16
BASIC RELATIONSHIPS BETWEEN
TRIGONOMETRIC FUNCTIONS
There are two trigonometric identities next
to it. These equations hold for all angles for
which appropriate trigonometric functions are
defined.
The first of the identities, called ”the trigonometric one”, holds for all angles. The second identity
is true for the angles whose tangent is determined.
sin2 α + cos2 α = 1
sin α
tan α = cos α
We will justify that for any angle α we have the equality sin2 α + cos2 α = 1.
Proof
Let P = (x, y) denote a point (different from the
point (0, 0)) lying on the end arm of angle α,
and r is the distance of P from the origin (0, 0).
sin α =
y
r
cos α = x ,
Therefore:
sin2 α + cos2 α =
2
y
r
r 2 = x2 + y 2
where
r
2
2
y 2 + x2
+ x =
= r2 = 1
2
r
r
r
EXERCISE Prove that for the angles for which tangent is determined, holds the
equality tan α = sin α .
cos α
EXAMPLE 1
Calculate tan α, knowing that sin α =
2
.
7
cos2 α = 1 − sin2 α
We use the identity
sin2 α + cos2 α = 1.
4
cos α = 1 −
49
45
cos2 α =
49
√
3 5
cos α =
or
7
2
tan α =
tan α =
2
7
√
3 5
7
√
2 5
15
PROBLEM
cos α = −
√
3 5
7
√
2
2 5
= √ =
15
3 5
or
tan α = −
2
or
tan α =
7√
−3 5
7
=−
√
2 5
15
We use the identity
sin α
tan α = cos
α.
√
2 5
15
Calculate tan α, when cos α = − 34 .
BASIC RELATIONSHIPS BETWEEN TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 17
17
Knowing that tan α = −5 and −270◦ < α < −180◦, calculate
sin α and cos α.
EXAMPLE 2
sin α
= −5, so sin α = −5 cos α
cos α
sin α
We use the identity tan α = cos
α.
(−5 cos α)2 + cos2 α = 1
We use the trigonometric one.
25 cos2 α + cos2 α = 1
cos2 α =
1
26
√
cos α =
√
26
26
or
cos α = −
26
26
Of the condition −270◦ < α < −180◦ it follows that the end arm of angle α lies in
the II quadrant of the coordinate system, so
cos α < 0.
Does not meet
the conditions.
√
26
sin α = −5 × −
26
=
√
5 26
26
We have previously established that
sin α = −5 cos α.
PROBLEM
Knowing that tan α = 1 and −180◦ < α < −90◦, calculate sin α and cos α.
EXAMPLE 3
Prove the identity.
3
tan2 α
+ cos2 α = 1
1 + tan2 α
L=
tan2 α + cos2 α(1 + tan2 α)
tan2 α + cos2 α + sin2 α
1 + tan2 α
=
=
=1=R
1 + tan2 α
1 + tan2 α
1 + tan2 α
PROBLEM
Prove the identity
1
tan2 α + 1
= cos2 α.
PROBLEMS
1. a)
Calculate sin α, when cos α = 3 .
4
b) Calculate cos α, when sin α = −0,8.
c) Calculate sin α, when cos α = − 1 and 180◦ < α < 270◦.
5
d) Calculate cos α, when sin α = 2 and −270◦ < α < −180◦.
3
2. a)
Calculate tan α, when sin α = −0,8 and cos α = 0,6.
√
b) Calculate cos α, when sin α = − 10 and tan α = 1 .
10
3√
√
2
5
c) Calculate sin α, when tan α = −
and cos α = − 5 .
5
3
18
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 18
3. Prove the identity.
a) tan2 α − sin2 α = tan 2 α × sin2 α
d)
1
= (1 − sin α)(1 + sin α)
1 + tan2 α
b) cos α + sin α =
e)
2
− tan α + 1
tan α
cos2 α
tan α
c)
1
sin α
cos α + tan α = − 1
sin α − 1
cos α
4. Determine
2
= tan2 α −
1
tan2 α
f) 1 − 2 sin2 α × cos2 α = sin4 α + cos4 α
what is the smallest and the largest value that can assume the ex-
pression
a) 2 sin2 α − cos2 α
b) 5 − 3 sin2 α − 2 cos2 α
c) 1 cos2 α + 3 sin2 α
2
Let sin α×cos α = a and 180◦ < α < 270◦. Express the value of the (sin α − cos α)2
and the expression sin α + cos α using a.
5. a)
b) Let sin β + cos β = b and 90◦ < β < 180◦. Express the value of sin β × cos β and
sin β − cos β using b.
FUNCTION GRAPH y = sin α
EXERCISE A The radius of the circle shown on
the right has the length 1. Determine the second coordinate of point A.
Angle α shown in the figure is the central
angle in a circle with radius 1 (called the
unit circle). Point P is the common point
of the end arm of this angle and the circle,
therefore the distance of point P from the
origin of the coordinate system is equal to
1. So, the second coordinate of point P is
equal to sin α. We will use this property to
plot the function graph y = sin α.
Since sin 0◦ = 0, sin 90◦ = 1 and sin 30◦ = 12 ,
the graph of the functiony = sinα includes the
points (0◦, 0), (90◦, 1) and 30◦, 12 .
FUNCTION GRAPH y=sin α
MLR3-1 [R] str. 19
19
To mark other points of the graph of function y = sin α, we will use the
so-called trigonometric circle.
We draw a circle with a radius of
1 and mark in it, for example, an
angle of 40◦ (see the drawing on
the left). The second coordinate
of point A is sin 40◦.
Acting in the same way as above,
we can mark subsequent points
on the graph of function y = sin α.
All points belonging to this graph,
whose arguments satisfy the condition 0◦ ≤ α ≤ 90◦, form the line
shown in the figure on the left.
Note. The unit on the x-axis has no relation to the unit on the y-axis — their
sizes are chosen arbitrarily.
Proceeding in a similar way, we can plot the function graph y = sin α for
angles greater than 90◦. The figure
below
shows a fragment of this graph
for arguments in the interval 0◦ ; 360◦ .
One can look at the formation of the graph in question a bit differently.
Imagine that the point moves counterclockwise on a circle with radius 1.
The fragment graph shown above corresponds to one complete loop.
20
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 20
It is easy to imagine that after each successive encirclement of the circle,
we will get a line of the same shape, which is the graph of the function for
angles greater than 360◦. If the point moves on the circle in the opposite
direction, i.e. clockwise, we will get a fragment of the plot for negative
angles. This creates the graph of the function y = sin α, called sine wave
(or sinusoid).
EXERCISE B
Using the graph of the function y = sin α, answer the questions.
a) Which number is greater: sin 200◦ or sin 202◦ ?
b) Which of the numbers: sin (−170◦), sin (−190◦), sin 600◦, sin 250◦ is positive?
The graph of function y = sin α is a good illustration of the function’s
various properties. E.g:
• The value of sin α cannot be greater than 1 or less than −1.
The range (set of values) of the function y = sin α is the interval −1; 1 .
−1 ≤ sin α ≤ 1
• When the arguments differ by 360◦, the function values are the same.
sin α = sin (α + 360◦) = sin (α − 360◦) = sin (α + 2×360◦) = . . .
sin α = sin (α + k ×360◦), where k ∈ Note. We say that the functions whose values are repeated regularly are periodic.
The function y = sin α is periodic with a period equal to 360◦. We’ll talk more
about periodic functions in one of the next topics.
• The function takes the greatest value (equal to 1) for the following arguments: 90◦, 90◦ + 360◦,
90◦ + 3×360◦ etc.
90◦ − 360◦,
90◦ + 2×360◦,
90◦ − 2×360◦,
sin α = 1, when α = 90◦ + k ×360◦, where k ∈ FUNCTION GRAPH y=sin α
MLR3-1 [R] str. 21
21
• The function has infinitely many zeros: 0◦, 180◦, −180◦, 2 × 180◦,
−2×180◦ and so on.
sin α = 0, when α = k ×180◦ , where k ∈ • The function assumes positive values when:
0◦ < α < 180◦ or 360◦< α < 540◦, or −360◦ < α < −180◦ etc.
sin α > 0, when α ∈ k ×360◦ ; 180◦ + k ×360◦ , where k ∈ • The function graph has an infinite number of symmetry axes that are
perpendicular to the x-axis. Several of them are shown in the figure. The
line segments marked in blue on the x-axis are of equal length. Thus, the
sine function takes the same value for the angles: α0 , α1 = 180◦ − α0 ,
α0 + 360◦, α1 + 360◦, α0 − 360◦, α1 − 360◦ etc.
sin α = sin α0 , when α = α0 + k ×360◦ or
α = 180◦ − α0 + k ×360◦, where k ∈ EXERCISE C
Using the graph of the function y = sin α, answer the questions.
a) For what arguments does the function take the value −1?
b) For which arguments are the function values negative?
PROBLEMS
1. Use the sine wave to determine which of the following numbers is greater.
a) sin 100◦ or sin 170◦
d) sin 351◦ or sin 622◦
b) sin 633◦ or sin 635◦
e) sin (−442◦) or sin (−277◦)
c) sin (−457◦) or sin (−460◦)
f) sin (−205◦) or sin 193◦
22
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 22
2. Sketch
the appropriate fragment of the sine wave, then read the largest and
smallest values and the zeros of y = sin α, when:
a) −360◦ ≤ α ≤ 0◦
b) 360◦ ≤ α ≤ 540◦
c) −450◦ ≤ α ≤ −270◦
3. Use the sine wave to answer which of the following numbers are negative.
sin(−50◦)
sin 121◦
sin 179◦
sin 194◦
sin(−320◦)
sin 451◦
sin(−721◦)
4. Which sentences are true?
1 Among the angles greater than 100◦ and less than 1000◦, there are three angles
who’s the sine is −1.
2 Among positive and less than 360◦ angles, there are two angles whose sine is
equal to − 13 .
3 For arguments greater than 810◦ and less than 900◦, the function y = sin α does
not take the value 1.
FUNCTION GRAPH y = cos α
The graph of function y = cos α we could obtain using (similarly to the
previous topic) the trigonometric circle. However, an appropriate graph
can also be obtained in another way — by using the sine wave. For this we
will use some relationship between sine and cosine.
The figure opposite shows the points
P and P lying at the same distance
from the origin of the coordinate system. Point P lies on the end arm of
angle α and point P — on the end arm
of the angle α + 90◦. The shaded triangles are congruent. So, if P = (a, b), then
P = (−b, a).
From the definition of sine and cosine, the following equations follow:
cos α = a
r
Hence, we get:
sin (α + 90◦) = a
r
cos α = sin (α + 90◦)
Note. The angle α shown in the figure is an acute angle. You can justify that the
above equality is true for any angle α.
FUNCTION GRAPH y=cos α
MLR3-1 [R] str. 23
23
The graph of function y = cos α is therefore the same as the graph of the
function defined by the formula y = sin(α + 90◦), and the graph of this
function is obtained when we move the sinusoid 90 units to the left along
the x axis.
The figure on the right shows a
fragment of the graph of function y = cos α, which was created after a suitable shift of the
sinusoidal fragment, in blue.
The graph of the function y = cos α, called a cosine wave, has the same
shape as the sine wave, but differs only in its position in the coordinate
system.
EXERCISE A
Using the graph of function y = cos α, answer the questions.
a) Which number is greater: cos 200◦ or cos 202◦ ?
b) Which of the following numbers: cos 300◦, cos(−170◦), cos(−370◦), cos 390◦
is negative?
Based on the cosine wave, various properties of the function y = cos α can
be determined. E.g:
• The range of function y = cos α is the interval −1; 1 .
−1 ≤ cos α ≤ 1
• When the arguments differ by 360◦, the function values are the same.
cos α = cos (α + k ×360◦), where k ∈ Note. Like the function y = sin α, function y = cos α is a periodic function with a
period of 360◦.
24
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 24
• The function takes the highest value (equal to 1) every 360◦. The function
takes the smallest value (equal to −1) with the same regularity.
cos α = 1 , when α = k ×360◦, where k ∈ cos α = −1, when α = 180◦ + k ×360◦, where k ∈ • The function has infinitely many zeros.
cos α = 0, when α = 90◦ + k ×180◦, where k ∈ • The function assumes positive values when:
−450◦ < α < −270◦ or −90◦ < α < 90◦, or 270◦ < α < 450◦ etc.
cos α > 0, when α ∈ −90◦ + k ×360◦ ; 90◦ + k ×360◦ , where k ∈ • The cosine wave has infinitely many axes of symmetry that are perpendicular to the x-axis. Several of them (one of which is the y-axis) are
shown in the figure. The line segments marked in blue on the x-axis are
of equal length. Thus, the cosine function takes the same value for the
angles: α0 , α1 = −α0 , α0 + 360◦, α1 + 360◦, α0 − 360◦, α1 − 360◦ etc.
cos α = cos α0 , when α = α0 + k ×360◦ or
α = −α0 + k ×360◦, where k ∈ EXERCISE B Using the graph of function y = cos α, answer for which arguments the values of this function are negative.
FUNCTION GRAPH y=cos α
MLR3-1 [R] str. 25
25
PROBLEMS
1. Using the cosine wave, determine which of the following numbers is smaller.
a) cos 14◦ or cos 62◦
c) cos (−83◦) or cos 192◦
b) cos (−342◦) or cos (−339◦)
d) cos (−176◦) or cos 98◦
2. Sketch the appropriate part of the cosine wave and read the largest value, smallest value, and zeros of the function y = cos α when:
a) −360◦ ≤ α ≤ 0◦
c) 450◦ ≤ α ≤ 630◦
b) 180◦ < α < 360◦
d) −90◦ < α < 540◦
3. Sketch the sine wave and cosine wave in one coordinate system. Using the figure,
compare the numbers:
a) sin 50◦ and cos 50◦
d) sin 170◦ and cos (−2◦)
b) sin 294◦ and cos 294◦
e) sin (−297◦) and cos (−157◦)
c) sin (−127◦) and cos (−127◦)
f) sin (−354◦) and cos 354◦
4. The
figures below show fragments of graphs of the function y = sin α and
y = cos α. Which of them are sine wave fragments, and which are cosine wave
fragments?
5. Explain why there is no angle that satisfies the given equality.
a) sin α + cos α = 2
b) 2 + cos α = −1
2 + sin α
c) sin α × cos α = 2
6. The graph of function y = cos α can be obtained from the graph of function
y = sin α by reflecting the sine wave symmetrically about the line perpendicular to
the x-axis, passing through the point (45◦, 0). Find equations of all lines having
the
same property and which intersect the x-axis in the interval −360◦ ; 360◦ .
26
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 26
FUNCTION GRAPH y = tan α
EXERCISE The line P Q shown in the figure is
perpendicular to the x-axis and passes through
the point (1, 0). Determine the coordinates of
point P and the coordinates of point Q.
The angle α shown in the figure is the central angle in the unit circle with the center
(0, 0). The line k is tangent to the circle and
perpendicular to the x-axis. On the end arm
of α we select point P = (1, b) which lies also on line k, The second coordinate b of
point P is therefore equal to tan α.
Note. The tangent owes its name to the described above property. In Latin, tangere means
touch.
The figure below shows how to
belonging to the graph of the
select points
◦
◦
function y = tan α, when α ∈ −90 ; 90 .
It is known that tangent is undefined for 90◦ and −90◦ angles, so the plot
will not cross the green lines (which are parallel to the y-axis). Those lines
are the (vertical) asymptotes of the function graph.
FUNCTION GRAPH y=tan α
MLR3-1 [R] str. 27
27
We already know what the graph of y = tan α looks like when −90◦ < α <
90◦. To plot the graph for remaining arguments, we’ll use some tangent’s
property.
The figure opposite shows points P and P lying at the same distance from the origin
of the coordinate system. Point P lies on
the end arm of angle α and point P — on
the end arm of angle α + 180◦. The shaded
triangles are congruent, so if P = (a, b), then
P = (−a, −b). Therefore:
tan (α + 180◦) = −b = b
tan α = b
a
−a
a
Thus, we get:
tan α = tan (α + 180◦)
Note. Angle α shown in the figure is an acute angle. It can be shown that the
above equality is true for each angle α for which tangent is a real number.
The equality tan α = tan (α + 180◦) means that the values of the tangent
function repeat every 180◦, which is used to plot the graph of y = tan α,
called the tangensoid (see the figure below).
Based on the tangensoid, different properties of the function y = tan α can
be determined. For example:
• The function is not defined for angles with measures 90◦ + k × 180◦ ,
where k ∈ . Lines parallel to the y-axis, crossing the x-axis at points
with the first coordinate α = 90◦ + k × 180◦ , where k ∈ , are asymptotes
of the function graph.
• The range of function y = tan α is the set of real numbers.
• The function has infinitely many zeros.
tan α = 0, when α = k ×180◦ , where k ∈ 28
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 28
• For arguments that differ by 180◦, the function values are the same.
tan α = tan (α + k ×180◦), where k ∈ Note. The function y = tan α is a periodic function with a period of 180◦.
• The function assumes positive values when:
0◦ < α < 90◦ or 180◦ < α < 270◦ , or −180◦ < α < −90◦ etc.
tan α > 0 , when k ×180◦ < α < 90◦ + k ×180◦, where k ∈ • The function is increasing in every interval −90◦ + k×180◦ ; 90◦ + k×180◦ ,
where k ∈ .
PROBLEMS
1. Sketch the appropriate fragment of the tangensoid and read:
a) the zeros of function y = tan α, when −180◦ ≤ α ≤ 270◦,
b) for which angles α satisfying the condition −360◦ ≤ α ≤ 540◦, tangent cannot be
specified,
c) for which arguments the function y = tan α takes negative values.
2. Using the
graphs of the trigonometric functions, arrange the given numbers in
order from lowest to highest.
a) sin 85◦, cos 85◦, tan 85◦
c) sin 80◦, cos (−85◦), tan (−80◦)
b) sin 92◦, cos 92◦, tan 92◦
d) sin (−10◦), cos 50◦, tan (−95◦)
3. Find all the angles that meet the given condition:
a) tan α = tan 78◦
c) tan α = tan (273◦)
b) tan α = tan (−3◦)
d) tan α = tan (−319◦)
FUNCTION GRAPH y=tan α
MLR3-1 [R] str. 29
29
4. Among
angles meeting the given
condition find those whose tangent is
equal to tan 72◦.
a) −360◦ ≤ α ≤ 0◦
b) 90◦ < α < 720◦
c) −180◦ ≤ α ≤ 90◦
tan α = tan 50◦
d) −90◦ < α < 450◦
α = 50◦ + k × 180◦, where k ∈ REDUCTION FORMULAS
Let’s take a look at sinusoid once
more. One of its axes of symmetry is marked in the picture. This
figure illustrates the equality:
sin α = sin (180◦ − α)
The sine wave also has many centers
of symmetry. One of them is the origin
of the coordinate system. For arguments with opposite signs, the function values are opposite. The picture
illustrates the equality:
sin (−α) = − sin α
Using the trigonometric tables, we can
find sines of acute angles. By using the
sine wave or the formulas presented in
the box besides, it is possible to reduce
calculation of sine of any angle to finding
the sine of an acute angle.
30
sin α = sin (α + k ×360◦),
where k ∈ sin α = sin (180◦ − α)
sin (−α) = − sin α
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 30
EXAMPLE 1
Calculate sines of the angles: 147◦, −57◦, 250◦, 315◦, −243◦.
Method I
Method II
We use the sine wave (we find appropriate
segments of the same length).
We use the formulas.
sin 147◦ = sin (180◦ − 33◦) =
= sin 33◦ ≈ 0,5446
sin (−57◦) = − sin 57◦ ≈ −0,8387
sin 250◦ = sin (180◦ − 250◦) =
= sin (−70◦) = − sin 70◦ ≈ −0,9397
sin 315◦ = sin (315◦ − 360◦√
)=
= sin (−45◦) = − sin 45◦ = −
2
2
sin (−243◦) = − sin 243◦ =
= − sin (180◦−243◦) = − sin (−63◦) =
= sin 63◦ ≈ 0,891
PROBLEM
Calculate sines of the angles: 102◦, −83◦, 200◦, 307◦, −210◦.
REDUCTION FORMULAS
MLR3-1 [R] str. 31
31
The graph of function y = cos α
has infinitely many axes of symmetry. One of them is the y
axis. The figure opposite illustrates the equality:
cos (−α) = cos α
The cosine graph also has many symmetry centers. One of them is point
(90◦, 0). The figure opposite illustrates
the equality:
cos α = − cos (180◦ − α)
As for the y = sin α function, with
the help of trigonometric tables, we
can calculate the cosines of any angles, using the cosine graph or the
formulas given here.
EXAMPLE 2
cos α = cos (α + k × 360◦),
where k ∈ cos α = − cos (180◦ − α)
cos (−α) = cos α
Find cosines of the angles −30◦, 159◦ and 312◦.
Method I
Method II
We use the cosine graph.
We use the formulas.
√
3
cos (−30◦) = cos 30◦ = 2
cos 159◦ = − cos (180◦ − 159◦) =
= − cos 21◦ ≈ −0,9336
cos 312◦ = cos (312◦ − 360◦) =
= cos (−48◦) = cos 48◦ ≈ 0,6691
PROBLEM
Calculate cosines of the angles: −70◦, 225◦, 350◦.
32
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 32
The graph of the function y = tan α is center symmetric. One of the centers of its symmetry is the
origin of the coordinate system (see the drawing on
the right).
So the following equality is satisfied:
tan (−α) = − tan α
tan α = tan (α + k × 180◦),
where k ∈ tan (−α) = − tan α
EXAMPLE 3
With the help of trigonometric tables, you can calculate the tangent of
any angle, using the function graph
or the formulas opposite.
Calculate tangents of the angles: −135◦, 215◦, 120◦.
Method I
Method II
We use the tangensoid.
We use the formulas.
tan (−135)◦ = tan (−135◦ + 180◦) =
= tan 45◦ = 1
tan 215◦ = tan (215◦ − 180◦) =
= tan 35◦ ≈ 0,7
tan 120◦ = tan (120◦ − 180◦) = tan (−60◦) =
√
= − tan 60◦ = − 3
PROBLEM
Calculate tangent of the angles: −100◦, 225◦, 150◦.
REDUCTION FORMULAS
MLR3-1 [R] str. 33
33
The formulas that allow to reduce the problem of calculating the values
of trigonometric functions to finding the values of these functions for
appropriate acute angles are called reduction formulas.
In addition to the formulas used so far, there are also other reduction
formulas. Below are formulas that we will use.
Basic reduction formulas
sin (−α) = − sin α
cos (−α) = cos α
tan (−α) = − tan α
sin (180◦ − α) = sin α
cos (180◦ − α) = − cos α
tan (180◦ − α) = − tan α
sin (180◦ + α) = − sin α
cos (180◦ + α) = − cos α
tan (180◦ + α) = tan α
sin (90◦ − α) = cos α
cos (90◦ − α) = sin α
sin (90◦ + α) = cos α
cos (90◦ + α) = − sin α
Note. Each of these formulas can be applied to any angle α for which the corresponding function is specified.
PROBLEMS
In the following problems, give the exact result (if possible) or use trigonometric tables to
approximate the result to the nearest hundredths.
1. Calculate.
a) sin 150◦, sin 120◦
c) sin (−80◦), sin (−10◦)
e) sin 303◦, sin 342◦
b) sin (−60◦), sin (−45◦)
d) sin 238◦, sin 254◦
f) sin (−222◦), sin (−316◦)
2. In the interval
−360◦ ; 360◦ , find all angles satisfying the condition.
a) sin α = sin 100◦
b) sin α = sin(−40◦)
c) sin α = sin 230◦
all the angles α that satisfy the condition −360◦ ≤ α ≤ 360◦ and the
equality:
3. Find
a) sin α =
√
2
2
√
b) sin α =
3
2
c) sin α = − 1
2
√
d) sin α = − 2
2
4. Find the angles that meet the given conditions.
a) sin α = 0,26 and −180◦ < α < 180◦
c) sin α = −0,1 and −360◦ < α < 0◦
b) sin α = 0,99 and −360◦ < α < 360◦
d) sin α = −0,53 and −90◦ < α < 270◦
34
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 34
5. Using the appropriate graphs, calculate:
a) tan 150◦, tan 135◦
c) tan (−17◦), tan (−82◦)
e) tan (−166◦), tan (−134◦)
b) tan 210◦, tan 225◦
d) tan 205◦, tan 197◦
f) tan 290◦, tan 324◦
6. Calculate using the reduction formulas.
a) sin 410◦, cos 375◦, tan 415◦
c) sin (−410◦), cos (−490◦), tan (−480◦)
b) cos 550◦, sin 600◦, tan 690◦
d) cos (−610◦), sin (−730◦), tan (−800◦)
7. Justify the identity.
a) sin (45◦ + α) = cos (45◦ − α)
b) cos (45◦ + α) = sin (45◦ − α)
8. Give the exact value of the expression:
a) sin 130◦ × cos 40◦ − cos 130◦ × sin 40◦
c)
cos 10◦ × cos 30◦ × cos 50◦
sin 40◦ × sin 60◦ × sin 80◦
b) tan 20◦ × sin 70◦ + cos 110◦
d)
sin 25◦ × sin 50◦ × sin 75◦
sin 105◦ × sin 130◦ × sin 155◦
9. Simplify the expression then calculate its exact value.
a) cos 225◦ × tan 315◦ − cos 135◦
b)
cos 225◦
× tan 135◦
sin 315◦× sin 240◦
10. Justify that the value of the expression:
tan(α − 180◦) × sin(α − 90◦) − cos(450◦ + α)
is the same for any angle α other than 90◦ + k × 180◦.
ARC MEASURE OF AN ANGLE
Until now, we used the degree measure when determining the size of an
1
angle. Its basic unit is 1◦ (such a measure has an angle, which is 90 of a
right angle.
The degree measure is not the only measure of angles used. For example,
in geodesy, the sizes of angles are often specified in terms of units called
gradians. A measure of 1 gradian (abbreviated as 1 grad) has the angle that
1
is 100
of a right angle.
ARC MEASURE OF AN ANGLE
MLR3-1 [R] str. 35
35
Worth knowing!
Gradians and (angular) degrees have an interesting relationship with units
of length. The distance of 1 km was originally defined by the length of the
meridian: it was assumed that 10 000 km is equal to half the length of the
meridian, i.e., it tells for example the distance measured along the meridian
between a pole and the equator (the difference in latitude of these points is
1
right angle, so an angle of 1 grad is equal to 100 km
90◦). The gradian is 100
(measured along the meridian).
Another unit of length, a nautical mile, was also defined by the length of
the meridian. It is defined as a distance along the meridian corresponding
to 1 (one minute) of latitude, i.e., an angle of 1◦ corresponds to 60 nautical
miles.
Maritime navigation often makes use of the relationship between the angle measure (geographical coordinates) and distance (voyage length). It
is then inconvenient to use degrees as the angle
unit and the distances in kilometers. Usually, angles are given in degrees and distances are given
in nautical miles, or angles are given in gradians
and distances are given in kilometers.
We will now discuss one more way of defining the measure of an angle.
The central angle marks an arc on the circle, the length of which depends
on the measure of the angle and the length of the radius of the circle.
EXERCISE A For each of the three
circles shown opposite, calculate the
ratio of the length of the marked arc
to the radius.
l1 =
α × 2πr
1
360◦
l2 =
α × 2πr
2
360◦
l1
l
= r2 = α ◦ × 2π
r1
360
2
The figure opposite shows the central angle α, which cuts arcs of different lengths from two circles. Although the lengths of these arcs depend on the length of the radii of
the circles, the ratio of the length of
the arc to the radius of the circle is
the same in each case.
The arc measure of the central angle in a circle is the ratio of the length
of the arc on which this angle is subtended, to the length of the radius of
the circle.
36
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 36
Notice that the central angle, which is a
straight angle, is based on the semi circle.
So, the arc measure of the angle of 180◦ is
1
2 ×2π r
r
. Hence we get that:
arc measure of the angle 180◦ is π
arc measure of the angle 90◦ is π
2
arc measure of the angle 360◦ is 2π
Calculate the arc measures of the angles: 30◦, 60◦ and 45◦.
EXERCISE B
The arc measures of some angles are summarized in the table below.
degree measures
arc measures
360◦ 180◦
2π
π
90◦
60◦
45◦
30◦
π
2
π
3
π
4
π
6
The notation sin π denotes the sine of an angle whose arc mea-
EXERCISE C
6
sure equals π , so sin π = sin 30◦ = 1 . Give the values of cos π , tan π , sin π .
6
6
2
4
3
2
Since the arc measure of the angle 180◦ is π , then if α is the measure
of an angle in degrees, and x — the arc measure of this angle, then the
relationship between α and x will be determined by the proportion:
α
x
=
180◦ π
α — degree measure of the angle
x — arc measure of the angle
We determined the arc measure of the angles for the centralangles in
◦
◦
a circle. The above equality assigns
each
angle in the interval 0 ; 360
some number from the interval 0 ; 2π . We can ”extend” this assignment
to angles greater than 360◦ and negative angles. We assume that the above
proportion determines the relationship between the degree measure of any
angle and the arc measure of that angle.
An angle with an arc measure of 1 is called a radian. So, 1 radian (abbreviated as 1 rad) is the central
angle that cuts an arc in a circle equal to the radius
of the circle. Thus, in an angle with an arc measure
a radians is composed of a angles of measure 1 rad.
If we say the angle has an arc measure of e.g., π , we
7
mean the angle π rad.
7
ARC MEASURE OF AN ANGLE
MLR3-1 [R] str. 37
37
It’s easy to calculate how many degrees there are in the angle of 1 rad.
◦
If x = 1, then transforming the above proportion we get α = 180 .
π
◦
1 rad = 180 ≈ 57◦17 45
π
So:
EXAMPLE
a) Express the angle of 40◦ in radians.
x — arc measure of the angle of 40◦
40◦
x
=
π
180◦
α
180◦
2
x
=
9
π
x=
3
c) Express the angle of 5 π radians
in degrees.
3
α — degree measure of the 5 π angle
=
3
π
5
π
3
=
5
180◦
α
2
π
9
α=
3
× 180◦
5
α = 108◦
b) Express the angle −400◦ in radians.
d) Express the angle −3π radians in degrees.
x — arc measure of the angle − 400◦
α — degree measure of the angle −3π
− 400◦
x
=
π
180◦
x=
α
180◦
α
−400◦×π
180◦
x =−
180◦
−3π
π
= −3
α = (−3) × 180◦
20π
9
PROBLEM
=
α = −540◦
a) Express the angle 420◦ in radians.
b) Express the angle of
5
4
radians in degrees.
PROBLEMS
1. a)
In a circle with radius 5, a certain central angle designates an arc of length 3.
Find the arc measure of this angle.
√
b) In a circle with a radius of 2 3, the central angle with arc measure of 3 π
4
designates an arc. Calculate the length of this arc.
c) Calculate the arc measure of the central angle supported by 1 of the circle.
10
2. The radius of each of the circles shown
here has a length of 4. The lengths of the
arcs are given. Calculate how many radians the angles α, β and γ have.
38
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 38
3. Determine:
a) measure in radians of angles:
720◦
15◦
10◦
22,5◦
75◦
9◦
b) measure in degrees of angles with arc measures:
2π
3
3π
4
5π
6
6π
π
8
7π
6
4. Give a value of:
cos π
6
sin π
4
tan π
6
cos 0
sin π
tan 2π
5. If a body moves in a circle, its angular velocity can be established, i.e., the angle
by which it rotates per unit time. Express the angular velocity in rad/h of:
a) minute hand of the clock,
b) the Earth rotating around its axis,
c) a carousel that performs 1 revolution in 5 seconds.
TRIGONOMETRIC FUNCTIONS
OF A REAL VARIABLE
The trigonometric functions, the properties of which we discussed earlier, to the measure of an angle (in degrees) assigned a certain number.
The properties of these functions (including trigonometric identities and
reduction formulas) are also satisfied when the angle measure is given in
radians instead of degrees.
So far, we have drawn graphs of trigonometric functions in the coordinate
system where the units on the horizontal and vertical axes were different.
EXERCISE A Redraw the axis below. Under the marked points, write down the
appropriate arc measures, and then mark the point 1.
If we convert the numbers of degrees to the corresponding numbers of
radians, we can consider trigonometric functions whose arguments are
real numbers.
TRIGONOMETRIC FUNCTIONS OF A REAL VARIABLE
MLR3-1 [R] str. 39
39
Fragments of graphs of the functions y = sin x and y = cos x are presented
below. The domain of both functions is the set of real numbers. Below each
graph, several properties that can be read from it are given.
• The zeros of the function y = sin x are numbers of the form kπ , where
k ∈ .
• The function y = sin x takes the smallest value −1 for numbers of the
form x = − π2 + 2kπ , where k ∈ .
• The
function y = sin x is increasing in each interval of the form
π
π
− 2 + 2kπ ; 2 + 2kπ , where k ∈ .
EXERCISE B
Using the graph of the function y = sin x, answer the questions.
a) Is sin 10 positive or negative a number?
b) Which number is greater: sin (−5) or sin (−6)?
c) Is sin (−1) + sin 3 positive or negative a number?
• The zeros of the function y = cos x are numbers of the form
where k ∈ .
π
2
+ kπ ,
• The function y = cos x takes the largest value of 1 when x = 2kπ , where
k ∈ .
•
The function y = cos x is decreasing in each interval of the form
2kπ ; π + 2kπ , where k ∈ .
EXERCISE C
Using the graph of the function y = cos x, answer the questions.
a) Is cos (−6) positive or negative a number?
b) Which number is greater: cos 12 or cos − 41 ?
c) Is cos (−3) + cos 4 positive or negative a number?
40
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 40
The figure below shows a fragment of the graph of the function y = tan x.
This function is not defined for numbers of the form π2 + kπ , where k ∈ .
π
Then, the domain of this function is the set \ {x : x = 2 + kπ, where k ∈ }.
• The asymptotes of the function y = tan x intersect the x-axis at points
with the abscissa π2 + kπ , where k ∈ .
• The zeros of the function y = tan x are numbers of the form kπ , where
k ∈ .
• The function y = tan x takes negative values for x ∈ − π2 + kπ ; kπ ,
where k ∈ .
• The function y = tan x is increasing in each interval − π2 + kπ ; π2 + kπ ,
where k ∈ .
There is some repeatability in the graphs of the trigonometric functions.
For example, for the function y = tan x the value for any argument a is the
same as for arguments a + π , a − π , a + 2π , a − 2π , etc. In other words,
the same value repeats regularly. Functions having this property we call
periodic.
We call a function f periodic function if there is such a number t > 0 that
for each argument x the number x + kt (where k ∈ ) also belongs to the
domain of the function and f (x) = f (x + kt). The number t is called a period
of the function. The smallest positive period of a function (if it exists) is
called the basic period.
The trigonometric functions are periodic because for any argument x and
for any integer k, the following equations hold:
sin x = sin (x + 2kπ)
cos x = cos (x + 2kπ)
tan x = tan (x + kπ)
The basic period of the function y = sin x and the function y = cos x is
number 2π . However, the basic period of the function y = tan x is number π .
TRIGONOMETRIC FUNCTIONS OF A REAL VARIABLE
MLR3-1 [R] str. 41
41
The y-axis is the symmetry axis of the graph of function y = cos x. Functions with this property are called even functions.
The function y = cos x is an even function, so cos (−x) = cos x.
The origin of the coordinate system is the center of symmetry of the graph
of function y = sin x and the graph of the function y = tan x. Functions
with this property are called odd functions.
The functions y = sin x and y = tan x are odd functions,
so sin (−x) = − sin x and tan (−x) = − tan x.
Worth knowing!
By treating the trigonometric functions as functions of a real variable, we can compare the shape of their graphs with other curves.
The figure opposite shows a fragment
of the sinusoid, a semicircle, and a frag
ment of parabola with vertex π , 1
2
and zeros 0, π. As you can see,
the
shape of sinusoid in the interval 0 ; π
differs significantly from the semicircle
and is also not identical to the shape of
the marked parabola.
The following figures compare the graphs of trigonometric functions with those of
other functions.
Graph of the function
y = sin x and the line
y = x have one common point.
Graph of the function
y = cos x and the parabola y = −x2 + 1 have one
common point.
Graph of the function
y = tan x and the curve
y = x3 have
in the interval − π ; π one common
2 2
point.
Sine and cosine graphs have the same shape. One of these curves can be
obtained from the other as a result of a π2 shift along the x-axis. This
relationship is expressed by the identities below.
sin x = cos x − π2
cos x = sin x + π2
By moving the graphs of trigonometric functions, we can create graphs of
other functions and read their properties.
42
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 42
EXAMPLE
.
Sketch the graph of the function y = 1 − sin x + π
4
First, we sketch the graph y = sin x. By
, 0 , we obtain
shifting it by vector − π
4
.
graph y = sin x + π
4
Reflecting the graph y = sin x + π
sym4
metrically about the x-axis, we get the
graph y = − sin x + π
4 .
After moving the graph y = − sin x + π
4
by vector [0, 1] we get the graph
y = 1 − sin x + π
4 .
PROBLEM
Sketch the graph of the function y = −2 + cos x − π .
3
PROBLEMS
1. Sketch the graph of y = sin x.
a) List all numbers from the interval
tion are equal to 1.
−2π ; 2π
for which the values of this func-
b) Determine how many zeros the function has in the interval 0 ; 8π .
c) Which of the following numbers are positive?
sin − 3 π
sin 3π
sin π
8
4
sin (2π + 1)
d) Order from smallest to largest number:
sin 3
sin 3,5
sin 3,14
e) Determine for how many arguments in the interval
function are integers.
sin π
0 ; 60 the values of this
2. Sketch the graph of y = cos x.
a) List the zeros of this function belonging to the interval −2π ; 2π .
b) Provide five arguments for which the function values are −1.
c) Order the given numbers from lowest to highest.
cos 2
cos 3
TRIGONOMETRIC FUNCTIONS OF A REAL VARIABLE
MLR3-1 [R] str. 43
cos 4
cos 5
43
d) Which of the following numbers are negative?
cos 10π
cos − π
cos 3
cos − π − 1
2
4
e) Determine how many numbers there are in the interval −10π ; 10π , for which
the function values are integers.
3
3. a)
Give the monotonicity intervals of the function y = sin x and the function
y = cos x.
b) Determine for which arguments the values of the function y = sin x are positive,
and for which — the values of the function y = cos x.
c) Find all arguments for which the function values y = sin x and y = cos x are
negative at the same time.
4. Sketch the graph of y = tan x.
a) How many zeros does this function have in the interval (−5π ; −π )?
b) How many numbers in the interval −3π ; 3π does not belong to the domain
of the function?
c) Give examples of arguments for which the function values are less than −1.
d) Order the given numbers from lowest to highest.
tan (−1)
tan 1
tan 2
tan 3
tan 4
5. Determine which of the given points belong to the graph of y
= sin x, which —
to the graph of y = cos x, and which — to the graph of y = tan x.
√
A = (0, 1)
C = −3 π, 0
E = π,1
G = −π , − 3
2
B = − π , −1
2
D = −3 π, 1
2
6 2
6
F = − π , −1
H=
4
π,
3
3
√
3
6. Define the domain and range.
a) y = sin2 x − 3 cos2 x
7. a)
c) y = sin x
b) y = tan x × cos x
Does the graph of the function y =
the sine wave or the cosine wave?
1
x
10
tan x
d) y = −2
cos x
have more points in common with
b) How many common points does the sinusoid with the function graph y = 4 x
have at least, and how many with the graph of function y = − 2 x?
5π
5π
8. Sketch the appropriate graph and solve the equation:
a) sin x = cos x
c) sin x + cos x = 0
e) |cos x| + sin x = 1
b) cos x = 2 − sin x
d) 1 + sin x − cos x = 0
f) |sin x| − cos x = 0
44
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 44
FUNCTIONS y = a sin x, y = sina x ...
EXERCISE Give the values of the function y = sin x and the function y =
2 sin x for arguments: 0, π , − π , π , −π. Do the graphs of these functions have
4
4 2
common points?
In the figures below, the black color shows part of the sinusoid, and the
blue color shows parts of graphs of four functions of the type y = a sin x.
The range of the function
y = 2 sin x
is the interval −2 ; 2 . This function has
the same zeros as the function y = sin x.
The range of the function
y = 32 sin x
3 3
is the interval − 2 ; 2 . This function has
the same zeros as the function y = sin x.
The range of the function
y = 12 sin x
1 1
is the interval − 2 ; 2 . This function has
the same zeros as the function y = sin x.
The range of the function
y = 34 sin x
3 3
is the interval − 4 ; 4 . This function has
the same zeros as the function y = sin x.
The graph of a function of the type y = a sin x, where a > 0, resembles the
sine wave ”compressed” or ”stretched” along the y-axis.
When a < 0, we additionally need to reflect the appropriate graph symmetrically with respect to the x-axis.
Similarly, graphs of the function y = a cos x can be obtained.
FUNCTIONS y= sin x, y=sin x...
MLR3-1 [R] str. 45
45
In the figures below, a part of the sine wave is drawn in black, and parts
of the graphs of y = sin bx are marked in blue.
The basic period of the function
y = sin 2x is 2 times smaller than the
period of the function y = sin x, so it
is π.
The basic period of the function
y = sin 3x is 3 times smaller than the
period of the function y = sin x, so it
is 23 π.
The basic period of the function
The basic period of the function
y = sin
1
x
2
is 2 times greater than the pe-
riod of the function y = sin x, so it is 4π.
y = sin 3 x is equal to the quotient of 2π
4
by 3 , so it is equal to 8 π.
4
3
Each of these graphs resembles a ”squeezed” or ”stretched” sine wave
.
along the x-axis. Basic period of the function y = sin bx is equal to 2π
|b|
The period of the function y = cos bx can be determined in the same way.
EXAMPLE
2
3
Sketch the graph of function y = −2 sin x.
First, we sketch the graph
2
of function y = sin 3 x. The
period of this function is
2π
2
3
= 3π .
Then we sketch the graph
of function y = 2 sin 23 x.
After the symmetric reflection of the graph of function y = 2 sin 23 x about the
x-axis, we get the graph of
function y = −2 sin 23 x.
PROBLEM
Sketch the graph of function y = 1 cos 2x.
46
2
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 46
Worth knowing!
Curves that resemble the sine wave appear when describing various physical
phenomena, especially those related to
rotation. For example, when we rotate
a metal frame at a constant speed in a
uniform magnetic field, an alternating
current is produced in it. The diagram
on the right shows how the intensity of
this current changes.
PROBLEMS
1. Sketch the graph and read the zeros of the function.
b) y = 5 sin x
a) y = 2 cos x
c) y = sin 1 x
4
d) y = cos 6x
4
2. Give the set of values and the basic period of the function.
a) y = 3 cos x
c) y = cos 3x
e) y = 5 sin 4x
g) y = 4 sin π x
b) y = 3 sin x
d) y = sin 1 x
f) y = 1 cos 3 x
h) y = 2 cos π x
4
3
2
2
2
7
3. Sketch the function graph and determine the set of values.
b) y = − 1 sin 1 x
a) y = 2 sin 3x
3
2
c) y = 3 cos 2 x
3
d) y = − 2 cos 3 x
3
2
4. The figures below show fragments of graphs of functions:
1 y = − cos 1 x
3 y = −2 cos 2x
5 y = 1 cos 2x
2 y = − 1 cos 2x
2
4 y = −2 cos 1 x
2
6 y = 2 cos 1 x
2
2
2
Match formulas to graphs.
FUNCTIONS y= sin x, y=sin x...
MLR3-1 [R] str. 47
47
A pen on a spring is attached to
pendulum (see figure). When the
then the curve the pen draws on
tape has the shape of the graph
y = a sin bx.
the weight of the
pendulum moves,
the moving paper
of a function like
The time of one swing (back and forth) is called the
pendulum period. The amplitude of the pendulum
motion is the greatest distance the pendulum weight
moves away from the vertical.
5. a) The figure opposite shows the line
drawn by the pen. Read the period
and amplitude of the pendulum motion. Give the formula of the function
whose plot is this line.
b) For another pendulum, the stylus
traced a curve that is the graph of the
function y = 3 sin 2π x . Give the period
4
5
and movement amplitude of this pendulum.
TRIGONOMETRIC EQUATIONS
AND INEQUALITIES
EXERCISE A The figure shows a fragment of the function graph y = sin x. Give
the first coordinates of the points marked with dots.
Let’s consider an equation of the type: sin x = a.
Using the sine wave, it is easy to see that such an equation has no solutions
when |a| > 1, while in other cases the equation has an infinite number of
solutions. To find all solutions of an equation of this type, it is enough
to find solutions
from the interval of length 2π , for example from the
interval −π ; π then use the periodicity of the function y = sin x.
48
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 48
When a = 1 or a = −1 or a = 0, the solutions of the corresponding equations
can be easily read from the graph.
sin x = 1, when x = π + 2kπ, where k ∈ 2
sin x = −1, when x = − π + 2kπ, where k ∈ 2
sin x = 0, when x = kπ, where k ∈ Suppose a ∈ −1 ; 0 or a ∈ 0 ; 1 . Then in the 2π long interval there
are two numbers x1 and x2 satisfying the equation sin x = a (see figures
below).
Thus, using the periodicity of the function y = sin x, we can define two
families of solutions to this equation:
x = x1 + 2kπ or x = x2 + 2kπ , where k ∈ Note that from the symmetry of the sinusoid it follows:
If a ∈ 0 ; 1 , then x2 = π − x1 (see the first figure).
If a ∈ −1 ; 0 , then x2 = π + |x1 | (see second figure). Because in this case
x1 < 0, so |x1 | = −x1 . Hence x2 = π − x1 .
So, we showed that two families of solutions can be defined in the way
written under the figures.
sin x = a, when x = x1 + 2kπ or x = π − x1 + 2kπ, where k ∈ TRIGONOMETRIC EQUATIONS AND INEQUALITIES
MLR3-1 [R] str. 49
49
Solve the equation.
EXAMPLE 1
√
3
2
a) sin x =
First we find all solutions in the interval 2π long, e.g. in the inter
3π
val − π
.
2 ; 2
x1 =
x=
π
3
π
3
,
x2 = π −
+ 2kπ
or
π
3
=
x=
sin π
3 =
2
π
3
2
π + 2kπ , where k ∈
3
√
3
2
√
b) sin x = −
π
x1 = − ,
x2 = π +
4
x=−
π
4
2
2
+ 2kπ
PROBLEM
√
a) sin x =
or
π
4
x=
=
5
π
4
5
π + 2kπ , where k ∈
4
Solve the equation.
2
2
√
b) sin x = − 3
2
Note. Notice also that in example 1 in point b) we could assume that that x2 =
π
3π
= −π + π = − 3 π, then the solution notation would be: x = − 4 + 2kπ or x = − 4 + 2kπ,
4
4
where k ∈ . These notations define the same set of numbers as the answer in
the example.
EXERCISE B Give numbers from the interval 4π ; 7π that belong to the sets
of solutions to the equations discussed in example 1.
Consider an equation like:
cos x = a, where a ∈ −1 ; 1
50
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 50
As in the case of a sine wave, when a = 0 or a = 1 or a = −1, the solution
of the corresponding equations can be easily read from the graph.
cos x = 1, when x = 2kπ, where k ∈ cos x = −1, when x = −π + 2kπ, where k ∈ cos x = 0, when x = π + kπ, where k ∈ 2
In the interval of length 2π there are two numbers x1 and x2 satisfying the
equation cos x = a. Using the periodicity of the function y = cos x and the
symmetry of its graph, we obtain two families of solutions to the equation
under consideration.
cos x = a, when x = x1 + 2kπ or x = −x1 + 2kπ, where k ∈ 1
2
Solve the equation cos x = − .
EXAMPLE 2
We find all the solutions to the equation in
an
2π long, e.g., in the interval
interval
−π ; π .
x1 = π −
x=
π
3
=
2π
,
3
2π
+ 2kπ
3
or
x2 = −
x =−
2π
3
cos π
=
3
2π
+ 2kπ , where k ∈
3
1
,
2
so cos π − π
= − 12
3
√
PROBLEM
Solve the equation cos x =
TRIGONOMETRIC EQUATIONS AND INEQUALITIES
MLR3-1 [R] str. 51
3
.
2
51
To solve the equation tan x = a, where a is any real number,
it is enough
π π
to find a solution in the interval of length π , for example − 2 ; 2 , then
use the periodicity of y = tan x.
tan x = a, when x = x0 + kπ, where k ∈ EXAMPLE 3
√
Solve the equation tan x = − 3.
π π
We find a solution in the interval − 2 ; 2 .
x0 = −
x=−
π
3
π
3
+ kπ , where k ∈ √
PROBLEM
Solve the equation tan x = − 3 .
EXERCISE C
3
Solve the equation.
a) tan x = 1
52
b) tan x = 0
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 52
Solve the equation.
EXAMPLE 4
sin x × (1 − 2 cos x) = 0
sin x = 0
or
1 − 2 cos x = 0
cos x =
1
2
We find all solutions in the interval −π ; π .
x = kπ or x = −
PROBLEM
π
3
+ 2kπ , or x =
π
3
+ 2kπ , where k ∈ Solve the equation cos x 2 sin x +
√
3 = 0.
The following examples show how to solve slightly more complicated equations.
Solve the equation.
EXAMPLE 5
√
π
3 − 2 cos 5x + 4 = 2
√
π
2
cos 5x + 4 = 2
√
π
2
Let 5x + 4 = t. Then cos t = 2 .
We find√ all solutions to the equation
cos t = 22 in the interval −π ; π .
π
π
t1 = 4 ,
t2 = − 4
π
t = 4 + 2kπ
π
π
π
π
or
π
t = − 4 + 2kπ , where k ∈ 5x + 4 = 4 + 2kπ
or
π
5x = 4 − 4 + 2kπ = 2kπ
2
x = 5 kπ
or
PROBLEM
π
π
5x + 4 = − 4 + 2kπ
or
π
π
π
5x = − 4 − 4 + 2kπ = − 2 + 2kπ
2
x = − 10 + 5 kπ , where k ∈ √
Solve the equation 2 sin 3x − π = 2.
TRIGONOMETRIC EQUATIONS AND INEQUALITIES
MLR3-1 [R] str. 53
4
53
Solve the equation.
EXAMPLE 6
4 sin2 x = 1
1
4
1
1
sin x =
or sin x = −
2
2
sin2 x =
We find all solutions
to the equation in
the interval −π ; π .
π
x1 = 6 ,
x=
π
6
π
x2 = − 6 ,
π
5π
x3 = π − 6 = 6 ,
+ 2kπ or x = −
π
6
+ 2kπ , or x =
5π
π
x4 = −π + 6 = − 6
5π
5π
+ 2kπ , or x = −
+ 2kπ , k ∈
6
6
Solve the equation 4 cos2 x = 3.
PROBLEM
The solutions of the equation from example 6 can be written
way. In
in a simpler
the figure below, solutions to this equation in the interval − 3π ; 9π have been
2
2
marked
The figure shows that each of the selected numbers belongs to one of two families:
x = π + kπ
6
or x = − π + kπ, where k ∈ 6
Trigonometric identities are helpful in solving some equations.
EXAMPLE 7
a) 3 sin x =
Solve the equation.
√
3 cos x
Case 1. Assumption cos x = 0
Case 2. Assumption cos x = 0
sin x = 0
√
sin x
3
=
cos x
√3
3
tan x =
3
Contradiction with assumption.
x=
3 sin x = 0
54
π
6
+ kπ
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 54
b)
sin x cos x
=1
tan x
Assumptions:
x = 2 + kπ , x = kπ , where k ∈ The equation makes sense for numbers where
tan x is a real number but not 0.
sin x cos x
sin x cos x
=
= cos2 x
sin x
tan x
We transform the left side of the equation
sin x
using the identity tan x = cos
x.
π
cos x
2
cos x = 1
cos x = 1
or
cos x = −1
x = 2kπ
or
x = π + 2kπ , where k ∈ Each of the numbers 2kπ and π + 2kπ is
a multiple of the number π , so it does not
satisfy the assumption x = kπ , where k ∈ .
The equation has no solutions.
c) 2 sin2 x + 7 cos x + 2 = 0
2(1 − cos2 x) + 7 cos x + 2 = 0
−2 cos2 x + 7 cos x + 4 = 0
Let cos x = t.
−2t 2 + 7t + 4 = 0
Δ = 49 − 4 × (−2) × 4 = 81
t1 =
−7 − 9
=4
2×(−2)
t2 =
−7 + 9
1
=−
2×(−2)
2
cos x = 4
We solve the quadratic equation.
1
or
The equation cos x = 4 has no solutions because −1 ≤ cos x ≤ 1.
cos x = − 2
contradictory
equation
We find all solutions of the
equation
cos x = − 12 in the interval −π ; π .
π
2
x = 3 π + 2kπ
or
2
x1 = π − 3 = 3 π ,
2
x2 = − 3 π
2
x = − 3 π + 2kπ , where k ∈ PROBLEM
Solve the equation.
√
b) tan x × cos x = 1
a) 2 sin x + 2 3 cos x = 0
TRIGONOMETRIC EQUATIONS AND INEQUALITIES
MLR3-1 [R] str. 55
c) cos2 x + sin x + 1 = 0
55
The knowledge in solving trigonometric equations can be used to solve
inequalities.
Solve the inequality.
EXAMPLE 8
√
3
a) sin x ≤ 2
√
We find solutions to the equation sin x = 23
in any interval
of 2π length, in this case
in − 32π ; π
2 , and mark on the x-axis those
numbers in this interval that satisfy the inequality.
π
π
4π
x1 = 3 ,
x2 = −π − 3 = − 3
4π
π
x ∈ − 3 + 2kπ ; 3 + 2kπ , k ∈ Every number belonging to any of the intervals of this form satisfies the inequality.
1
2
b) tan x > 1
1
Let 2 x = t. Then tan t > 1.
π
4
π
4
π
2
π
+ kπ < t < 2 + kπ
1
π
+ kπ < 2 x < 2 + kπ
We read from the graph
the solution of the inequality
in the interval
π
;
−π
2
2 .
+ 2kπ < x < π + 2kπ
x∈
π
2
+ 2kπ ; π + 2kπ , k ∈ c) 2 cos2 x ≤ 1
cos2 x ≤
√
√
1
2
2
, then −
≤ cos x ≤
2
2
2
In an interval 2π long, for example, −π ; π ,
we mark the √set of all numbers
satisfying the
√
inequality − 22 ≤ cos x ≤ 22 .
π
3
π
3
x ∈ − π + 2kπ ; − + 2kπ or x ∈
+ 2kπ ; π + 2kπ , k ∈ 4
4
PROBLEM
Solve the inequality.
√
a) cos x ≥
4
2
2
b) tan 3x <
√
3
4
c) 4 sin2 x ≥ 3
Note.
point c), one could write the solution shorter:
In the above example,
π
3π
+ kπ ;
+ kπ .
x∈
4
4
56
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 56
PROBLEMS
1. Provide all solutions belonging to the interval
−2π ; 2π .
√
a) sin x = 1
b) sin x = −1
2
c) cos x = − 3
2
√
d) tan x = − 3
3
2. Solve the equation.
√
a) sin x = − 2
b) cos x = − 1
2
2
√
3
d) tan x = −1
c) 4 cos2 x = 3
d) tan 2 x = 3
c) tan x =
3. Solve the equation.
b) 2 sin2 x = 1
a) tan2 x = 1
4. Solve the equation.
a) sin x × sin x − 1 = 0
c) cos x tan x + tan x = 0
√ d) (cos x − 2) cos x + 2 = 0
2
b) cos2 x − cos x = 0
2
5. Find zeros of the given function.
a) y = −2 sin x + 1
1
b) y = 1 cos x + √
3
2 3
√
c) y = 5 tan x + 5 3
√
d) y = 3 tan x − 1
6. Solve the equation.
√
a) tan x − π = 3
5 √
b) cos 3x − π = − 2
4
c) 2 sin (3x + 1) = 1
√
d) tan (x − 5) − 3 = 0
2
7. Solve the equation.
c) 4 cos2 x + π = 3
2
3
d) tan2 3x + π = 1
a) 2 sin2 2x + π − 1 = 0
2
b) tan2 3x − π + 2 = 5
6
4
8. Find coordinates of the points of intersection of the graphs of f
a) f (x) = sin (x − π )
b) f (x) = sin x
g(x) = − cos x
9. Solve the inequality.
a) sin x ≤ 1
2√
b) sin x > − 3
2
g(x) = tan x
√
2
2
d) cos x ≥ − 1
2
c) cos x <
TRIGONOMETRIC EQUATIONS AND INEQUALITIES
MLR3-1 [R] str. 57
and g functions.
e) tan x < 1
√
f) tan x > − 3
3
57
SINE, COSINE AND TANGENT OF THE SUM
AND DIFFERENCE OF ANGLES
You already know the basic relationships between trigonometric functions.
We will now discuss some other trigonometric identities. The formulas for
sine of the sum and difference and cosine of the sum and difference of
two angles are given below. These identities, of course, hold both when
the angles are given in degrees and when they are given in radians.
sin (α + β) = sin α cos β + cos α sin β
sin (α − β) = sin α cos β − cos α sin β
cos (α + β) = cos α cos β − sin α sin β
cos (α − β) = cos α cos β + sin α sin β
We will prove the formula for cosine of the difference of angles.
cos(α − β) = cos α cos β + sin α sin β
Proof
The figure shows two angles α and β placed
in the coordinate system. On the end arms,
points A and B were selected, the distances
of which from the origin of the coordinate
system are equal to 1.
Therefore:
A = (cos α, sin α), B = (cos β, sin β)
The length of the line segment AB can be calculated in two ways — using the
formula for the distance between points in the coordinate system and using
the law of cosines.
|AB|2 = (xB − xA )2 + (yB − yA )2 = (cos β − cos α)2 + (sin β − sin α)2 =
= cos2 β − 2 cos α cos β + cos2 α + sin2 β − 2 sin α sin β + sin2 α =
= 2 − 2 cos α cos β − 2 sin α sin β
|AB|2 = |OA|2 + |OB|2 − 2|OA| × |OB| × cos BOA =
= 12 + 12 − 2 × 1 × 1 × cos (α − β) = 2 − 2 cos (α − β)
Comparing the two results leads to the equality:
2 − 2 cos (α − β) = 2 − 2 cos α cos β − 2 sin α sin β
Here from:
cos (α − β) = cos α cos β + sin α sin β
58
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 58
Using the identity proved on the previous page and the reduction formulas, one can justify that the other three identities are true.
cos (α + β) = cos α − (−β) =
We take advantage of
the fact that
cos (−β) = cos β and
sin (−β) = − sin β.
= cos α cos (−β) + sin α sin (−β) =
= cos α cos β − sin α sin β
sin (α + β) = cos π − (α + β) =
2
= cos π − α − β =
2
= cos π − α cos β + sin π − α sin β =
2
We use theequality
sin x = cos π
2 −x .
2
= sin α cos β + cos α sin β
EXERCISE A
Prove the identity: sin (α − β) = sin α cos β − cos α sin β
Calculate sin 75◦.
EXAMPLE 1
√
2
√
3
√
2
1
sin 75◦ = sin (45◦+30◦) = sin 45◦×cos 30◦ + cos 45◦×sin 30◦ = 2 × 2 + 2 × 2 =
PROBLEM
√
√
6+ 2
4
Calculate cos 15◦.
The next box gives the formulas for the tangent of the sum and difference
of angles. Of course, they hold when tangents of the angles appearing in
the formulas are determined.
tan (α + β) =
tan α + tan β
1 − tan α tan β
tan (α − β) =
tan α − tan β
1 + tan α tan β
We will now prove the formula for tangent of the sum of angles.
tan (α + β) =
tan α + tan β
1 − tan α tan β
Proof
sin α cos β + cos α sin β
L = tan (α + β) =
sin α cos β + cos α sin β
sin (α + β)
cos α cos β
=
= cos α cos β − sin α sin β =
cos (α + β) cos α cos β − sin α sin β
cos α cos β
tan α + tan β
=R
=
1 − tan α tan β
EXERCISE B
Prove the formula for the tangent of difference of angles.
SINE, COSINE AND TANGENT OF THE SUM AND DIFFERENCE OF ANGLES
MLR3-1 [R] str. 59
59
The box gives formulas for the sine, cosine and tangent of a double angle.
Using the trigonometric one, the cosine
of a double angle can be written in two
more other ways:
2
cos 2α = 2 cos α − 1
sin 2α = 2 sin α cos α
cos 2α = cos2 α − sin2 α
tan 2α =
2 tan α
1 − tan2 α
cos 2α = 1 − 2 sin2 α
EXERCISE C
Prove the formulas for sine, cosine and tangent of a double angle.
EXAMPLE 2 Find all
solutions to the equation sin 2x + sin x = 0 belonging to
the interval −π ; 2π .
sin 2x + sin x = 0
2 sin x cos x + sin x = 0
sin x × (2 cos x + 1) = 0
We find solutions of the
re equations
ceived in the interval −π ; 2π .
1
sin x = 0
or
cos x = − 2
sin x = 0 and x ∈ −π ; 2π
x1 = 0,
x2 = π
cos x = −
x3 = π −
1
and x ∈ −π ; 2π
2
π
3
=
2
π
3
2
3
x4 = −x3 = − π
2
3
x5 = x4 + 2π = − π + 2π =
2π
2π
4π
x∈ −
, 0,
, π,
3
3
3
4
π
3
From the set of all solutions to the
equation, we choose only those
that
belong to the interval −π ; 2π .
P R O B L E M
Find
all solutions to the equation sin 2x − cos x = 0 belonging to the
interval −3π ; 0 .
60
TRIGONOMETRIC FUNCTIONS
MLR3-1 [R] str. 60
PROBLEMS
1. Find, using the formulas for the sine and cosine of a sum or difference of angles.
a) cos 75◦
c) sin 15◦
e) cos (−105◦)
g) tan 15◦
b) sin 105◦
d) cos (−15◦)
f) sin 165◦
h) tan (−105◦)
2. Calculate without using a calculator or trigonometric tables.
a) sin 10◦ cos 50◦ + cos 10◦ sin 50◦
c) cos 140◦ cos 40◦ − sin 140◦ sin 40◦
b) cos 110◦ cos 20◦ + sin 110◦ sin 20◦
d)
tan 40◦ + tan 20◦
1 − tan 40◦ × tan 20◦
3. Sketch the graph of the function.
a) y = sin x − cos x
b) y = sin x +
√
3 cos x
4. Prove the identity.
a) sin (α + β) = cos α cos β(tan α + tan β)
b) tan α = sin α + β + sin α − β
cos α + β + cos α − β
π
c) tan
+ α + tan π − α =
4
4
2
cos2 α − sin2 α
5. Sketch the graph of the function.
a) y = sin x cos x
b) y = cos4 x − sin4 x
d) y = − 1 sin2 x
c) y = cos2 x
3
6. Let sin 17◦ = a. Express the following values with the number a.
a) cos 34◦
b) sin 68◦
c) sin 13◦
d) cos 28◦
7. Prove the formula.
a) cos 3α = 4 cos3 α − 3 cos α
b) cos 4α = 8 cos4 α − 8 cos2 α + 1
8. In the following formulas t = tan α2 . Prove these formulas.
a) tan α =
2t
1 − t2
b) sin α =
2
c) cos α = 1 − t 2
2t
1 + t2
1+t
9. Prove the identity.
3α
a) tan 3α = 3 tan α − tan
2
1 − 3 tan α
b) sin 3α − cos 3α = 2
sin α
cos α
SINE, COSINE AND TANGENT OF THE SUM AND DIFFERENCE OF ANGLES
MLR3-1 [R] str. 61
61
Table of values for trigonometric functions
α
sin α
cos α
tan α
α
sin α
cos α
tan α
0◦
1◦
2◦
3◦
4◦
5◦
6◦
7◦
0
0,0175
0,0349
0,0523
0,0698
0,0872
0,1045
0,1219
1
0,9998
0,9994
0,9986
0,9976
0,9962
0,9945
0,9925
0
0,0175
0,0349
0,0524
0,0699
0,0875
0,1051
0,1228
46◦
47◦
48◦
49◦
50◦
51◦
52◦
53◦
0,7193
0,7314
0,7431
0,7547
0,7660
0,7771
0,7880
0,7986
0,6947
0,6820
0,6691
0,6561
0,6428
0,6293
0,6157
0,6018
1,0355
1,0724
1,1106
1,1504
1,1918
1,2349
1,2799
1,3270
8◦
9◦
10◦
11◦
12◦
13◦
14◦
15◦
16◦
17◦
18◦
19◦
20◦
21◦
0,1392
0,1564
0,1736
0,1908
0,2079
0,2250
0,2419
0,2588
0,2756
0,2924
0,3090
0,3256
0,3420
0,3584
0,9903
0,9877
0,9848
0,9816
0,9781
0,9744
0,9703
0,9659
0,9613
0,9563
0,9511
0,9455
0,9397
0,9336
0,1405
0,1584
0,1763
0,1944
0,2126
0,2309
0,2493
0,2679
0,2867
0,3057
0,3249
0,3443
0,3640
0,3839
54◦
55◦
56◦
57◦
58◦
59◦
60◦
61◦
62◦
63◦
64◦
65◦
66◦
67◦
0,8090
0,8192
0,8290
0,8387
0,8480
0,8572
0,8660
0,8746
0,8829
0,8910
0,8988
0,9063
0,9135
0,9205
0,5878
0,5736
0,5592
0,5446
0,5299
0,5150
0,5000
0,4848
0,4695
0,4540
0,4384
0,4226
0,4067
0,3907
1,3764
1,4281
1,4826
1,5399
1,6003
1,6643
1,7321
1,8040
1,8807
1,9626
2,0503
2,1445
2,2460
2,3559
22◦
23◦
24◦
25◦
26◦
27◦
28◦
29◦
30◦
31◦
32◦
33◦
34◦
35◦
0,3746
0,3907
0,4067
0,4226
0,4384
0,4540
0,4695
0,4848
0,5000
0,5150
0,5299
0,5446
0,5592
0,5736
0,9272
0,9205
0,9135
0,9063
0,8988
0,8910
0,8829
0,8746
0,8660
0,8572
0,8480
0,8387
0,8290
0,8192
0,4040
0,4245
0,4452
0,4663
0,4877
0,5095
0,5317
0,5543
0,5774
0,6009
0,6249
0,6494
0,6745
0,7002
68◦
69◦
70◦
71◦
72◦
73◦
74◦
75◦
76◦
77◦
78◦
79◦
80◦
81◦
0,9272
0,9336
0,9397
0,9455
0,9511
0,9563
0,9613
0,9659
0,9703
0,9744
0,9781
0,9816
0,9848
0,9877
0,3746
0,3584
0,3420
0,3256
0,3090
0,2924
0,2756
0,2588
0,2419
0,2250
0,2079
0,1908
0,1736
0,1564
2,4751
2,6051
2,7475
2,9042
3,0777
3,2709
3,4874
3,7321
4,0108
4,3315
4,7046
5,1446
5,6713
6,3138
36◦
37◦
38◦
39◦
40◦
41◦
42◦
43◦
44◦
45◦
0,5878
0,6018
0,6157
0,6293
0,6428
0,6561
0,6691
0,6820
0,6947
0,7071
0,8090
0,7986
0,7880
0,7771
0,7660
0,7547
0,7431
0,7314
0,7193
0,7071
0,7265
0,7536
0,7813
0,8098
0,8391
0,8693
0,9004
0,9325
0,9657
1
82◦
83◦
84◦
85◦
86◦
87◦
88◦
89◦
90◦
0,9903
0,9925
0,9945
0,9962
0,9976
0,9986
0,9994
0,9998
1
0,1392
0,1219
0,1045
0,0872
0,0698
0,0523
0,0349
0,0175
0
7,1154
8,1443
9,5144
11,4301
14,3007
19,0811
28,6363
57,2900
—
119
MLR3-1 [R] str. 119