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```Chapter I. INTRODUCTION
Trigonometry (which means "triangle measurement") is a branch of
mathematics which is concerned with the properties and applications of the (six)
circular or trigonometric functions. These functions have many applications in
algebra, geometry, calculus and in applied mathematics. To be able to study these
functions, we need to review some basic concepts encountered in your earlier
mathematics courses.
1.1 The Cartesian Plane and the Distance Formula
1.1.1 The Cartesian Coordinate System
Defn. A Cartesian Coordinate System (named after the French philosopher
and mathematician, Rene' Descartes [1596-1650]) is formed in a plane
by two perpendicular number (or real) lines that intersect at the point
corresponding to the number 0 (zero) on each line.
Remarks:
1.
2.
3.
4.
The point of intersection of the two number lines is called the origin and is
denoted by O.
The horizontal real line is called the abscissa or the x-axis while the
vertical line is called the ordinate or the y-axis.
The axes divide the plane into four regions called quadrants.
The coordinate system is also called the rectangular system and the
plane containing it is called a cartesian plane, a coordinate plane, or an
xy-plane.
y
II
I
x
III
IV
Defn. A point P on a Cartesian plane is an ordered pair of real numbers,
denoted by P(a,b) or simply (a,b) where a is the projection of P on the xaxis, and b is the projection of P on the y-axis.
Remarks:
1.
2.
The projection of a point P on a line R is the point of intersection of l with a
line through P perpendicular to R.
There exists a 1-1 correspondence between the points on the Cartesian
plane and the ordered pairs of numbers belonging to ú 2, where ú2 is the
Cartesian product R x R defined as:
R2 = R x R = { (x,y)* x  R, y  R }.
This implies that the Cartesian coordinate system is a method of
associating with each point in a plane a unique ordered pair of numbers.
Conversely, to every ordered pair of real numbers there corresponds a single
point in the plane.
To illustrate, let a point P be in the plane, say, in the first quadrant. We
construct perpendiculars from P to the coordinate axes.
y
B
P
A
x
The intersection with the x-axis is at the point A, and the intersection with tye
y-axis is at point B. The distance from the origin to A is a, that is, x = a.
Similarly the distance from the origin to B is b, that is, y = b. Then the ordered
pair associated with P is the pair (a, b) called the coordinates of the point P.
The number a is called the x-value of P and the number b is the y-value. In
general the coordinates of the point P in the plane is designated by an ordered
pair of numbers (x, y). Thus we can easily locate or plot the points in the
plane if we are given the coordinates of the point. The symbol P(x, y) is read
to be "the point P with coordinates (x, y)".
Defn.
Two points A and B are said to be
a) symmetric with respect to (wrt) a third point P if the lengths of
segments AP and PB are equal, that is, AP  PB; and,
b) symmetric with respect to a line R if R is the perpendicular bisector of
line segment AB.
Remarks:
1.
2.
3.
4.
P(a,b) and Q(-a,-b) are symmetric wrt the origin.
P(a,b) and R(a,-b) are symmetric wrt the x-axis.
P(a,b) and S(-a,b) are symmetric wrt the y-axis.
P(a,b) and T(b,a) are symmetric wrt the line y = x.
y
S(-a, b)
P(a, b)
x
Q(-a, -b)
R(a, -b)
1.1.2 Exercise Set 1.1.1
A.
1.
2.
3.
4.
5.
B.
Plot each point with the given coordinates:
A(5, 0)
6. F(-3, -4)
B(3, 4)
7. G(0, -5)
C(0, 5)
8. H(2, %2)
D(-3, 4)
9. I(2, -%3)
E(-5, 0)
10. J(-3/4, -%2)
Find the coordinates of the point symmetric to the given point with respect
to
a) the origin
c) the y-axis
b) the x-axis
d) the line y = x
1. A(4, 7)
2. B(-3, 2 1/3)
3. C(-1 1/4, -3)
4. D(2%2, -5)
5. E(-7, 0)
1.2
The Distance Formula
Defn. (Distance Formula) The (undirected) distance between any two points
in the plane P1(x1,y1) and P2(x2,y2), or the length of the line segment
P1P2 is
d = P1P2  x 2  x1 )2  ( y2  y1 )2
Remark: The length of the line segment from the origin to any a given point
P(x,y) is called the radius vector (r) and is always positive. Applying
the distance formula, we obtain
r=
x 2  y2
Example 1. What is the distance of the point A(3, 4) from the origin?
Given: x = 3, y = 4
Required: r = ?
Solution: r =
x 2  y2
= 32  42
= 9  16
= 25
r = 5 Ans.
Example 2. Given the points P1(1,3), P2(2,1) and P3(-3,2), find the undirected
distance between the following points:
a) P1 and P2
b) P3 and P1
y
P3
3
2
1
P1
P2
x
-3
1 2
3
a. P1P2  (2  1)2  (1  3)2  12  (2)2  5
b. P3P1  (1  (3)2  (3  2)2  42  12 )
= 17
Example 3. If A(4, 3) and B are on a line through the origin, and the x-coordinate
of B is 8, find the y-coordinate and the radius vector of B.
y
Given:
B
A
3
y
B
x
0
4
8
Required: 1. y = ?
2. OB= ?
Solution:
1. Since OCA - ODB (The symbol - means "similar"), hence
corresponding sides are proportional.
y 3

8 4
y = 6 Ans.
2. OB =
x 2  y2
=
82  y 2
= 64  36
= 100
OB = 10 Ans.
1.2.1 Exercise Set 1.1.2
A.
Find the radius vector of each of the following points:
1.
2.
3.
4.
5.
B.
A(6, 8)
B(9, 12)
C(5, 12)
D(12, 2)
E(3/4, 1)
6. F(23, 3)
7. G(2, %5)
8. H(-3, 4)
9. I(0, 7)
10. J(-12, -9)
Find the undirected distance between each of the given points:
1.
2.
3.
A and B
C and D
E and F
4. G and H
5. I and J
6. D and F
C.
Two of the three numbers x, y and r that are associated with any point P are
given. Find the value of the third number and locate the point.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
x = -4, y = 3
x = 4 , y = -3
x = -5, y = -12
x = 2 , y = -5
x = 5 , r = 34, P in QI
x = -7, r = 58, P in QIII
y = 8 , r = 73, P in QII
y = -2, r = 53, P in QIV
x = 4 , r = 5 , P in QI and QIV
x = -5, r = 61, P in QII and QIII
D.
Points A and B are on a line through the origin. Draw the figure for each
problem and find the unknowns:
1.
2.
3.
4.
The coordinates of A are (3, -4) and the x-coordinate of B is 6. Find the ycoordinate and radius vector of B.
The coordinates of A are (-5, 12) and the y-coordinate of B is 24. Find the
x-coordinate and radius vector of B.
The x-coordinate and radius vector of A are 3 and 5. If the y-coordinate of
B is -4, find its x-coordinate and radius vector.
The y-coordinate and radius vector of A are 6 and 10. If the x-coordinate
of B is 4, find the y-coordinate and radius vector of B.
1.3
Angles and their Measures
Defn. An angle is formed by 2 rays that have a common endpoint. This
common endpoint is called the vertex of the angle, and the 2 rays are
called the sides of the angle.
Remark: If one ray can be designated as the initial side and the other the
terminal side, thus the angle is formed by a rotation from the initial
side to the terminal side.
Defn. A positive angle is formed when rotation of the ray is in a
counterclockwise direction. When the rotation is in a clockwise direction,
a negative angle is formed.
Defn. The unit angle is called the degree (o). It can be considered as being
divided into 60 equal parts called minutes (N) and each minute is divided
into 60 equal parts called seconds (O).
When we talk of an angle, of say 45o, we could be interested in any one (or all)
of the following closely related properties:
1.
Arc of a circle which is 45o/360o or 1/8 of the entire circumference. Thus, a
specific part of a circumference may be expressed as an angle.
2.
Relative position of 2 rays having a common point at the center of a circle
such that the 2 rays subtend an angle or an arc of 45o on the circle.
3.
Amount of turning (about a point) required for a line (or object) to move
from one position to another position.
e.g. Ray OA would have to turn through (or generate) a 45o angle to reach the
position of ray OB.
Notice that the measure of a rotation is always in reference to the circle. A
rotation of one-fourth of a circle is called a right angle (90o), one-half of a
circle is called straight angle (180o), less than one-fourth of a circle is called
an acute angle, more than one-fourth but less than one-half of a circle is
called an obtuse angle.
Example 1: Convert 15.48o to degrees, minutes and seconds (DMS).
Solution:
.48o = .48o x
60'
 28.8'
1o
.8’ = .8’ x
60"
 48"
1'
Thus, 15.48o = 15o 28' 48"
Example 2: Convert 23o 30' 36" to degrees only.
Solution:
23o
1o
1o
+ 30' x ----- + 36" x -------- = 23o + 0.5o + 0.01o
60'
3600"
Thus, 23o 30' 36" = 23.51o
Defn. An angle is said to be in standard position when a Cartesian
coordinate system is superimposed so that the vertex is at the origin and
the initial side is along the positive axis. If the terminal side coincides
with one of the axes, the angle is a quadrantal angle. Furthermore,
angles in standard position that have the same terminal sides are
coterminal angles.
y
y
 in standard
position
0o + 360o are
coterminal

x
x
Coterminal angles can be found either by adding 360 o to the given angle or
substracting 360o from the given angle. That is, if  is the given angle,
and  is its coterminal angle, then
 =  " 360o
This method can be applied repeatedly to find other coterminal angles.
Example 3. Sketch the following angles in standard position:
a) 60o
b) the quadrantal angles 90o and 180o .
c) 45o and -45o
y
y
180o
60o
90o
x
x
(a)
(b)
y
45o
x
-45o
(c)
Example 4. Determine the values of two angles which are coterminal with an
angle of 120o and sketch the angles.
Given:
Find :
Solution:
 = 120o
1 = ?
2 = ?
1 = 120 + 360 = 480o
2 = 120 - 360 = -240o
y
480o
-240o
120o
x
Defn. The reference angle (also called related) of a given angle is the acute
angle formed by the terminal side of the angle and the x-axis.
Example 5. Determine the reference angle of each given angle and make a
sketch of the angles.
a. 120o
b. 200o
y
y
60o
c. 310o
120o
y
200o
x
310o
x
20o
(a)
 + 120o = 180o
 = 180o = 120o
 = 60o
50o
(b)
(c)
 + 180o = 200o
 = 200o - 180o
 = 20o
310o +  = 360o
 = 360o - 310o
 = 50o
Defn. Two Angles are complementary if their measures add up to 90o. Two
angles are supplementary if their measures add up to 180o.
Example: Finding complement and supplement of angles:
Complement of 15o24’: 89o60’
- 15o24’
---------74o36’
1.4
Supplement of 15o24’15”: 179o59’60”
- 15o24’15”
-----------164o35’45”
Defn. The radian is the angle measured by a circular arc whose length is
equal to the radius of the circle.
Remarks:
1.
2.
Compared to the degree, a radian is a rather large unit
Since C = 2r, there will be 2 radians in one circle
C
e.g. C = 2r, Y --- = 2
r

180

180o
 57.3o

Examples:
1.
270o x
2.

3

180
2
8 x
180
 1440o

Express angle in terms of degrees only before changing to
24o42’
=
24o
= 24.7o x
1o
+ 42’
 24.7o
60'
180
4.
Express 4.563 radians - degrees, minutes and seconds.
180o
 261.4589o

= 261o26’31”
1.4.1 Exercise Set 1.2
A.
Draw the given angles using protractor and ruler
1.
2.
3.
4.
5.
B.
86.23o
186.27o
2.
5.
42.17o
218.34o
3.
6.
115.52o
-315.78o
45o
170o
85o48'
163.4o
239 2o
6. 86 3/4o
7. 208 5/6o
8. 137o26'14"
9. 326o58'34"
10. 209o3'56"
-690o
-135o
725o
1100o
-630o
Determine the reference angle of each angle given
1.
2.
3.
4.
F.
, -90o
, 450o
, -30o
, -250o
, -400o
Find the smallest positive the angle that are coterminal with the following
angles:
1.
2.
3.
4.
5.
E.
120o
-150o
-360o
225o
300o
Determine one positive and one negative coterminal angle for each angle given
1.
2.
3.
4.
5.
D.
,
,
,
,
,
Convert the following to degrees, minutes and seconds:
1.
4.
C.
60o
330o
50o
45o
100o
130o
210o
305o
140o26'
5.
6.
7.
8.
225o31'
295o37'
143 2/3o
168 5/6o
9. 287 3/4o
10. 110o14'27"
11. 203o29'54"
12. 347o27'4"
Convert the following to decimal notation and find the complement and
supplement in degrees, minutes and seconds.
1.
4.
17o47N13O
27o40N33O
2. 38o48N
5. 68o39N
3. 110o12N39O
6. 143o11N
G. Find the radian measure of each of the following.
1.
2.
64o40’
86o
7.
8.
35.8o
38o16’
3.
4.
5.
6.
24o25’
48o.9’
25.8o
16.7o
9. 43.2o
10. 32o42’52”
11. 12o46.8’
H. Find the degree measure of each angle:
1.
11
6
4.

12
3
2.
4
10
5.
3
9
3.
2
6. 3
1.5
Central Angles, Arc Lengths and Sector Areas
Defn. An angle with its vertex at the center of a circle is called a central angle
and the segment of the circle it intercepts is called the intercepted arc.
The length of the intercepted arc is called the arc length on the circular
distance and the area bounded by the 2 sides of the central angle and
its intercepted arc is called a sector.
Remark: The sector area and the arc length are proportional to their central
angles. Thus, when the radius of the circle and the central angle are
known, it is possible to compute the arc length and the sector area.
Arc length: S = r
  
 9 
or S = 
2r   
r
 360 
 180 
Sector Area: AB = ½ r2
   2
or AB = 
r 
 360 
Examples:
1.
How long is an arc of 35o if the diameter of the circle is 10 ft?
175
 35 
 3.0543 ft .
S= 
5 
180
 180 
2.
Find the radius of a circle if the circular distance of 24.36 cm subtends a
central angle of 65o35N.
85o35’ = 65.583o x
180 o
r=
S
24.36 cm

= 21.2 cm.
3.
Find the angle in degrees, minutes and second of a circle with radius 15.6
cm and subtended by a circular distance of 27.42 cm.
S 27.42
 = --- = ------r 15.6
 = (1.758)(57.296) = 100.726o
= 100o + (.726)(60N)
= 100o43’ + (.56)(60”)
= 100o43’34”
4.
Given a circle of radius 8 cm and a central angle of 42o. Find the area of
the sector.
42o = -----------------180
A2 = 2r2
= 2 (8)2(.7319)
= 23.42 sq.cm.
5.
What is the area of the sector of a circle with a diameter of 12.8 ft. and a
central angle of 60o?
   2  80 
2
2
2
AB = 
r   
6.4  21.4466 ft  21.45ft
 360 
 360 
Some applications of arc lengths and sector areas:
1.
How far does the tip of an exhaust fan blade move if the blade has a radius
of 8.0 ft and turns through 45 radians?
S = r = 45(8) = 360 ft.
2.
Through what angle must pulley A with an 8-ft radius be turned to move
point P on the belt a distance of 960 ft?
S = r - 9 =
S 960

r
8
180 S (180)(960)
  
S= 

 6875.48o
r   
r
(3.1416)(8)
 180 
3.
A 12-in diameter pie is sliced equally into 12 pieces. What is the area of
each pie?
   2
AB = 
r 
 360 
36
 360 / 12  2
 3(3.1416)  9.42sq.in .
=
6  
12
 360 
1.6
Applications in Angular and Linear Velocities
To compute the arc length, we use the formula
S = r
Dividing both sides by t, we then have
S r


r
t
t
t
Let
S

 V,   , hence, we can say
t
t
V = r
where V = linear velocity and  = angular velocity.
Defn. The velocity of a point along a circular path is equal to the product of the
radius of the path and the angular velocity , [(in radians/unit of time)]
i.e.
V = r.
Examples:
1.
Find the surface velocity of a steel shaft 20 in. in diameter if the shaft
has an angular velocity w of 250 radians/sec.
Solution:
V = (10)(250)
= 2,500 in/sec.
2.
A boy rides a bicycle with a speed of 22 ft/sec. If the cycle has wheels
of 28-in diameter, what is the angular velocity of the wheels?
Solution: V = r   =
 
3.
V 22 ft / sec
1 ft

, r  14 in x
 7/6
r
7/6
12 in
132
7
Find the angular speed of the wheel which makes 25 rev in 5 seconds.
Solution:  = 25/5 = 5 rev/sec
 = 360o x 5 = 1800o/sec
4.
A wheel with radius 12 in. turns with a speed of 120 rev/min. Find the
linear speed in ft/min?
Solution:  = 120 rev/min x 2 rad
v = r
= 1 ft x 240  rad
= 173.6 ft/min
1.6.1 Exercise Set 1.3
A. Solve the following:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
B.
Find , if S = 15 m
,r=8m
Find , if S = 34.6 cm
, r = 12 cm
Find S, if  = .45 rad
, r = 9 cm
Find S, if  = 1.23 rad
, r = 18 cm
Find S, if  = 2.6 rad
, r = 14 m
o
Find r, if  = 38 20’
, S = 36 cm
o
Find r, if  = 72
, S = 24 cm
Find r, if  = 47o45N
, S = 54 cm.
Find As, if  = /4
, d = 10 in.
o
Find As, if  = 1 30’
, r = 16.3 ft.
Find As, if  = 20o
, C = 16.8 in.
Find As, if  = /3
, r = 5 cm
Problem Applications: Circular distance and its central angle.
1.
A given circle has a radius of 16.42 meters and a circular distance of 37.15
meters. Find the central angle subtended by the arc:
b. in DMS notation
2.
In Luneta, there is a circular garden with two paths that radiate from the
center. the angle between the two paths is 142o. Each path is 130 ft. long.
Find the length of fence necessary to enclose this sector of the garden.
3.
Find the radius of a circle whose arc subtended by an angle of 20 o is 15
cm. in length.
4.
A circle of diameter 12 cm has a circular distance subtending an angle of
40o. Find the circular distance.
5.
A circle of radius 6 cm has an arc subtending an angle of 18 o. Find the
length of the arc.
6.
If the radius of the earth is 3,960 miles, what it the circular distance
between two places on the surface of the earth making a central angle of
23o?
7.
Find the diameter of a circle whose central angle of 10o intercept an arc of
length 18 cm.
8.
An arc of 18 inches subtends an angle of 38o. How long is the radius?
9.
The radius of a circle is 15 ft. and the intercepted arc is 27.2 ft. Find the
angle in degrees and minutes.
10. Find the circular distance, if the diameter of a circle is 24 cm. and the
subtended angle is 56o.
11. Find the central angle of a circle subtended by an arc 32 inches long and
12. Find the distance around a sector bounded by two radii, each 63 cm, long
and the angle between them is 76o.
13. The front wheel of a bicycle has a diameter of 16 in and the back wheel
has a diameter of 26 in. through what angles (in radians and degrees)
does the front wheel turn if the rear wheel turns through 24 radians?
14. If 88 ft of belt pass around a wheel of 9 in. in diameter, through what angle
does the wheel turn?
C.
Problem Applications: Area and Sector
1.
Find the area of a sector of a circle of radius 12 inches with an arc of
length 26 inches.
2.
Find the area of a sector of a circle of diameter 18 cm, with an angle of
32o26N.
3.
A circular garden of radius 15 ft. is to be planted with different growing
plants. The two radii bounding the space for roses make an angle of 85 o;
for gumamela 52o; for dahlia 80o; for camia 83o and for rosal 60o. Find the
area of each sector.
4.
In front of the ViSCA Laboratory High School is a circular garden. A
portion of it is planted with San Francisco plants. If the two radii bounding
the San Francisco plants are 15 ft. each in length, and the angle between
them is 53o, what is the area of the portion planted with San Francisco?
5.
Find the arc of a sector of a circle whose radius is 8 cm. and area of 512
sq.cm.
6.
Find the radius of the circle whose sector ares is 426 sq.cm. and central
angle 24o.
7.
A sector of a circle has an angle of 58.2o and an area of 149 sq. inches.
Find the diameter.
8.
A sector of a circle has an angle of 32o28N and the diameter is 18 inches.
What is the area of the sector?
9.
In a circle of radius 8 ft. what is the central angle of the sector between the
two radii if the area is 2016 sq.ft.?
10. Find the area of a sector of a circle whose diameter is 20 inches and the
subtended angle is 63o12N.
C.
11. The area of the sector of a circle is 4.4 x 102 sq.ft. If the arc of the sector
is 44 ft, what is the radius of the circle?
Problem Applications: Angular Velocity and Linear Speed
1.
How many revolutions per minute will an automobile tire make at the rate
of 45 miles per hour, if the radius is 1.5 feet?
2.
The radius of the earth is assumed to be 3960 miles. What is the velocity
in miles per hour at which a point on the equator moves, if it moves
through one complete revolution every 24 hours?
3.
A bicycle wheel has a diameter of 2 feet and turns at the rate of 70
revolutions per minute. How fast does the wheel move in feet per hour?
4.
The Merry-go-round is 40 feet in diameter. What is the angular velocity in
degrees per minute and the linear velocity in feet per minute of a rider on
the outer most part of the merry-go-round, if it makes 3 complete
revolutions per minute?
5.
A fly wheel has a diameter of 2.5 feet and spins about its axis at the rate of
50 revolutions per minute. How fast does the wheel move in feet per
second?
6.
How many revolutions will the bicycle wheel make to reach a distance of
60 yards, if the diameter is 1.5 feet?
7.
The velocity of a roller is 80 revolutions in one minute. If the roller is 20
inches in diameter, what is the distance covered by the roller in one
second?
8.
An automobile tire is 2.5 feet in diameter. What is its angular velocity in
radians per second and the linear speed in feet pr second, if it makes 50
revolutions in one second?
9.
Find the angular velocity of a fly wheel 3 ft. in diameter, when it turns a
distance of 840 ft. per minute?
10. Find the linear speed of a truck tire 3 ft. in diameter, if it makes 90
revolutions per minute?
11. If the velocity of sound in air is 950 ft/sec, find the angular velocity in
radians/sec at which the tip of a propeller blade (r = 8 ft) enters the
supersonic range.
12. What is the linear velocity in ft/sec of the tip of a form blade (r = 6 ft) when
the blade is turning 60 revolution per minute?
1.7
Functions and their Graphs
Recall:
Defn.
A function is a rule of correspondence between two non-empty sets of
elements, called the domain and the range of the function, such that to
each element of the domain there corresponds one and only one element
of the range, and each element of the range is the correspondent of at
least one element of the domain.
Remarks:
1.
A function is also called a mapping and is said to map its domain one its range,
and f(x), the value of f at x is also called the image of x in the mapping f.
2.
A function can also be considered as a set of ordered pairs (x,y) or (x,f(x)) in
which no two pairs have the same first element. The set of all first elements is
the domain and the set of all second elements is the range of the function.
3.
A function is a one-to-one correspondence if for each element of the domain
these corresponds one and only one element of the range, and each element
of the range is the correspondent of exactly one element of the domain.
Example:
V = s3 the volume V of a cube is a function of the side s
V = f(s) = s3
, for every s, there is a unique value of V
e.g. f(1) = 13 = 1 corresponds to the ordered pair (1,1)
f(2) = 23 = 8 corresponds to the ordered pair (2,8)
f(3) = 33 = 27 corresponds to the ordered pair (3,27)
and so on.
Here, V is called the dependent variable because the value of V
depends on the value of s, whereas, s is called the independent
variable because the choice of s is not dependent on V.
Defn.
Consider the set of ordered pairs {(x,f(x)*x  Df, x, f(x)  ú}. Each ordered
pair than be interpreted as the Cartesian coordinates of a point in the
plane. Then, the totality of such points of any given function is called the
Cartesian graph of that function.
Remark: Since to each element of the domain of a function, there corresponds
exactly one element of the range, no vertical line can intersect the
cartesian graph of a function at two points.
Example: Graph the linear function f(x) = 2x + 1.
Graph
x
0
1
2
…
F(x)
1
3
7
…
1.7.1 Exercise Set 1.4:
1.
Determine whether the correspondence given by the following set of ordered
pairs (x,y) is a function:
a) {(1,2), (2,-3), (3,4), (-4,-1), (1,5)}
b) {(-1,5), (7,2), (3,2)}
c) {(-2,2), (-1,1), (0,0), (-1,-1)}
d) {(0,1), (1,1), (2,1)}
e) {(3,2), (-6,2), (-3,9), (-6,9)}
2.
Evaluate the given functions at the given choices for the independent variable.
a) f(x) =
2x  4; x  1, x  2
x2
b) f(x) = 4
; x  0; x  1; x  2
x  16
3. Given r =
4.
2m  5
, express m as a function of r.
3m
Express the radius of a circle as a function of
a) the arc length and the central angle in degrees and in radian;
b) the sector area and the central angle in degrees and in radian.
5. Express the area A of a circle as a function of its circumference.
6.
Express the area A of a square as a function of its perimeter.
Chapter 2
2.1
TRIGONOMETRIC FUNCTIONS - THE RIGHT TRIANGLE
Right Triangles and their Properties
Recall:
Defn. A right triangle is a three-sided polygon with a right angle. The side
opposite the right angle is called the hypotenuse and the perpendicular
sides are called the legs of the right triangle.
Thm. (The Pythagorean Theorem).
The sum of the squares of the perpendicular sides of a right triangle is
equal to the square of the hypotenuse. That is, if a and b are the length
of the perpendicular sides and c is the length of the hypotenuse, then
c2 = a2 + b2.
Remark: This equation is used in finding the length of one side when two other
sides are given.
Example: Given: a = 4m, b = 3m, c = ?
Solution: c2 = a2 + b2
c2 = (4m)2 + (3m)2
c2
=
16m2
+
2m2
c
a
c = %25m2
c =5m
b
Special Types of Right Triangles:
a.
30o-60o right triangle
In a 30o - 60o right triangle, the side opposite the 30o angle is one-half the
length of the hypotenuse. The sides are in the ratio 1: 3 : 2
b.
Isosceles right triangle
In an isosceles right triangle (45o - 45o - 90o) the length of the hypotenuse
is equal to x%2 where x is the length of the equal sides. The sides are in
the ratio 1: 1: %2
Example: Solve for the unknown sides of each right triangle ABC where:
1. C = 90o, . A = 60o, b = 7 in.
2. . C = 90o, . B = 45o, c = 12 in.
Solutions:
1.
From the given information, we can conclude that . B = 30E. Since
= 7, then c, being the hypotenuse must be twice that of side b.
That is,
c = 2b = 2(7) = 14 in.
Also,
a=
2.
3(7) = 12.12 in.
From the given information, this triangle must be an isosceles right
triangle. Thus, . A = 45E and
c=
2a a = c/ 2 = 12/ 2 = 8.49 in.
Also, a = b,  b = 8.49 in.
b
2.1.1 Exercise 2.1
1.
Solve for the unknown side of the following right triangle
ABC ( C = 90o)
2.
a. a = 6, b = 8
d.
a = 1.09, b = 3.14
b. c = 13, a = 5
e.
a = 93.2, b = 21.6
c. b = 6.3, c = 8.9
f.
a=5 , b=4
Solve for the unknown sides of the following isosceles right triangle ABC.
( C = 90o)
a. c = 8.2
3.
d.
b = 0.633
b. a = 419
e.
c = 11.2
c. b = 0.105
f.
a = 212
Solve for the length of the other sides of the following 30 o - 60o right ABC
where:
C = 90o, pB = 60o
a. a = 5.9
d.
c = 609
b. c = 123
e.
a = 0.333
c. b = 0.163
f.
b = 4.6
2.2
Trigonometric Functions of Angles on a Right Triangle
Defn. Let  be the measure of an acute angle in a right triangle. Then, the six
trigonometric functions can be defined on this triangle as follows:
sin  
opp
hyp
cot  
opp
cos  
hyp
sec  
hyp
tan  
opp
csc  
hyp
opp
where: hyp - is the hypotenuse of the right triangle
opp - is the side opposite angle 
Example:
5
3

4
sin  = 3/5
cot  = 4/3
cos  = 4/5
sec  = 5/4
tan  = 3/4
csc  = 5/3
Exercise 2.2:
Given the following right triangles, give the values of the 6 trigonometric functions of
:
1.
2.
3.
1.2
5
13

3
6


12
2.3
Fundamental Trigonometric Identities
Defn. (Fundamental Identities)
For any angle , the other four trigonometric functions can be defined in
terms of the sine and cosine functions as follows:
Quotient Identities:
tan  
sin 
cos 
cot  
cos 
sin 
Reciprocal Identities:
sec  
1
cos 
csc  
1
sin 
cot  
1
tan 
These identities can be verified using the earlier definition of these
functions. For example,
sin  opp / hyp opp


 tan 
The others can be verified in a similar manner. Using these identities, we
can find the values of all six trigonometric functions once we know the
values of sin  and cos .
Examples:
1.
Find the remaining trigonometric functions if sin A = 5/13.
Solution: From the given that sin A = 5/13, we draw the
triangle.
B
13
5
C
A
By Pythagorean Theorem
b2 + 52 = (13)2
b2 + 25 = 169
b2 = 169 - 25
b2 = 144
b = 144
b = 12
2.
cos A =
12
13
sec A =
13
12
tan A =
5
12
csc A =
13
5
cot A =
12
5
If  is an acute angle and sec  = 7/2, find the values of the other
trigonometric functions of .
Solution: We can sketch a right triangle with an acute angle  satisfying
sec  = 7/2, that is, adj = 2 and hyp = 7.
7

2
Using the Pythagorean Theorem, we can find opp as follows:
22 + (opp)2 = 72

(opp)2 = 72 - 22 = 49 - 4 = 45
 opp =
45 = 3 5 .
The remaining trigonometric function values are thus:
sin  
3 5
7
cot  
2/7
2
2 5


15
3 5 /7 3 5
cos  
2
7
csc  
1
7
7 5


15
3 5 /7 3 5
tan  
3 5 /7 3 5

.
2/7
2
Given: sin  = 4/5, tan x = 5/12, cot y = 24/7
3.
Find : (cos )2 cos x sec y = ?
Solution: sin  = 4/5, tan x = 5/12, cot y = 24/7
5
S
4
12
t

9
5
24
By Pythagorean Theorem we get
p2 + 42 = 5 2
s2 = 52 + (12)2
t2 = 72 + (24)2
p2 + 16 = 25
s2 = 25 + 144
t2 = 49 + 576
p2 = 25 - 16
s2 = 169
t2 = 625
p2 = 9
s = 169
t =
p = 9
s = 13
625
t = 25
7
(cos )2 cos x sec y = 3/5 ! 12/13 ! 25/24
= 9/25 ! 12/13 ! 25/24
= 9/26 Ans.
2.3.1 Exercise Set 2.3:
A.
Use the fundamental identities to find the values of the remaining trigonometric
functions of :
1. sin  = 2/ 13 ; cos  = 3/ 13
2. sin  = 1/ 65 ; tan  = 1/8
3. cos  = 5/ 29 ; cot  = 5/2
4. csc  = 5/3 ; sec  = 5/4
B.
Find the values of the remaining trigonmetric functions by drawing an
appropriate triangle:
1. sin  = 12/13
2. sec  = 2/ 3
C.
3. tan  = 2/5
4. csc  = 10
Find the remaining trigonometric functions given that:
1.
4
65
sin A = ---
6. csc x = -----
5
16
8
n2 - 1
2. cos B = -----
7. sin y = ------
17
n2 + 1
7
2mn
3. tan  = -----24
12
4. cot  = -----
8. cos y = -------m 2 + n2
2n
9. tan A = -------
35
n2 - 1
41
2mn
5. sec x = ---9
10. cos B = --------m 2 - n2
D.
Given that sin A = 4/5, tan  = 7/24, sec x = 41/9, find
cos2A cot  cscx
1.
6. cos x sec x
2. sec A sin  cos x
7. tan  cot 
3. sin x csx  tan A
8. sin2 A + cos2 A
4. sec2x cos A tan 
9. sec2 x - tan2 x
10. csc2 - cot2 
5. sin A csc A
2.4
Solutions to Right Triangles and Applications
Given a right triangle with a hypotenuse of 25, and an acute angle of 33 o,
find the length of the 2 other sides and the measure of the other acute
angle.
Ex. 1:
let  the other acute angle
a
sin
33o
 + 33o + 90o = 180o
= ----
 = 180o - (33o + 90o)
25
= 180o - 123o
a = 25 sin 33o
= 57o
= 25 (.5446)
= 13.62

a
25
33o
c
c2 = a2 + b2
b 2 = c2 - a 2
b =
=
c2  a 2
252  (13.62) 2  439.4956
= 20.96
Ex. 2: Given  = 35o and b = 8.19, solve the right triangle below.
b
cos  = --c

b
c
b
8.19
8.19
c = ----- = ------- = ------ 9.998 . 10
cos
a=
cos 35o .8192
c 2  b 2  (10) 2  (8.19) 2
= 100  67.0761  32.9239  5.74
 = 90o
 +  +  = 180o
 = 180o - ( + )
= 180o - (35 + 90o) = 55o
2.5
Application Problems involving Right Triangles
Defn.
The angle of elevation and the angle of depression of an object is the
angle between the horizontal and the line of sight from the viewer to the
object. The angle of elevation is above the viewer while the angle of
depression is below the viewer.
Remark:
In air navigation, bearing is called course and is the clockwise angle less
than 360o measured from due north to the direction of flight. On the other
hand, in surveying, the direction or bearing of a line can be determined by
knowing the acute angle that it makes with the north south line and by telling
whether it is east or west of that line.
Ex: 55o east of north, N 47o W, S 15o W or S 62o E
Examples:
1.
The angle of elevation of the top of the tree is 60o as seen from a point 42 ft
(measured horizontally) from its base.
a) How high is the tree?
b) What is the distance between the eyes of the viewer and the top of the
tree?
c
a
60o
a
a. tan 60o = ---
a 2  b2
b. c =
b
=
(72.75) 2  (42) 2
= 42 (1.732)
=
7056.5625
= 72.75 ft.
= 84 ft.
a = b tan 60o
2.
In the figure, an observer at O is at an elevation of 1000 ft above the level of
the point P. The angle of depression of P as seen from O is 20 o41N. What is
the straight-line distance of P from O?
a
sin  = --c
0
a
=30o41’
1000
c = ------ = ---------sin 
a
sin 20o40N
1000
p
b
= -----.3529
= 2833.42 ft.
2.5.1 Exercise Set 2.4
A. Solve each of the following triangle:
1.
 = 56o, a = 73
2.
 = 63o10’, c = 17.9
3.
a = 17, b = 24
4.
 = 43o15’, a = 15.
5.
c = 7.5, a = 5.083
6.
a = 31.5,  = 29o32’
3.
a = .013,  = 52o11’
B. Application Problems:
1.
In making a dive, the course of a submarine made an angle of 10o20' with the
surface of the water. What depth had it attained after traveling 1000 m?
2.
A jet cruising at 870 km/hr. climbs at an angle of 15o30'. What is its given
altitude in 4 min.?
3.
From the top of a cliff 250 m. high the angle of depression to a boat is 23o.
How far is the boat from the foot of the cliff?
4.
From points on opposite sides of a river the angles of elevation to the top of a
65-m. tree are 68o and 15o. The points and tree are on the same straight line,
which is perpendicular to the river. How wide is the river?
5.
A flagstaff breaks off 22 m. from the top and, the parts still holding together, the
top of the staff reaches the earth 11 m. from the foot. What angle does it make
with the ground?
6.
A flagpole is atop a building. From a point on the ground 700 m. from the
building, the angles of elevation of the top and bottom of the flagpole are 33 o30'
and 30o31', respectively. How tall is the flagpole?
7.
The diameter of a one-peso coin is 1 5/16 cm. If the coin is held so that it
subtends an angle of 40o at the eye, what is its distance from the eye?
8.
A rectangle has a base 10 cm. long, and the diagonal makes an angle of 12 o30'
with the base. Find the width of the rectangle and the length of the diagonal.
9.
A rectangle 83 cm. long has an area of 492 sq. cm. Find the angle which the
diagonal makes with the base.
10. The radius of a circle is 30 cm., and the length of a chord is 44 cm., find the
angle subtended at the center.
11. Find the radius of a circle if a chord whose length is 5 cm. subtends at the
center an angle of 133o.
13. Find the height of a utility pole which casts a shadow 70 ft. long when the angle
of elevation of the sun (angle of elevation of the top of the pole as observed
from the tip of the shadow) is 47o.
14. If a rectangle is 7 m. x 11 m. find the length of the diagonal and the angles it
makes with the sides of the rectangle.
15. From a position 30 ft high, an object can be viewed and the veiwer's sight
makes an angle of depression of 46o with the horizontal. What is the straight
line distance between the viewer's sight and the object?
16. A man standing 50 ft from a 20-ft tall house looks up at a TV antenna located
on the edge of the roof. If the angle between his line of sight to the edge of the
roof and his line of sight to the top of the antenna is 12o, how tall is the
antenna?
17. A device for measuring cloud height at night consist of a vertical beam of light
which make a spot on the clouds. The spot is viewed from a point 135 m. away.
The angle of elevation is 67o40N. Find the height of the clouds.
18. From an observation tower two markers are viewed on the ground. The
markers and the base of the tower are in line and the observers eye is 65.3 m.
above the ground. The angle of depression are 53o10N and 27o50N. How far
is it from one marker to another?
19. From the top of a building 85 m. high the angle of depression of an automobile
is 29o10N. How far is the automobile from the foot of the building measured
horizontally.
20. A farmer wishes to fence a field in a form of a right triangle. If one angle is
43.2o and the hypotenuse is 200 m. Find the amount of fencing needed.
21. A ladder is leaning against the side of the house. If the ladder is 26 m. long
and Cos A = 1/4 how far is the foot of the ladder from the house?
22. The sides of a parallelogram are 18.75 and 32.68 m. respectively and the
angle between the sides is 52o18N. Find the area of the parallelogram.
23. A tree is broken off by a storm so that the top of the tree reaches a point on the
ground 40 m. from the remaining part of the tree. The angle between the upper
part of the tree and the ground is 32o. Find the height of the tree before the
storm.
24. From a roof of a building 520 m. high the angle of depression of two points P
and Q in the ground and in line with the base of the building are 28 o52N and
47o33N respectively. Find the distance from P to Q.
25. Calculate the area in square meters and find the 3rd side of the land bounded
as follows. Beginning at a cypress tree that is marked with two X's the south
402 m, then N 30 10N E 464 m then due west to the starting point.
26. An airplane took off from field A and flew from 3 hours at 180 kph in a direction
125o and landed at field B. After refueling if flew on a course 270o and landed
at C directly south of A. Find the distance between C and B.
27. If a ball is thrown due east at 15 m/sec from a car traveling due north at the rate
of 20 m/sec. Find the velocity of the ball and the direction of its path.
28. An airplane is headed due noth at a speed of 184 meters per hour. If the wind
is bloiwng due west with a velocity of 27.3 meters per hour. Find the direction
and speed of the plane?
29. At a certain instant, a ship was 4 km south of the light house, the ship was
travelling west ward and after 10 minutes its bearing was S 21 o 15N W from the
light house. Find the speed of the ship.
30. A regular polygon of seven sides (hexagon) is inscribed in a circle of radius 3.
Find the length of one side of the regular polygon.
Chapter 3
3.1
CIRCULAR FUNCTIONS
The Unit Circle and Circular Functions
Defn. The unit circle is the circle with center at the origin and radius equal to 1.
It has the equation x2 + y2 = 1.
Remark:
The unit circle, therefore, is the set of points
{(x, y) x2 + y2 = 1}
y
1
x
1
-1
-1
3.2
The Circular Functions
Recall the definition of arc length
S = r ( is radians)
Consider an angle  = t radians, hence
S = rt  t =
S
.
r
Thm. In the unit circle, (since r = 1) an angle of t radians, subtends an arc of
length t units.
1 + 1 units
unit
a
t
Remark: An angle of 1 radian intercepts an arc of length 1 in the unit circle.
Let Pt be the point of intersection of the terminal side of the angle of t radians
with the unit circle.
If we form a right triangle with an axis as side, we then have
y
Pt(x, y)
t
sin t =
y
y
r
cot t =
x
, y0
y
cot t =
x
x
r
sec t =
1
, x0
x
tan t =
y
x0
x
csc t =
1
, y  0.
y
3.2 Trigonometric Functions of any Real Number
Defn. The value of each trigonometric function at a real number t is defined to
be the value of the function at an angle of t radians, provided that value
exists.
y
(cos t, sin t)
t
x
Remarks:
1.
For any real number t, cos t and sin t are the x- and y-coordinates, respectively,
of the point of intersection of the terminal side of the angle of t radians (in
standard position) with the unit circle.
2.
Because of the role played by the unit circle in the derivation of the values of
trigonometric functions, the latter are also referred to as the circular functions.
3.3
Quadrantal angles are angles whose terminal sides lie on the coordinate axes.
Hence, we can readily see the value of the circular functions at these angles.
/2

(0, 1)
0, 2
(-1, 0)
(1, 0)
(0, -1)

0
/2

3/2
2
Cos 
1
0
-1
0
1
P(x, y)
(1, 0)
(0, 1)
(-1, 0)
(0, -1)
(1, 0)
3/2
Sin 
0
1
0
-1
0
Tan 
0
undefined
0
undefined
0
Cot 
Undefined
0
undefined
0
undefined
Sec 
1
undefined
-1
undefined
1
Csc 
1
-1
-
Exercise Set 3.3:
Find the values of all the circular functions at the following given angles and draw
these angles in standard position.
1. 5/2
4. 11/2
2. -7
5. -9/2
3. -12
6. 13
3.4
Trigonometric Functions of Special Angles: The Circular Approach
Let P1(x1,y1) be the point of intersection of the terminal side of 1 = /4 with the
unit circle. Note that 1 = /4 = ½(/2). Hence, P1 lies on the line that bisects
the first quadrant which is the line y = x. Since P1 also lies on the unit circle, it
follows that
x2 + y2 = 1  x2 + x2 = 1
 2x2 = 1  x = 

P1(x,y)
1
2

2
2
 2
2
. Thus,
,
 P1 = 

2
2


sin

2

4
2
and
cos

2

4
2
Let P2(x2,y2) be the point of intersection of the terminal side of 2 = /6 with the
unit circle; P4(x4,y4) be that of 4 = /2; and P5(x5,y5) be that of 5 = - /6.
Then,
 1    
2 = /6 implies that 2 =     . Also,
 3  2  6
P2P42 = P2P52
 (x22 - x4)2 + (y2 - y4)2 = (x2 - x2)2 + [(y2 - (-y2)]2
 (x2 - 0)2 + (y2 - 1)2 = (2y2)2
/2
/6
P4(x4,y4)
2
5
 x22 + y22 - 2y2 + 1 = 4y22
2
P2(x2, y2)  1 - 2y2 + 1 = 4y2
 4y22 + 2y2 - 2 = 0
-/6

2y22 + y2 - 1 = 0
 (y2 + 1)(2y2 - 1) = 0
 y2 = -1 or y2 = ½ since P2  Q1
Then, since P2 is in the unit circle, x22 + y22 = 1, x22+(1/2)2 = 1
 x 22  1 
1
4
 x2 
3
4
 3 1
 P2  
,  . Hence, sin /6 = ½ and cos /6 =
2
2

3 /2.
Let P3(x3,y3) be the point of intersection of the terminal side of 3 = /3 with the
unit circle. Then,
3 
  3   2 
 


2 6
6
6
3
P1P2= P1P3 so that P2 and P3 are symmetric with respect to a line
passing
thru P1 and the origin which is the line
y = x.
/4
P3(x3, y3)
P2(x2, y2)
 From discussion on symmetry of points,
1 3
. Hence,
P3 =  ,

2
2


sin /3 =
3
, cos /3 = ½
2
3.4.1 Exercise Set 3.4
1.
Fill up the following table:

sin 
cos 
tan 
cot 
sec 
Csc 
0
/6
/4
/3
/2
2. Find the values of each of the following trigonometric functions and draw the
angle in
standard position.
a) sin 11/6
f) csc -21/6
b) cos -7/3, sec 7/3, tan (7/3)
g) tan 11/4
c) tan 13/6
h) csc 18/6
d)
cot 13/4
i) cos 15/3
e)
sec -10/3
j) sin 9/4, cot 9/4, csc 9/4
3. Find the values of the following:
a. sin (9) + tan (9)
b. sec (11/6) + tan (11/6)
c. 2 sin /6 - (cos /3)/3
d.
4 cot 2 / 3  7  / 6
3 cos 
Chapter 4
4.1
TRIGONOMETRIC FUNCTIONS THE GENERAL ANGLE
Trigonometric Functions of General Angles
Defn.
Let  be an angle in standard position with a point P(x, y) on the terminal
side a distance of r = x 2  y 2 from the origin (r  0). Then the six circular
functions are defined as follows:
y

x
r
P(x,y)
=
sin  
y
r
cos  =
x
r
sec  =
r
( x  0)
x
tan  
y
( x  0)
x
csc  =
r
( y  0)
y
cot  
x
( y  0)
y
x 2  y2
Note that the radius vector r is always positive while x and y have the sign
that goes wih the quadrant. Considering the definitions of the trigonometric
functions, we see that the sign of the trigonometric functions depend only on
the values of x and y. The table below shows the sign of each function
depending on where the terminal side of angle  lies.
Signs of the Trigonometric functions
the terminal side of 
I (0 <  < /2)
II (/2 <  < )
III ( <  < 3/2)
IV (3/2 <  < 2)
sin 
+
+
-
Cos 
+
+
Tan 
+
+
-
Cot 
+
+
-
Sec 
+
+
Csc 
+
+
-
Examples:
1. In which quadrant does the terminal side of  lie if sin  > 0 and tan  < 0?
2.
In what quadrant does the terminal side of angle 230 o lie?
Ans. Q III.
3.
Construct two angles A given sin A = 3/5.
Solution: Note that the sine function is positive in the first and second quadrant.
Thus, y = 3 and r = 5.
r=
x 2  y2
P(-4, 3)
A
3
P)4, 3)
r2 = x2 + y2
A
x2 = r2 – y2 = 25 – 9 = 16
x=
4.
16  4
Determine the algebraic sign of cos 130o.
Solution: Since the terminal side of 130o is in the second quadrant and in said
quadrant cosine is negative,  the sign of cos 130o is -.
5.
Find the exact values of the six trigonometric function of the angle  if  is in
standard position and the terminal side of  contains the point (-3, 1).
Solution.
r=
x 2  y 2  (3) 2  12  10
sin  =
1
10

10
10
cot  = -3
cos  =
 3  3 10

10
10
sec  = -
tan  =
1
1

3
3
10
3
csc  = 10
4.1.1 Exercise Set 4.1:
A.
State the quadrant in which the terminal sides of the given angles lies.
1.
2.
3.
4.
5.
B.
11. 749o
12. 830o
13. 936o
14. 995o
15. 1080o
sin A = 1/2
cos A = 1/2
tan A = 1/2
cot A = 1/2
sec A = 2
6. csc A = -5/3
7. sin A = -4/5
8. cos A = -3/5
9. tan A = -2/3
10. cot A = -4/5
Determine the algebraic sign of the given trigonometric functions:
1.
2.
3.
4.
D.
6. 318o
7. 394o
8. 485o
9. 575o
10. 656o
Construct two angles A, given the following:
1.
2.
3.
4.
5.
C.
47o
80o
125o
175o
205o
sin 125o
cos 155o
tan 320o
cot 290o
5.
6.
7.
8.
sec 115o
csc 140o
sin 250o
cos 110o
9. tan (-15o)
10. cot (-100o)
11. sec (-470o)
12. csc (-370o)
Find the trigonometric functions of  (in standard position) and whose terminal
side contains each of the given points below. Sketch the angles and give the
1.
2.
3.
4.
5.
(-2, -2)
(2, 5)
(-1, 4)
(-3, -7)
(0, -5)
4.2
Trigonometric Functions of Special Angles
Consider the isosceles right
trigonometric functions of an angle
size of the triangle, we may choose
the equal sides are of length 1 unit.
triangle.
Because the values of the
depend only on the angle and not on the
to use the isosceles right triangle for which
That is,
a=b=1
Using the Pythagorean Theorem, we have
45o
c
a
c2 = a2 + b2
= 1 + 1 = 2.
c =
45o
2 2.
b
-
-
e we
have the following trigonometric functions for 
sin 45o =
1
2

2
2
cot 45o = 1
cos 45o =
1
2

2
2
sec 45o =
2
csc 45o =
2
tan 45o = 1
-right triangle with a hypotenuse of length 2
units. Next to this triangle, we can place another triangle congruent to it. Note
that the two triangles will give us an equilateral triangle whose angles are each
e is of length 2 units. In particular, the base is 2a = 2, so that
a = 1.
30o
65o
c
65o
65o
a
65o
a
Using the Pythagorean Theorem, we have
b 2 = c2 - a 2
= 2 2 - 12 = 4 - 1 = 3
b =
3.
We thus have the follow
as follows:
Sin 30o = cos 60o =
a 1

c 2
Cos 30o = sin 60o =
b
3

c
2
sec 30o = csc 60o =
Tan 30o = cot 60o =
a
1
3


b
3
3
csc 30o = sec 60o = 2
cot 30o = tan 60o =
2
2 3
3
Examples:
1. Using the values of the functions of special angles, find the values of each.
a. sin 60o tan 30o sec 30o
b. sin 60o cos 30o + cos 60o sin 30o
c.
2 tan 60o
1  cot 2 30o
1  sin 30o
d.
2
Solutions:
a) sin 60o tan 30o sec 30o =
=
3 3 2 3
3 2 3
,
,


2 3
3
23 3
3
. Ans.
3
b) sin 60o cos 30o + cos 60o sin 30o =
3 3 1 3

 
2 2 2 2
= 3/4 + 1/4
= 4/4 = 1 Ans.
c)
2 tan 60o
2 3

2
o
1  cot 30
1  ( 3 )2
=
2 3
1 3
=
2 3
2
=-
1  sin 30

2
o
d)
1
3 Ans.
1
2
2
 1
21  
2
= 
2(2)
=
2 1

4
1 1
 . Ans.
4 2
2. Given the following triangle, find , a and b.
Solution:
a)  = 180o - (90o + 30o)
= 180o – 120o
= 60o
b) a/20 = sin 30o
a = 20 sin 30o
a = 20 (
1
)
2
a = 10 Ans.
c) b/20 = cos 30o
b = 20 cos 30o
b = 20 ( 3 /2) = 10 3
b = 17.32 Ans.
4.3
The Trigonometric Table
The Table of Trigonometric Functions contains four-decimal-place
approximations to sin , cos , tan , cot, sec  and csc  for values of the angle
 in quadrant 1. For values of  between 0o and 45o, we use the left-hand column
together with the headings at the top of the table. For an angle  between 45o and
90o, we use the right-hand column with the headings at the bottom of the table. The
table can be arranged in this fashion, since
sin (90o - ) = cos  and cos (90o - ) = sin ,
and, therefore,
tan (90o - ) =
sin( 90o  ) cos 

 cot ,
cos(90o  ) sin 
and sec (90o - ) =
1
1

 csc .
o
cos(90  ) sin 
Example 1:
Approximate (a) sin 37o50’ and (b) cot 51o20’.
Solution:
a.
Since 37o50’ is between 0o and 45o, we locate the angle in the left-hand
column and then read down the column labeled sin in the table until we
reach the row in which the angle appears.
We see that
sin 37o50’  0.6134.
b.
Since 51o20’ is between 45o and 90o, we locate the angle in the right-hand
column and read up the column labeled cot. From the table we find that
cot 51o20’  0.8002.
The table lists angles from 0o to 90o in increme
approximate the value of a trigonometric function of an angle not listed in
the table, we use linear interpolation as illustrated in the examples below.
Linear Interpolation
Examples:
1. Determine tan 30o33’
From the table,
tan 30o30’ = .5890
tan 30o33’ = x
tan 30o40’ = .5930
33  30
x  .5890

40  30 .5930  .5890
 tan 30o30’ = .5902

3 x  .5890

10
.004

(.004)(3)
 .5890  x
10
 x = .5902
2. Determine cot 35o16’
From the table,
cot 35o10’ = 1.459
cot 35o16’ = x
cot 35o20’ = 1.111
35o16'  35o10'
x  1.419

o
o
35 20'35 10' 1.411  1.419
6
x  1.419

10
 .008
6(.008)
 1.419  x
10
- .0048 + 1.419 = X
 x = cot 35o16’ = 1.4142
3. Determine  if sin  = .5404
Solution: sin 32o40’ = .5398
sin 
sin
= .5404
32o50’
= .5422

  32o 40'
.5404  .5398

o
o
33 50'32 40' .5422  .5398

  32o 40'
.0006

10
.5422  .5398
  = (.25)(10) + 32o40’
= 2.5 + 32o40’= 32o42.5’
= 32o42’0”
4.3.1 Exercise Set 4.3:
A.
Determine the values of the given trigonometric functions by using the
trigonometric tables:
1. cos 28o17’
8.
csc 356o27' =
2. csc 48o3’
9.
3. tan 39o12’
10. cos 108 3/4o =
4. cot 27o4’
11. tan 496 5/6o =
5. sin 62o45’30”
12. cot 1047 2/3o =
6. cot 126.9o =
13. sec 73o36’10”
sin 139o52' =
7. sec 247.8o =
B. Use the trigonometric table to find the angles to the nearest minute
1.
sin  = -0.493187
2.
cos  = 0.666313
3.
sin  = .1340
4.
tan  = .3830
5.
sec  = 1.0535
6. cot  = -0.622045
7. sec  = 2.18769
8. tan  = 2.83175
Chapter 5
5.1
TRIGONOMETRIC IDENTITIES
Pythagorean Identities
Let  be an angle in standard position, and let P(x, y) be a point on the
terminal side of . Then
( x  0) 2  ( y  0) 2  x 2  y 2
r = d(0, P) =
 x2 + y2 = r2


x 2  y2 r 2
 2
r2
r
x 2 y2

1
r2 r2
2
2
x  y
     1
r r

Cos2 + sin2 = 1
Identity 1
Dividing both sides of identity 1 by cos2 , we then have
cos2 + sin2
1
------------------ = --------cos2
cos2
cos2
sin2
1
--------- + --------- = -------cos2
cos2
cos2
 sin  
 1 
 1 
 

 cos  
 cos  
2
1 + tan2 = sec2
2
Identity 2
Dividing both sides of identity 1 by sin2 , we then have
cos2 + sin2
1
------------------- = ------sin2
sin2
cos2
sin2
1
--------- + ---------- = -------sin2
sin2
sin2
 cos  
 1 

 1  

 sin  
 sin  
2
2
Cot2 + 1 = csc2
Identity 3
Defn: The following identities are called the Pythagorean identities:
1.
2.
3.
cos2  + sin2  = 1
 cos2  = 1 - sin2 ,
sin  = 1 - cos2 
1 + tan2  = sec2 
 tan2  = sec2 - 1 ,
sec2  - tan2 ,= 1
cot2  + 1 = csc2 
 cot2  = csc2 - 1 ,
csc2 - cot2  = 1
Example 1: Given tan  = -2 and sin  > 0, find the exact values of the remaining
5 trigonometric functions of .
Solution:
Using Identity 2, we have
1 + tan2  = sec2 
 1 + (-2)2 = sec2 
 sec2  = 5
 sec  = 5
Since tan  < 0 and sin  > 0, and because tan  =
which implies sec  < 0.
sin 
, then cos  < 0,
cos 
hence sec  =  5
1
 sec  = ------cos
- 5
1
= -------cos
1
- 5
 cos  = - ----- = ------5
5
also, tan =
sin 
cos 
-2=
sin 
 5
5

5 2 5

 sin  = (-2)  

5
5


1
1
1
Also cot  = -------- = --------- = - ------tan
-2
2
and csc  =
1
1
5


sin  2 5 2 5
5
Hence, the 6 trigonometric functions are:
2 5
sin  = ------5
5
cos  = - -----5
tan  = -2
cot  = ½
sec  = -  5
5
csc  = -----2
Example 2: Show that cos u + tan u sin u2 = sec u
sin u
Proof: cos u + tan u sin u = cos u + -------- sin
cos u
cos2 u + sin2 u
= ----------------cos u
1
= -------- = sec u
cos u
tan x
Example 3: Prove ----------- = sin x cos x
1 + tan2x
Proof:
sin x
tan
tan x

 cos x  sin x cos x
2
1
1  tan x sec 2 x
cos 2 x
Example 4: Prove: sin2x - sin4x = cos2x - cos4x
Proof:
sin2x - sin4x = sin2 x (1 - sin2x)
= sin2 x (cos2x)
= (1 - cos2x)(cos2x)
= cos2x - cos4x
cos x csc x
Example 5: Prove the identity ----------- = tan x
cot2x
cos x csc x
1
tan 2 x
tan2x cos x csc x = tan x
tan x (tan x cos x csc x) = tan x
sin x
1
tan x (--------- cos x -------) = tan x
cos x
sin x
tan x = tan x
Example 6: Prove the identity sec2x + csc2x = sec2x csc2x
sec2x + csc2x = (1 + Tan2x) csc2x
=
csc2x
sin2x
1
+ ---------  -------cos2x
sin2x
1
= csc2x + -------cos2x
sec2x + csc2x = csc2x + sec2x
csc x
Example 7: Prove the identity --------------- = cos x
tanx + cotx
csc x
1
tan x 
tan x
 cos x
csc x
 cos x
tan 2 x  l
tan x
csc x tan x
 cos x
sec 2 x
1 sin x

sin x cos x  cos x
1
cos 2 x
1 sin x cos 2 x


 cos x
sin x cos x
1
cos x = cos x
5.1.1 Exercise Set 5.1
A.
Given the following functions, give the quadrant where the terminal side of each
 lies and find the values of the remaining trigonometric functions:
1. cot  = 4, cos  < 0
2. cos  = 3/4,  in QII
3. csc  = -5, sec  > 0
4. tan  = 3, sin  < 0
B. Prove the following:
cos x
cos x
1. ------------ + ---------- = 2 sec x
1 + sinx
1-sin x
21. cos B + tan B sin B = sec B
2.
csc u cos u = cot u
3.
cos2x (1 + tan2x) = 1
sin A
23. ------------------ = 1 + cos A
csc A - cot A
4.
sin u + cos u cot u = csc u
24. cot  + tan  = cot  sec2
5.
cot u + tan u = sec u csc u
6.
1 - 2 sin2  = 2 cos2  - 1
1 - sin2  cos 
25. ------------ = -------sin 
tan 
tan x - 1
7. ---------- = tan x
1 - cot x
22. cos4x - sin4x = 2 cos2 x - 1
26. 1 - 2 sin2x = 2 cos2x - 1
27. sin y tan y + cos y = sec y
8.
sec  - cos  = sin  tan 
28.sec x csc x - cot x = tan x
9.
cot2  - cos2  = cot2 cos2
29.
cos2 (1 - cos2) = 1
csc2x - csc2x cos2 x = 1
30.
sin x (1 + cot2x) = csc x
10.
31. sin x (1 + cot2x) = csc x
sin x
11. --------- = csc x + cot x
1 - cos x
32. sin x (csc x - sin x) = cos2x
1 + cos 
sin 
12. ------------- = ------------sin 
1 - cos 
sin x
33. --------- = csc x + cot x
1 - cosx
cot x + 1
sec  + csc 
13. ------------- = 1 + Tan x
cot x
34. ------------------- = csc 
1 + tan 
14. Tan2x cos2x + Cot2x sin2x = 1
4 + sec x
15. 4 sin x + Tan x = ----------csc x
35. 2 sin4y - 3 sin2y + 1 = cos2y
(sec x - 1)2
sec x - 1
36. --------------- = -----------tan2 x
sec x + 1
(Sec x - 1)2
sec x - 1
16. ----------------- = ----------Tan2x
Cos2x
17.
Sec4x
-
Tan4x
1-sin
37. sec  + tan  + cot  = -------------cos sin
1 + sin x
= -----------Cos2x
38. cot  sec2 - cot  = Tan 
39. sin  sec  - sin  tan2 = sin 
18. (Tan x + Sec
x)2
1 + sin x
= ----------1 - sin x
19. Sin2 Sec2 = Sec2  - 1
Tan2x + 1
20. -------------- = tan2x
Cot2x + 1
5.2
Sum and Difference Identities
Proof of the formula for cos ( - ):
to find AP:
y
P(cos, sin)
AP =
P(x, y)
A(2, 0)
x
(cos   l) 2  (sin   0) 2
=
(cos 2   2 cos   l)  sin 2 
=
cos2   sin 2   2 cos2   l
= 1  2 cos   l
=
2  2 cos 
P(cos-sin)

Using the result above, we can say that
PaP =

2  2 cos(  )
But PaP using the distance formula is also that equal to:
PaP=
P(cos - sin)
(cos   cos ) 2  (sin   sin ) 2
=
cos 2   2 cos  cos   cos 2   sin 2   2 sin  sin   sin 2 
=
cos 2   sin 2   cos   sin 2   2 cos  cos   2 sin  sin 
= 1  1  2 cos  cos   2 sin  sin 
=
 PaP =
2  2(cos  cos   sin  sin )
2  2 cos   2  2(cos  cos   sin  sin )
 2-2 cos ( - ) = 2 – 2 (cos cos + sin sin)
 -2 cos ( - ) = -2(cos cos + sin sin
cos ( - ) = cos cos + sin sin
5.3
Special Reduction Formula
Let  = /2
 cos (/2-) = cos /2 cos + sin /2 sin  = sin 
Let  = /2 - 
 sin (/2-) = cos [(/2 - (/2 - )]
= cos [/2 - (/2 - ]
= cos [0 + ]
= cos 
sin (/2 - )
cos 
 tan (/2-) = ------------------ = ----------- = cot 
cos (/2 - )
sin 
cos (/2 - )
sin 
 cot (/2-) = ----------------- = ---------- = tan 
sin (/2 - )
cos 
Remark: Because of the above properties, the sine and cosine functions are
called cofunctions, likewise, the tangent and cotangent functions. It
is also easy to show that secant and cosecant are cofunctions.
Consider  = -
Then cos [ - (-)] = cos ( + )
 cos( + ) = cos cos(-) + sin sin(-)
= cos cos + sin (-sin)
= cos cos - sin sin
Note that here we used the following reduction identities:
cos(-) = cos  and sin(-) = - sin .
These identities can be easily verified graphically. These types of
identities are discussed further in the next section.
Let us now consider sin ( + ):
sin ( + ) = cos[(/2 - ( + )]
= cos(/2 -  - )
= cos[(/2 - ) + ]
= cos(/2 - ) cos  + sin (/2 - ) sin
= sin cos + cos sin
sin ( - ) = sin [ + (-)]
= sin cos(-) + cos sin(-)
= sin cos + cos (-sin )
= sin cos - cos sin
1
sin(   ) sin  cos   cos  sin  cos  cos 


tan ( + ) =
1
cos(  ) cos  cos   sin  sin 
cos  cos 
sin  cos  cos  sin 

tan   tan 
cos  cos  cos  cos 
=

cos  cos  sin  sin  1  tan  tan 

cos  cos  cos  cos 
tan ( - ) =
tan   tan 
1  tan  tan 
Proof is left as an exercise.
Defn. The following identities are called the Sum and Difference Identities.
1. cos( + ) = cos cos - sin sin
2. cos( - ) = cos cos + sin sin
3. sin( + ) = sin cos + cos sin
4. sin( - ) = sin cos - cos sin
tan + tan
5. tan( + ) = ------------------1 - tan tan
tan - tan
6. tan( - ) = -----------------1 + tan tan
Example:
Find the exact values of the 6 trigonometric functions of 15o
cos 15o = cos (45o - 30o) = cos 45o cos 30o + sin 45o sin 30o
= (.7071)(.866) + (.7071)(.5) = .9659
or:
 4 3 
 
   cos   
cos 15o = cos /12 = cos 
 12 12 
3 4
= cos /3 cos /4 + sin /3 sin /4
2
6
 1  3   3  2 




=  
2
4
 2  2   2  2 
sin 15o =
1 - cos215o =
1 - (.9659)2 =
2 6
 .9659
4
.0670 = .2589 = sin /12
tan
15o
sin 15o
.2589
= ----------- = -------- = .2680
cos 15o
.9659
cot
15o
1
1
= --------- = --------- = 3.731
tan 15o .4687
1
1
sec 15o = --------- = --------- = 1.035
cos 15o .9659
csc
15o
1
1
= --------- = --------- = 3.862
sin 15o .2589
5.3.1 Exercise Set 3.2
A.
Using sum and difference identities, find the exact values of the 6 trigonometric
functions of each angle below:
1.
2.
3.
4.
B.
5.
6.
7.
8.
-345o
-105o
15/12
/12
9.
5/12
Change each of the following expressions to a function of  only.
10.
11.
12.
13.
C.
-15o
195o
75o
165o
cos (30o + )
cos ( - 45o)
sin ( - 45o)
sin ( - ) = sin 2
14. tan(/4 + )
15. tan 2 = tan ( + )
Evaluate the following based on the given information:
a.
12
Sin A = 4/5 , A in QI, Cos B = - ----- , B in QII
13
1. sin (A + B)
2. cos (A + B)
3.
cos (B - A)
4. Tan (A + B)
6.
Tan (A - B)
7. Sin (A - B)
8. cos (A - B)
5.4
Reduction Formulas
Defn. The following identities are called the Reduction Formulas:
1.
2.
3.
4.
5.
6.
sin(-) = -sin 
cos(-) = cos 
tan(-) = -tan 
cot(-) = -cot 
sec(-) = sec 
csc(-) = -csc 
7.
8.
9.
10.
11.
12.
cos (/2-) = sin 
sin (/2-) = cos 
tan (/2-) = cot 
cot (/2-) = tan 
sec (/2-) = csc 
csc (/2-) = sec 
13. sin (-) = sin 
14. cos (-) = -cos 
15. tan (-) = -tan 
16. cot (-) = -cot 
17. sec (-) = -sec 
18. csc (-) = csc 
These formulas are used to express the function of one angle as a function
of another angle and they can be divided into three groups. The first group
(formulas 1-6) are called functions of negative angles, the second (formulas 7 12) are called functions of complementary angles, and the third, functions of
supplementary angles. The proofs of these identities are similar per group and
can easily be done using the sum and difference identities. Some of these
proofs are given below. The rest are left as exercise.
Proof:
1. Reduction Formula #3:
sin(0-)
sin 0 cos  - cos 0 sin 
tan(-) = tan(0-) = ----------- = -------------------------------cos(0-) cos 0 cos  + sin 0 sin 
0  cos  - 1  sin 
- sin 
= ------------------------- = --------- = - tan 
1  cos  + 0  sin 
cos 
2. Reduction Formula #10:
cos(/2-)
cos /2 cos  + sin /2 sin 
cot(/2-) = --------------- = ----------------------------------------sin(/2-)
sin /2 cos  - cos /2 sin 
0  cos  + 1  sin 
sin 
= ------------------------- = --------- = tan 
1  cos  - 0  sin 
cos 
3. Reduction Formula #15:
1
1
sec(-) = ---------- = --------------------------------cos(-) cos  cos  + sin  sin 
1
1
= ------------------------- = ---------- = - sec 
-1  cos  + 0  sin  - cos 
The reduction formulas can also be easily verified graphically. Let P 1(x1,
y1) and P2(x2, y2) be points on the terminal sides of  and - respectively (see
figure) and r the radius vector of P1.
y
P1(x1, y1)
(/2-)
B
x
-B
P2(x2, y2)
It is apparent that x2 = x1 , y2 = - y1.
Then by definition of trigonometric functions of general angles, we have
y2
- y1
y1
tan (-) = ------ = -------, tan  = -----x2
x1
x1

tan (-) = - tan 
Also, by definition of trigonometric functions on right triangles, we have
y1
y1
cot (/2-) = ------ , tan  = -----x1
x1
 cot(/2-) = tan 
Let P1(x1, y1) and P2(x2, y2) be the respective points on the terminal sides
of  and - in standard position. Using the circular function approach,
cos  = x1
and sin  = y1.
Similarly,
cos (-) = x2 and sin (-) = y2.
-

From the figure, we can see that y1 = y2 and x2 = - x1. Thus,
1
1
1
sec (-) = ----------- = ---- = ----- ,
cos (-) x2
-x1
1
1
sec  = ------- = ----cos  x1
 sec (-) = - sec 
Example 1: Express each as function of a positive acute angle less than 90o:
a. sin (-60o) = ?
b. cos (-176o) = ?
c. tan (-100o) = ?
Solution:
a.
sin (-60o) = - sin 60o
b.
cos (-176o) = cos 176o (RF #2)
= - cos (180o – 176o) = - cos 4o
cos (-176o) = - cos 4o
c.
(RF #1)
(RF #14)
tan (-100o) = - tan 100o
= - [- tan (180o – 100o)] = tan 80o (RF #15)
tan (-100o) = tan 80o
In the previous section, it was mentioned that the sine and cosine functions
are cofunctions, likewise, the tangent and cotangent functions, and the secant
and cosecant functions. If fn stands for any trigonometric function, then,
reduction formulas 7-12 can be generalized by the following expression:
fn of A = co-fn of B
where A + B = 90o
That is, any trigonometric function of an acute angle is equal to the co-function
of its complement.
Example 2. Express each as functions of its complementary angle
a. sin 38o
c. tan 26o 24'
b. cos 73o
d. cot 54o32'48"
Solution:
a) sin 38o = cos (90o - 38o) = cos 52o
b) cos 73o = sin (90o - 73o) = sin 17o
c) tan 26o 24' = cot (90o - 26o24') = cot 63o36'
d) cot 54o 32' 48" = tan (90o - 54o 32' 48")
= tan 35o 27' 12"
Example 3. Express each as functions of angle less than 45 o
a. sin 68o
Solution:
b. tan 54o12'
c. sec 61 3/4o
a) sin 68o = cos (90o - 68o) = cos 22o Ans.
b) tan 54o 12' = cot (90o - 54o 12') = cot 35o 48'
c) sec 61 3/4o = csc (90o - 61 3/4o) = csc 28
Example 4. Find A which satisfies the equation
sin 3A = cos (2A + 10o)
Solution: sin 3A = cos (2A + 10o)
cos (90o - 3A) = cos (2A + 10o)
90o - 3A = 2A + 10o
-3A - 2A = 10o - 90o
- 5A = - 80o
-80o
A = ----------5
A = 16o
Ans.
o
Example 5: Express each of the following as a function of its supplement:
a) sin 30o
b) cos 60o
Solution:
a) sin 30o = sin150o = .5
b) cos 60o = -cos120o = .5
c) tan 75o =-tan 105o = 3.7321
c) tan 75o
5.4.1 Exercise Set 5.3
A.
Express each as function of a positive acute angle less than 90 o:
1.
2.
3.
4.
5.
sin (-78o) =
cos (-127o) =
tan (-223o) =
cot (-304o) =
sec (-481o) =
6. csc (-112o 13') =
7. sin (-236o 48') =
8. cos (-331 2/3o) =
9. tan (-478 5/6o) =
10. cot (-757 1/4o) =
B.
Express each as functions of its complement:
1. sin 27o = ________________
6. csc 42.8o = __________________
o
2. cos 83 = ________________
7. sin 15 2/3o = ________________
3. tan 19o13' = _____________
8. cos 61 4/5o = ________________
o
4. cot 54 27' = _____________
9. tan 14o29'42" = ______________
o
5. sec 36.4 + ______________
10. cot 26o15'23" =_______________
C.
Express each as functions of angle less than 45o:
1.
2.
3.
4.
5.
D.
6. csc 51 3/5o = _________________
7. sin 78 56' = ________________
8. cos 69o 41' = ________________
9. tan 55o29'49" = ______________
10. cot 62o53'26" =_______________
Find A which satisfy the given equations
1.
2.
3.
4.
5.
E.
sin 52o = ________________
cos 63o = ________________
tan 72.4o = _____________
cot 85.8o = _____________
sec 48 ½o = ______________
90o - A = A
cos A - sin 2A
sin 3A = cos 7A
cos 6A = sin (2A + 18o)
tan (40o + 3A) = cot 2A
6. sec (25o + 3A) = csc (35o + 7A)
7. cot (65o - 3A) = tan (5o + 7A)
8. csc (2A + 60o) = sec (4A + 36o)
9. sin 5A = cos (2A + 41o)
10. tan (90o - A) = cot (3A - 90o)
Express each as function of a positive acute angle less than 45 o:
1.
2.
3.
4.
5.
sin (-83o) =
cos (-112o) =
tan (-157o) =
cot (-169o) =
sec (-237o15') =
6. csc (-268o 58') =
7. sin (-343 3/4o) =
8. cos (-479 5/7o) =
9. tan (-273.6o) =
10. cot (-638.4o) =
5.4
Double-Angle Formulas
Thm. Double-Angle Formulas
1. cos 2 = cos2 - sin2 = 1 - 2 sin2 = 2 cos2 - 1
2. sin 2 = 2 sin  cos 
2 tan 
3. tan 2 = ------------1 - tan2
Proof:
1. cos 2 = cos( + ) = cos cos - sin sin = cos2 - sin2
= cos2 - (1 - cos2) = cos2 - 1 + cos2 = 2 cos2 - 1
or cos2 - sin2 = (1 - sin2) - sin2 = 1 - 2 sin2
2. sin 2 = sin ( + ) = sin cos + cos sin = 2 sin cos
tan + tan 
2 tan 
3. tan 2 = ----------------- = ----------1 - tan tan 1 - tan2
Example 1:
a.
b.
cos 120o = cos2(60o) = cos2(60o) - sin2(60o) = (.5)2 - (.866)2
= 0.25 - 0.75 = -0.5
or:
cos 120o = cos 2/3 = cos2(/3) = cos2(/3) - sin2(/3) = (1/2)2 - ( 3 /2)2
= 1/4 - 3/4 = -2/4 = -1/2 or -0.5
sin 60o = sin 2(30o) = 2 sin(30o) cos (30o) = 2(.5)(.866) = .866
= sin /3 = sin 2(/6) = 2 sin /6 cos /6 = 2(1/2)( 3 /2)
= 3 /2 = 0.866
Example 2. Express sin 3x in terms of sin x only
sin 3x = sin (2x + x)
= sin 2x cos x + cos 2x sin x
= (2 sin x cos x) cos x + (1 - 2 sin2x) sin x
= 2 sin x cos2 x + sin x - 2 sin3 x
= 2 sin x (1 - sin2 x) + sin x - 2 sin3 x
= 2 sin x - 2 sin3x + sin x - 2 sin3x
= 3 sin x - 4 sin3 x
5.4.2 Exercise Set 5.4
In numbers 1-4, find the values of each angle by expressing the angle as
double or twice a special angle.
1.
2.
3.
4.
cos 180o
sin 120o
tan 60o
csc 90o
In numbers 5-9, use a double angle formula to write each of the given
expressions as a single trigonometric function of twice the angle.
5. cos22t - sin22t
tan 3t
6. -----------1 - tan23t
y
y
7. 2 sin ---- cos ----2
2
8. 1 - sin2 /5
 19 
9. 2 cos2  x   1
 2 
In numbers 10-13, verify each of the given identities:
10.
sin 4u = 4 cos u (sin u - 2 sin3u)
11.
cos 3v = 4 cos3v - 3 cos v
12.
(sin t + cos t)2 = 1 + sin 2t
13.
cot 2 = ½ (cot  - tan )
5.5
Half-Angle Formulas
Thm. (Half-Angle Formulas)
1. cos
a
1  cos 
= 
2
2
2. sin
a
1  cos 

2
2
3. tan
a 1  cos 
sin 


2
sin 
1  cos 
Proof:
1.
cos 2x = 2 cos2 x - 1

let x = ----2
cos 2x = 1 - 2sin2x

let x = ----2
 cos 2 (/2) = 2cos2(/2) - 1
 cos 2 (/2) = 1 - 2 sin2(/2)
 cos  = 2cos2(/2) - 1
 cos  = 1 - 2 sin2(/2)
 2 cos2 (/2) - 1 = cos 
 2 sin2/2 = 1 - cos 
cos2(a/2) =
1  cos 
2
cos(a/2) = 
3.
2.
1  cos 
2
sin2a/2 =
1  cos 
2
sin a/2 = 
1  cos 
2
sin /2
2 sin /2
2 sin2 /2
tan /2 = -----------  -------------- = -------------------------cos /2 2 sin /2
2 sin /2 cos/2
Recall: cos 2x = 1-2 sin2x
2 sin2x = 1 - cos 2x
let x = /2 , then
2 sin2(/2) = 1 - cos2(/2)
sin 2x - 2 sin x cos x
let x = /2
sin 2(/2) = 2 sin (/2)cos (/2)
sin  = 2 sin /2 cos /2
= 1 - cos 
1 - cos 
tan /2 = -----------sin 
Also,
sin /2
2 cos /2
2 sin /2 cos /2
tan /2 = -----------  -------------- = -----------------------cos /2 2 cos /2
2 cos2/2
Recall:
cos 2x = 2 cos2 x - 1
2 cos2 x = 1 + cos 2x
let x = /2 , then 2 cos2(/2) = 1 + cos 2(/2) = 1 + cos 
sin 
 tan /2 = -------------1 + cos 
Example 1: Use the half-angle formulas to compute sin /12 and tan /12 cos
5/12.
1  cos  / 6
1  3 / .2
 /6 
a. sin /2 = sin 



2
2
 2 
=
2 3
4
.06699  .2588
 180o
x
 15o
12

 sin 15o  .2588


1  cos 5 / 6
1 3 / 2
 5 / 6
b. cos 5/2 = cos 




2
2
 2 
2 3
4
= .2588
/6
1 - cos /6
1c. tan /12 = tan (----) = ---------------- = ------------ = ------------4
2
sin /6
1/2
1/ 2
= 2
3 = .2679
tan /12 = tan 15o = .2679
/6
sin /6
1/2
1/2
1
or tan /12 = tan (------) = ----------------- = --------------- = ------------ = ----------2
1+cos /6
1 + 3/2
2+ 3
2+ 3
1
= --------- = .2679
3.732
5.5.1 Exercise Set 5.5
In problems 1-5, find the exact value of each of the given expression.
1.
2.
3.
cos /12
sin 3/8
sin -/8
4. cot 105o
5. csc 22.5o
In problems 6-10, verify the given identities.
6. tan

 csc   cot 
2
7. cot2
 sec   1

2 sec   1
8. csc2

2

2 1  cos 
9. cos  
1  tan 2 ( / 2)
1  tan 2 ( / 2)
10. sec2

2

2 1  cos 
5.6
Sum and Product Formulas
5.6.1 Thm. Product Formulas
1.
2.
3.
4.
1
[cos (u - v) - cos (u + v)]
2
1
cosu cos v = [cos (u - v) + cos (u + v)]
2
1
sinu cos v = [sin (u + v) + sin (u - v)]
2
1
cosu sin v = [sin (u + v) - sin (u - v)]
2
sinu sin v =
Proof:
1.
2.
1
1
[cos (u-v) - cos (u+v)] = [cosu cosv + sinu sin v - (cosu cosv - sinu sinv)]
2
2
1
= [cosu cosv + sinu sin v - cosu cosv + sinu sinv)
2
1
= [2 sinu sinv] = sinu sinv
2
1
1
[sin (u+v) + sin (u-v)] = [sinu cosv + cosu sin v + sinu cosv - cosu sinv]
2
2
1
= [2 sinu cosv] = sinu cosv
2
Proofs of 2 & 4: left as exercise
Example: Use a product formula to rewrite each of the following:
1.
2.
3.
sin 45o cos 15o
sin 75o sin 15o
cos 2x cos 3x
Solution:
1.
2.
3.
5.6.2
sin 45o cos 15o =
1
[sin (45o + 15o) + sin (45o - 15o)]
2
=
1
[sin 60o + sin 30o]
2
=
1
1 1  3 1
3/ 2    


2
2 2  2 
sin 75o sin 15o =
1
[sin (75o - 15o) - cos (75o + 15o)]
2
=
1
[cos 60o - cos 90o]
2
=
1
1
[1/2 - 0] =
= .25
2
4
cos 2x cos 3x =
3 1
 .683
4
1
[cos (2x - 3x) + cos (2x + 3x)]
2
=
1
[cos (-x) + cos 5x]
2
=
1
[cosx + cos 5x]
2
Thm. Sum formulas
1.
x  y
x  y
sin x + sin y = 2 sin 
 cos 

 2 
 2 
2.
x  y
x  y
sin x - sin y = 2 cos 
 sin 

 2 
 2 
3.
x  y
x  y
cos x + cos y = 2 cos 
 cos 

 2 
 2 
x  y
x  y
cos x - cos y = -2 sin 
 sin 

 2 
 2 
4.
Proof:
Let x = u + v
y=u-v
 x + y = u + v = u - v = 2u and
x - y = u + v - (u - v) = u + v - u + v = 2v
xy
 x + y = 2u  u =
and
2
x - y = 2v  v =
xy
2
substituting in Product formula 1:
1
[cos (u - v) - cos (u + v)] = sin u sin v
2
1
x  y
x y
(-2) [cos y - cos x] = sin 
  sin 
(2)
2
 2 
 2 
x  y
x  y
 - cos y + cos x = -2 sin 
  sin 

 2 
 2 
x  y
x  y
 cos x - cos y = -2 sin 
 sin 
 (sum formula 4)
 2 
 2 
in Product formula 3,
[sin (u + v) + sin (u + v)] = sin u cos v
x  y
[sin x + sin y] = sin 

 2 
x  y
cos 

 2 
x  y
x  y
sin x + sin y = 2 sin 
 cos 
 (sum formula 1)
 2 
 2 
Example:
1.
Use a sum formula to write each of the following:
 75  15 
 75  15 
sin 75o + sin 15o = 2 sin 
 cos 

 2 
 2 
= 2 sin 45o cos 30o = 2( 2 /2)( 3 /2) =
2.
cos
90o
+ cos
30o
90o + 30o
90o - 30o
= 2 cos (-------------) cos (-------------)
2
2
= 2 cos 60o cos 30o
3
 1  3 

 .866
= 2  
2
 2  2 
3.
t + 5t
t - 5t
cos t - cos 5t = - 2 sin (----------) sin (----------)
2
2
= -2 sin 3t sin (-2t)
= -2 sin 3t(-sin 2t) = 2 sin 3t sin 2t
6 /2 = 1.2247
5.6.3 Exercise Set 5.6
In problems 1-5, evaluate the following using the sum and product formulas
1.
2.
3.
4.
5.
cos 75o - cos 15o
sin 90o + sin 30o
sin 3t - sin t
sin 75o cos 15o
cos 45o cos 15o
In problems 6-8, express each product as a sum containing only sines and/or
cosines.
6. sin 4 cos 2
3

7. sin ------ cos ----2
2
8. cos 3 cos 4
In problems 9-12, express each sum or difference as a product of sines and/or
cosines.
9. sin 4 - sin 2

3
10. cos ------ - cos -----2
2
11. cos  + cos 3
In problems 13-16, establish each identity.
sin  + sin 3
13. -------------------- = cos 
2 sin 2
cos  + cos 3
14. -------------------- = cos 
2 cos 2
cos  - cos 3
15. ---------------------- = tan 
sin  + sin 3
sin 4 - sin 8
16. ----------------------- = - cot 6
cos 4 - cos 8
5.7
A.
Chapter Exercise:
Determine the values of the given function using identities
1. sin 105o
11. cos 15o
2. cos 75o
12. sin 165o
3. tan 210o
13. tan 105o
4. sin 15o
14. cos 112.5o
5. cot 165o
15. tan 67.5o
6. sin 60o
13
16. sin ------12
7. tan 120o
8. cos 240o
17
17. sin ------8
9. sin 300o
10. cos 300o
7
19. cos ----12
17
20. sin -----12
11
21. cos ------12
13
22. cos ------8
3
18. cos ------8
B. Based on the given information, evaluate the expressions below:
a. Sin  = 4/5
12
Cos B = - ----13
where angle  is in QI and angle B in QII
1. sin (A + B)
2. cos (A + B)
b.
3. cos (B - A)
4. Tan (A + B)
1. sin 2x if cos x = 4/5 (QI)
12
6. Tan (A - B)
7. Sin (A - B)
8. cos (A - B)
2. cos 2x if sin x = ------ (QIII)
13
3
3. tan 2x if tan x = ----- (QII)
4
3
4. sin 4x if sin x = ----- (QI)
5
4
5. cos 4x if tan x = ----- (QIV)
4
c.
x
12
1. sin ----- if cos x = ------ (QI)
2
13
x
4
2. cos ---- if sin x = - ----- (QIII)
2
5
x
3
3. tan ----- if sin x = ----- (QII)
2
5
x
5
4. sin ---- if tan x = - ----- (QIV)
2
12
C.
Prove the given identities:
1.
sin (270 - x) = cos x
2.
cos (3/2 - x) = sin x
3.
1 - tan x
tan (45o - x) = ----------1 + tan x
4.
sin (x + y) sin (x - y) = sin2x - sin2y
5.
cos (x - y) + sin (x + y) = (cos x + sin x)(cos y + sin y)
6.
cos2x - sin2x = 2 cos2x - 1
7.
cot x - tan x
cos 2x = ----------------2
2 - sin2 x/2
13. cos x = ---------------sec2 x/2
8.
2 csc 2x Tan x = sec2x
9.
2 Tan x
------------------ = tan 2x
sec2x - 2 tan2x
1 - cos B
14. --------------- = sin B/2
2 - sin B/2
10.
x
sin2 ------ = 1 - cos x
2
11.
sec2
12.
x
sin x
Tan ------ = -------------2
1 + cos x
15. 2 cos x/2 = sin (1 = cosx) sec x/2
x
2(1 - cos x)
----- = ------------------2
sin2x
D. Evaluate each expression using sum/product formula:
1.
sin /12 cos /4
7.
sin 3  cos 
2.
cos (5/12)2
8.
cos 3 cos 5
3.
cos 5/12 cos /4
9.
sin 2 - sin 
4.
sin 7/12 + sin /12
10. cos 3 + cos 9
5.
sin 3/4 - sin /4
11. sin 3 - sin 5
6.
cos 13/12 + cos /4
Chapter 6
GRAPHS OF TRIGONOMETRIC FUNCTIONS AND THEIR
INVERSES
To sketch the graph of a trigonometric function, let us consider the various
positions of the point P on the unit circle.
2
2

3
3 3

4
4
5

6
6
0
7/6
11/6
5/4
7/4
4/3
3/2
5/3
For y = sin x, as x varies from 0 to /2, the value of y increases from 0 to its
maximum value 1. But as x varies from /2 to 3/2, the value of sin x decreases
from 1 to its minimum value -1. Note that at x = , sin x changes from positive to
negative. For values of x between 3/2 and 2, we see that the corresponding
values of sin x increase from -1 to 0.
Also, let us recall that any angle  is coterminal with   n(2) for any
integer n. Hence, sin (  2) = sin . Similarly, sin (  4)= sin , and so on.
Thus the graph of y = sin x for 2  x  4 will be the same as the graph for 0  x 
2. We can then sketch the graph as follows:
xy
-2

3
2
-


2

4

4
Y = sin x
x

2

3
2
2
This sine curve illustrates the following characteristics:
1.
It is periodic with a period of 2. The period for a cycle from [0, 2] is
repeated for interval [2, 4], [4, 6], etc. and also for [-2, 0], [-4, -2], etc.
2.
The values of sin x lie between -1 and 1, that is, the range of sin x is the set
of values in the interval [-1, 1]. The domain of sin x is the set of real
numbers.
3.
It is an odd function because it has the property that f(x) = -f(x), that is
sin(-x) = -sin x.
Following the procedure we did for sin x, the graph of y = cos x can be
sketched in the same manner. We then have the graph of y = cos x as follows:
y
y = cos x
-2
3
2
-

2
0

2

3
2
2
This cosine curve illustrates the following characteristics:
1.
It is periodic with a period of 2. The period for a cycle from [0, 2] is
repeated for intervals [2, 4]. [4, 6], etc and also for [-2, 0], [-4, -2],
etc.
2.
The value of cos x lies between -1 and 1, that is, the range of cos x is the set
of values in the interval [-1, 1]. The domain if cos x is the set of real numbers.
3.
It is an even function because it has the property that f(-x) = f(x), that is
cos(-x) = cos x.
Another interesting property that can be observed from these 2 curves is that
the sine function is identical to the cosine function but shifted /2 units to the right.
This is a consequence of the property
sin  = cos (/2 - ) = cos ( - /2).
Let us now consider the graph of the tangent function, h(x) = tan x. Again, by
considering its values for the special angles and plotting the corresponding points,
we can sketch the graph as follows:
This tangent curve illustrates the following characteristics:
1.
It is periodic with a period of . The period for a cycle from [-/2,/2] is
repeated for intervals [/2,3/2], [3/2,5/2], etc. and also for [-3/2, -/2],
[-5/2, -3/2], etc.
2.
The value of tangent x can be any real number. Thus its range is the set of
real numbers. Also, since tan x = sin x / cos x, its domain is the set of real
3.
It is an odd function because tan (-x) = - tan x.
Let us consider next the graph of y = csc x. Since csc x = 1 / sin x, the graph
of the cosecant function will have vertical asymptotes wherever sin t = 0, namely, t =

, ... Also, by taking the reciprocal of the non-zero values of sin x, we can
then sketch the graph as follows:
y=cos
x
x
-2
-
0
-/2
-1


2
3/
2
2
Using the fact that cot x = 1 / tan x, and sec x = 1 / cos x, we can obtain the graphs
of the cotangent and secant functions as follows:
Y = cot x
0

2
x

3
2
2
Y = sec x
1
-
0


-1 2
3
2
2
Example 1: Graph (a) y = - tan x and (b) tan (x+/2).
Solution:
(a)
(b)
The graph of y = - tan x is the reflection of the graph y = - tan x in the x-axis.
The graph of y = tan (x+/2) can be obtained by shifting the graph of y = tan
x to the left /2 units.
y
Y
0
x
 3
2

2

2
3
2
x
-
0
a. Y = tan x
b. Y =tan (x +


)
2
3
2
Example 2: Graph y = 2 + sec (t - ).
Solution:
The graph of y = 2 + sec (t - ) can be obtained by shifting the graph
of y = sec t  units to the right and 2 units up.
3
Y = 2 + sec (t )
2
1
-

2
0

2
3
2
2
5
2
t
Exercises:
1.
2.
Determine whether the cot, sec, and csc functions are odd or even. Justify
Give the periods of the cot, sec and csc functions and give the intervals at
which their graphs are repeated.
3.
Sketch the graphs of the following functions:
a. y = 1 + sin x
e. y = 2 + cos(x + )
b. y = cos (x - /4)
f. y = tan(x - /2) - 2
c. y = - csc (x + )
g. y = sin(x + /4) - 3
d. y = - (3 + cos x)
6.2
Inverse Trigonometric Functions and Their Graphs
Defn. The inverse sine function, denoted by sin-1 or arcsin, inverse cosine
function, denoted by cos-1 or arccos, and inverse tangent function,
denoted by tan-1 or arctan are given by:
y = sin-1x , if and only if x = sin y for -1  x  1 and -/2  y  /2
y = cos-1x , if and only if x = cos y where -1  x  1 and 0  y  
y = tan-1x , if and only if x = tan y where - < x <  and -/2  y  /2
Ex:
sin-1 (1) = /2 because sin /2 = 1
sin-1 (0) = 0 because sin 0 = 0
cos-1 (1) = 0 because cos 0 = 1
cos-1 (-1) =  because cos  = -1
tan-1 (1) = /4 because tan /4 = 1
cos-1 (1/2) = /3 because cos /3 = ½
Recall that the graph of an inverse function is the reflection of the given
function on the line y = x. Thus, we can sketch the graph of y = sin-1x or y =
arcsin x as follows:
y

2
Y = sin-1x
x
-1
1
-

2
Similarly the graphs of y = cos-1 x and y = tan-1x can be sketched as follows:

y
y
Y = cos-1x
Y = tan-1x

2
-1

2
0
x
1
-1
1
-
6.1

2
Arccsc Arcsec and Arccot
The functions cot x, sec x and csc x also have inverses when
restricted to the domain (0, ), [0, /2) U (/2, ] and [-/2, 0) U (0, /2],
respectively.
[Note that other domain restrictions can be defined for the inverse of these
functions]. These function called arccot, arcsec, and arccsc respectively are
not used as often as arctan, arcos, and arcsin. However we can obtain their
values by using the latter functions. But, unlike the fact that
1
1
sec x = -------, sec-1  --------- , rather,
cos x
cos-1 x
sec-1 x = cos-1 (1/x) for x  1. Similarly, csc-1x = sin-1(1/x)
and cot-1x = /2 - arctan x.
Ex:
1. Find a) sin-1(- ½)
3
b) arccos (- -----)
2
Solution:
a.
If we let y = sin-1
-1/2.. Then, y = /6. Note that we could have chosen y = 11/6 or 7/6 because sin
11/6 = sin 7/6 = 1/2, but these values are not in the range of y = sin1x.
b.
If we let y = arccos x, then we must findly of such that
3
5
cos y = - ------, then y = -----2
6
2. Find tan (sin-1 1/4).
Solution:
We must find tan t where sin t = ¼
sin t
¼
 tan t = -------- = --------cos t
cos t
but, cos t =
 tan t =
1
1 - sin2t = 1 - ( )2 = 15 / 16 = 15/4
4
1
4  1  15
15
15
16
4

 5 
3. Without using a calculator, find sin  tan 1   
 3 

Solution:
Let t = tan-1 - 5/3
 tan t = - 5/3
Using the identify 1 + tan2 t = sec2t, we have
sec t =  1  tan 2 t
2
25
34
 5

=  1      1 
9
3
 3
We chose the positive root, because the range of y = tan-1t is [/2, /2], and for angles in these interval, the secant function is
never negative.
 cos t =
t
3 34

sec t
34
Sin tan t =
sin t
cos t
 sin t = (tan t)(cos t)
 5  3 34  5 34

=   
34
 3  34 
6.1.1 Exercises 6.2
A.
Find each of the indicated value without using a calculator.
1. arcsin
3
2
2. arctan (- 3 )

2

3. sin-1  

2


4. arcos (-1/2)
5. arctan 0
6. sin[cos-1(3/5)]
7. sin (arctan ¼)
8. tan [sin-1(-1/2)]
9. sec [tan-1 4]
 5 
10. sin-1  sin

6 

11. cos(arctan [-2])
Chapter 7
SOLUTIONS TO OBLIQUE TRIANGLES AND APPLICATIONS
Recall:
Defn. A triangle which has no right angle is called an oblique triangle.
7.1
Law of Sines
Consider the following oblique triangle:
A

c
b
h
’
B

0
Let h be the altitude from vertex A to side BC.
Then, it follows that
h
Sin  = -----  h = c Sin 
c
Similarly,
h
Sin  = -----  h = b Sin 
b
Equating these 2 expressions for h, we have
c Sin  = b Sin 

Sin  Sin 

b
c
In a similar manner, we can prove that
Sin 
Sin 
--------- = ----------
C
a
b
Thus, combining these 2 equations above, we have
Sin  Sin 
Sin 
-------- = -------- = --------a
b
c
Thm. (Law of Sines)
The ratio of the length of any side of a triangle to the sine of the angle
opposite is the same as the corresponding ratio for any other side and angle
opposite. That is, given any triangle,
a
b
c
--------- = -------- = --------sin 
sin 
sin 
Example 1. given b = 17,  = 38o and  = 47o, solve the triangle.
 +  +  = 180o
  = 180o - ( + )

b
a
= 180o - (38o + 47o)
= 95o


c
b
a

sin  sin 
a=
b sin  17 sin 38o 17(.8157)


 14.31
sin 
sin 47o
.7314
Also,
c
b

sin  sin 
b sin  17 sin 95o 17(.9962)


 23.16
sin 
sin 47o
.7314
c=
Example 2. Suppose a tree grows vertically on a hillside having an inclination of
15o with the horizontal. If the tree casts a shadow 150 ft long directly
down the slope of hillside when the angle of elevation of the sun is
36o, how tall is the tree?
let  = 36o
Solution:
 = 180 - (90 + 36)
= 54o
h
 = 36o - 15o = 21o
36
h
150

sin  sin 
15o

B
h
h=
36o150’
15o
7.2
150 sin 21 150(.3584)

sin 54o
8090
= 66.45 ft.
SOLVING TRIANGLES: FOUR CASES
In general, we can use the law of sines to solve triangles for which we know (i)
two angles and any side or (ii) two sides and an angle opposite one of these sides.
Triangles for which we know either (iii) three sides or (iv) two sides and the included
angle cannot be solved directly using the law of sines. In the next section we will
consider a method for solving the latter two cases.
In Example 1, where we were given two angles and a side (case (i)), the
triangle had a unique solution. However, this may not always be true for case (ii)
triangles, that is, for triangles where we know two sides and an angle opposite one
of these sides. For instance, suppose that the sides b and c and the angle  in
triangle ABC are specified. As shown in figure (1) below, we draw angle  and
mark off side c to locate the vertices A and B. The third vertex C is located on the
base by drawing an arc of a circle of radius b with center A. As illustrated in Figure
2, there are four possible outcomes for this construction:
A
A
A
b
c
c
c
b

B
B

B
b
A
A
c
b

(c)
c.
d.
b
B 
B
a.
b.
c
(d)
The arc does not intersect the base and no triangle is formed.
The arc intersects the base in two distinct points C1 and C2 and two triangles
are formed.
The arc intersects the base in one point and one triangle is formed.
The arc is tangent to the base and a single right triangle is formed.
Because these various possibilities exist, case (ii) is called the ambiguous
case. The following three examples illustrate two-solution, single-solution, and nosolution outcomes for the ambiguous case.
Example 1: Find the remaining parts of a triangle with  = 50o, b = 5, and c = 6.
Solution: From the law of sines, we have
sin 50o sin 

5
6
or
 sin 50o 
 0.7660 
  6
sin   6
  0.9133.
 5 
 5 
This gives us the value of   66.82o. At this point in the solution it is essential to
recall that the sine function is also positive for quadrant II angles. Hence there is
another angle satisfying 0o    180o for which sin   0.9193. Using 66.82o as a
reference angle, we find the quadrant II angle:
180o - 66.82o = 113.18o
Therefore, the two possibilities for  are
1  66.82o and 2  113.18o.
Thus, as shown in the figure below, there are two possible triangles ABC 1 and ABC2
satisfying the given conditions.
A
A
6
2
1
6
5
5
1 = 66.52o
B
50o
a1
C
1
(a)
50o
2 = 112.15o
B
a2
C2
(b)
To complete the solution of triangle ABC1 shown in Figure (a), we first find 1:
1 = 180o - y1 - 
 180o - 66.82o - 50o = 63.18o.
To find a1, we use
sin 63.18o
sin 50o
--------------- = ----------a1
5
which gives
 sin 63.18o 
 0.8925 
  5
a1 = 5 
  5.83.
o
 0.7860 
 sin 50 
We complete the solution of triangle ABC2 shown in Figure (b) in a similar
manner. Since 2  113.18o,
1  180o - 113.18o - 50o = 16.82o.
We find a2 from the law of sines:
sin 16.82o sin 50o

a2
5
 sin 16.82o 
 0.2894 
  5
a2 = 5 
  1.89.
o 
 0.7660 
 sin 50 
7
Example 2: Determine the remaining parts of the triangle shown:
Solution: Let  = 80o, b = 4, and c = 3. From the law of sines, it follows that
sin  sin 80o sin 


,
a
4
3
and so
sin 80o sin 

4
3
sin  
3
(0.9848)  0.7386.
4
A

4
8

80o
B
a
There are two possible angles between 0o and 180o satisfying sin   0.7386. We
obtain the angle   47.61o. Using 47.61o as a reference angle in the second
quadrant, we find the other possibility to be 180 o - 47.61o = 132.39o. However, the
value  = 132.39o must be rejected, since the given triangle contains the angle  =
80o, and the therefore the sum of the angles would exceed 180 o.
Now we determine the angle :
 = 180 -  -   180o - 47.61o - 80o  52.39o
Substituting this value in (*) gives
sin 52.39o sin 80o

a
4
 sin 52.39o 
 0.7922 
  4
a = 4 
  3.22
o
 0.9848 
 sin 80 
Example 3: Find the remaining parts of the triangle with  = 40o, b = 9, and c = 5.
Solution: From the law of sines, we have
C=5
b=9
sin  sin 40

9
5
o
40o
 sin 40o 
 0.6428 
  9
sin   9
  1.1570
5
5




We know, however, that the sine of any angle must be between - 1 and 1. Thus, sin
 = 1.1570 is impossible, and this triangle has no solution. As the figure shows,
there is no solution because side c is not long enough to reach side a.
Note: Of the four cases described for solving triangles, only the ambiguous case (ii)
may yield more than one solution, and only cases (ii) and (iii) may yield no solution.
Sketching the triangle also helps in determining whether we have an ambiguous
case.
7.2.1 Exercise Set 7.1
A.
Solve the indicated triangles. The relative positions of , 6,  (or A,B,C), a, b
and c are shown below:
1.
 = 80o,  = 20o, b = 7
2.
3.
4.
5.
6.
7.
 = 37o,  = 51o, a = 5
 = 72o, b = 12, c = 6
a = 45.7, A = 65o12’, B = 49o4’
b = 3.07, A = 26o4’, B = 120o
a = 438, A = 37o3’, B = 34o5’
a = 4.6, b = 3.10, A = 18o31’
8.  = 13o20’,  = 102o, b = 9
9.  = 27.3o, b = 3, c = 2
10. a = 89, c = 37, C = 15o42’
B. Application Problems:
1.
A telephone pole makes an angle of 82o with the level ground. The angle of
elevation of the sun is 76o (see figure (a) below). Find the length of the
telephone pole if its shadow is 3.5 m. (Assume that the tilt of the pole is away
from the sun and in the same plane as the pole and the sun.)
2.
A man 5 ft 9 in. tall stands on a sidewalk that slopes down at a constant angle.
A vertical street lamp directly behind him causes his shadow to be 15 ft long.
The angle of depression from the top of the man to the tip of his shadow is 31 o.
Find the angle , as shown in figure (b), that the sidewalk makes with the
horizontal.
82o
76o
31o
3.5 m
Figure (a)
Figure (b)
3.
Angles of elevation to an airplane are measured from the top and the base of a
building that is 20 m tall. The angle from the top of the building is 38o, and the
angle from the base of the building is 40o. Find the altitude of the airplane.
4.
A diagonal of a parallelogram is 200 cm. Find the sides if the angles between
o10’.
them and the diagonal are 28o
5.
Observers A and B are 362 m apart and looking at point C. How far is C from
each observer of angle CBA is 37o20’ and angle CAB is 68o30’.
6.
A ship leaves port and travel due west. At a certain point it turns 30 o north of
west and travels an additional 42 km to a point 62 km from the port. How far
from the port is the turning point?
7.
A plane traveling at 650 km/hr with respect to the air is headed 30 o east of
north. The wind is blowing from the south causes the actual course to be 27o
east of north. Find the velocity of the plane and the wind with respect to the
ground?
8.
A garden has a triangular shape with two of the angles measuring 22o10’ and
98o50’ respectively. If the sides opposite the 22o10’ angle measures 124 m.
How many meters of fence will be needed to inclose the garden.
9.
The base of a triangle is 540 ft and the angles at the base are 40 o20’ and 130o.
Find the other sides and the remaining angle.
10. The angles of elevation of an airplane measure from points A and B are 32o12’
and 43o10’. If points A and B is 7450 m apart on the same horizontal plane.
How far is point A from the airplane?
11. Two lighthouses A and B are 76 miles apart. Their direction finders determine
that a ship is at point C such that angle ABC = 53o and angle BAC = 81o. Find
the distance of the ship from each lighthouse.
C.
Derive the law of sines for a triangle with an obtuse angle as shown in the
figure below:
A

c
B

h

C
7.3
Law of Cosines
A generalization of the Phythagorean Theorem
Consider an arbitrary triangle, not necessarily a right triangle, and superimpose
on it the Cartesian coordinate system, such that vertex C coincides with the
origin, and one of the sides lie on the x-axis.
A

c
b

C
c
B
a
Then, from the definition of general angles, A will have coordinates (x, y)
such that
cos  
x
 x  b cos y
b
and sin  
y
 y  b sin y
b
Thus, A has coordinates (b cos , b sin ). Using the distance formula, side c
will have length equal to,
c=
(b cos   a ) 2  (b sin   0) 2
 c2 = b2cos2  - 2ab cos  + a2 + b2 sin2
= a2 + b2 (cos2  + sin2 ) - 2ab cos 
= a2 + b2 - 2ab cos 
Note that this expression reduces to the Pythagorean Theorem if  = 90o.
If we reorient this triangle so that vertex A is at the origin, in a similar manner
we will derive
a2 = b2 + c2 - 2bc cos .
Using the same argument, we can also prove that
b2 = a2 + c2 - 2ac cos .
Thm. (Law of Cosines)
The square of the length of any side of a triangle is equal to the sum of the
squares of the other 2 sides less twice the product of their lengths and the
cosine of the included angle; that is
a2 = b2 + c2 - 2bc cos 
b2 = a2 + c2 - 2ac cps 
c2 = a2 + b2 - 2ab cos 
Example 1. Solve the triangle for which  = 69o, b = 31 and c = 23.

b = 31
a

69o
C = 23
a2 = b2 + c2 - 2bc cos 
= (31)2 + (23)2 - 2(31)(23) cos 69o
= 961 + 529 - (1426)(.3584)
a2 = 1490 - 511.033
= 978.97
a = 31.29
To find  and , we can use the law of sines. Hence, we have,
b
a

sin  sin 
a sin  = b sin 
sin  
=
b sin 69o
a
(31)(. 9336)
31.29
sin  = .9249
 = 67.66
 = 180o - ( + )
= 180o - (69 + 67.66)
= 44.34o
Remark:
The law of cosines are also useful in the following alternative forms.
b2 + c2 - a2
1. cos  = ---------------2bc
a2 + c2 - b2
2. cos  = ---------------2ac
a2 + b2 - c2
3. cos  = ---------------2ab
Example 2. Calculate the angles of a triangle whose sides are 5, 12 and 13.
Solution.
5
12
1
122 + 132 - 52 144 + 169 - 25
288
cos  = ------------------ = -------------------- = -------- = .9231
2(12)(13)
312
312
 = cos-1 (.9231)
= 22.62o
52 + 132 - 122
25 + 169 - 144
cos  = -------------------- = --------------------- = .3846
2( 5)(13)
130
 = arccos (.3846)
= 67.38o
 = 180o - ( + )
= 180o - (22.62 + 67.38o)
= 180o - 90o
= 90o
In general, to solve a triangle, it is important to determine which technique
to use first. The following table summarizes the various types of problems and
gives a correct approach for each.
----------------------------------------------------------------------------------------------------7.4 SOLVING A TRIANGLE
-----------------------------------------------------------------------------------------------------Type of Triangle
Information Given
Technique
-----------------------------------------------------------------------------------------------------Right
One angle and a side or any two
Basic definitions for
sides
sine, cosine, or tangent;
the Pythagorean theorem
-----------------------------------------------------------------------------------------------------Oblique
Three sides
The law of cosines
-----------------------------------------------------------------------------------------------------Oblique
Two sides and the included angle The law of cosines
----------------------------------------------------------------------------------------------------Oblique
Two angles and a side
The law of sines
-----------------------------------------------------------------------------------------------------Oblique
Two sides and an angle opposite The law of sines (If the
one of the sides
given angle is acute, it
is an ambiguous-case
problem.)
-----------------------------------------------------------------------------------------------------
Note of Caution:
i.
When three sides are given, if the length of the longest side is greater than or
equal to the sum of the lengths of the other two sides, there is no solution.
(This is because the shortest distance between two points is the length of the
line segment joining them.)
ii.
In applying the law of sines, if you obtain a value greater than 1 for the sine of
an angle, there is no solution.
iii.
In the ambiguous case, when solving for the first unknown angle, you must
consider both the acute angle and its supplement as possible solutions. The
supplement will be a solution if the sum of the supplement and the given angle
is less than 180o.
7.4.1 Exercise Set 7.2
A. Solve each of the following triangles
1.
a = 3.6, c = 12.5,  = 46.9o
5.
 = 36o, a = 10, b = 12
2.
a = 6, b = 7.56 c = 54o10’
6.
a = 0.24, b = 0.26, c = 0.153
3.
b = 87.3, c = 34.2, A = 130o4’
7.
b = 18.3, c = 27.1, A = 58.7o
4.
a = 39.1, b = 27.2, C = 29.1
8.
a = 14.3, b = 18.6, c = 17.1
B. Problems:
1.
To determine the distance across a lake, AB in the figure, a surveyor made the
indicated measurement. Use these data and the law of cosines to calculate
AB.
A
525o
62o
B
12o15’
2.
Two boats leave a point at the same time in directions making an angle of 65 o
with each other. If one boat move 19 mph and the other 23 mph, how far apart
are they after 2 hours.
3.
A rhombus has sides of length 10 cm. If the angle at one of the vertices is 50 o,
find the lengths of the diagonals.
4.
A ship sails 32 km from east and then turns 40o north of east. After sailing an
additional of 16 km, where is it with reference to the starting point?
5.
A pilot flew for 210 km. at 21o and then 142 km at 162o. How far is he from the
starting point.
6.
Trees A and B are on opposite sides of the pond. A third tree C is measured to
the 35 m from tree A and 54 m from tree B. Angle BAC is found to be 48 o.
Find the width of the pond?
7.
A body hikes 187 paces due north, then 226 paces n 81 o
steps would be make by taking a short out straight between the two points?
8.
The intersecting sides of a parallelogram are 50 cm and 40 cm. Their included
angle is 123o
9.
A surveyor measures side xy to be 142 m yz to be 200 m. If the angle xyz is
134o
10. The distance from a marker to two mango trees are 1.34 km and 2.68 km. The
angle made by the lines of the marker to the mango trees is 43 o
apart are the mango trees?
11. A slanted roof makes an angle of 35o with the horizontal and measures 28 ft
from base to peak. A television antenna 16 ft high is to be attached to the peak
of the roof and secured by a wire from the top of the antenna to the nearest
point at the base of the roof. Find the length of wire required.
Area of a triangle
Thm. The area of a triangle is given by the formula
a. A = ½ ab sin , when 2 sides and the included angle are known, and
a2 sin  sin 
b. A = ------------------, when 2 angles and the included side are known.
2 sin ( + )
c. (Heron's Area Formula)
A=
S(S  a )(S  b)(S  c).
where S = ½(a + b + c), when 3 sides are known.
Proof:
Recall: A = ½ bh (Eq. 1)
B


a. Given a, b, and 
C
h
sin  = ----a
 h = a sin 
substituting in Eq. 1, we then have
A = ½ b (a sin ) = ½ ab sin , (eq. 2)
b. Given , , a
b
a
-------- = --------sin 
sin 

A
a sin 
a sin 
 b = ----------- = -----------------------sin 
sin [180 - ( + )]
substituting in Eq. 2, we then have
a sin 
A = ½ a (---------------------) sin 
sin[180o-(+)]
a2 sin  sin 
= ----------------2 sin ( + )
c. Given a, b, c, we can use the law of cosines of the form
c2 = a2 + b2 - 2ab cos 
 2 ab cos  = a2 + b2 - c2
From area formula (a) (Eq. 2), we have
1
A = ---- ab sin 
2
1
4
2
2
2
2
2
2
2
 A = ----- a b sin  = [1/4 a b (1 - cos )  ----4
4
1
= ----- (2ab)2 (1 + cos )(1 - cos )
16
1
= ----- [(2ab)(1 + cos )][2ab (1 - cos )]
16
1
= ----- (2ab + 2ab cos )(2ab - 2ab cos )
16
1
= ----- [2ab + a2 + b2 - c2][2ab - a2 - b2 + c2] using law of cosines
16
1
= ----- [(a + b)2 - c2][c2 - (a - b)2]
16
=
a bc a bc ca b ca b



2
2
2
2
 a  b  c   a  b  c 2a   a  b  c 2b   a  b  c 2c 
 
 
 
= 

2
2
2 
2
2 
2
21 


a  b  c a  b  c
 a  b  c
 a  b  c 
 a 
 b 
 c
= 


2
2
2
2





les S =
abc
2
then A =
s(s  a)(s  b)(s  c)
Example 1. Find the area of a triangle with sides 12 in, 8 in and an included angle
of 53o.
A = ½ ab sin 
Solution:
= ½ (12)(8)(.7986)
8
= 38.33 sq.in.
53o
12
Example 2. Find the area of a triangle if 2 angles and the included side are 40o,
60o and 120 ft.
a2 sin  sin 
A = -------------------2 sin ( + )
(120)2(sin 40o)(sin 60o)
= -------------------------------
120 ft
40o
Solution:
60o
2 sin 100o
(120)2 (.6428)(.866)
= ---------------------------2(.9848)
8015.97
= ----------- - 4069.82 sq.ft.
1.9696
Example 3. Solve the following triangle and find its area.

16 cm.

56o
18cm.
Solution: Using the law of sines,
16
18
----------- = -----------.
sin 56o
sin 
18 sin 56o
18(.829)
 sin  = -------------- = -------------- .9327
16
16
 1
= 68.85o
2 = 111.15o
 1
= 180 - (56o + 68.85o) = 55.15o
2
= 180 - (56o + 111.15) = 12.85o
A1
= ½ ac sin 1
= ½ (18)(16)(.8207)
= 118.17 sq.cm.
A2 = ½ ac sin 2
= ½ (18)(16)(.2224)
= 32.03 sq.cm.
7.4.2 Exercise Set 7.3
A. Solve the following triangles and find their areas:
1.  = 61o,  = 24o, b = 305
3.  = 70o, b = 46, c = 28
2.
B.
 = 30o20’,  = 50o30’, c = 11.3
Find the area of the triangle given:
1. a = 16, b = 19, C = 27o10’.
4. b = 25.2, a = 24.3,  = 30o50’
4.
a = 278, b = 341, c = 143
2.
b = 5.4, c = 17.2, A = 130o4’.
5.
a = 82, b = 130, c = 71
3.
a = 32, B = 68, C = 53o15’.
6.
b = 12, A = 27o4’, C = 82o15’
4.
c = 81, B = 62o31’, C = 73o21’
C. Application Problems:
1.
Find the area of a triangular piece of land that has two sides 9.0 m and 2.5 m
intersecting at an angle of 50o21’.
2. A triangular garden has two angles 49o4’ and 75o3’, and the included side is 43
m. Find the area.
3. One side of an equilateral triangle is 46 m. Find the area of the triangle?
4. A triangular lot has two angles 36o48’ and 56o18’. The side between these
angles is 1054 m. What is the area of the lot?
5. Two sides of a triangular lot measures 500 m and 513 m with an angle of 123 o.
What is the area of the lot?
6. A tract of land is bounded by a fence and two street that met at an angle of
78o
What is the area of the tract?
7. Find the area of the triangle whose base is 24.68 cm and whose altitude is
18.44 cm.
8. If the sides of a triangle are 11, 24 and 17 units. What is the area of the
triangle?
9. A parallelogram has two sides 69.4 cm and 75.8 cm forming an angle of 95 o15’.
Find the area.
10. A surveyor measures one side of a triangular lot and finds it to be 315 m, long
and the adjacent angles are 69o45’ and 80o10’. What is the area of the lot?
Chapter 8
8.1
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Exponential Functions
Recall:
Defn.
For any a  r, n  Z+ (positive integers), then the product a  a  a ...  a,
where a occurs n times could be written in exponential notation as an.
That is,
an =
a  a  a  ...  a
n factprs
Thm. (Laws of exponents)
For any a, b  Z+, m, n  R, then
1.
aman = am+n
2.
am
------ = am-n
an
7.
3.
(am)n = amn
8.
a1/n = n a
4.
(ab)n = anbn
9.
am/n = (a1/n)m = (n a )m or
6.
a1 = a, a0 = 1
a-n
1
= -----an
am/n = (am)1/n = n a m
a
an
5. (-----)n = ------b
bn
1
10.
a-m/n
= ----am/n
Example:
(3xo) = 1
1.
2.
2-3
1
1
= ------- = -----23
8
2
3.
82/3 = 3 8 = 3 64 = 4
4.
4-3/2
1
1
1
1
= ------ = ------- = ------- = ----43/2
Defn.
4
3
64
8
The function f defined by
f(x) = bx
where b > 0, b  1, and the exponent x is any real number, is called an
exponential function to the base b.
Example:

2x,
3x,
1
1
  =  are all exponential functions.
2
2
Remarks:
1. The domain of an exponential function is the set of all real numbers.
2. The range is all positive real numbers.
3. Since bo = 1 for any b  R, the graph of bx for  bR, has y-intercept (0, 1).
There is no x intercept.
4. The graph of bx, b > 1 rises to the right and is asymptotic to the negative
x-axis.
5. The graph of bx, 0 < b < 1 rises to the left and is asymptotic the positive xaxis.
Table of values:
x
2x
3x
4x
(1/2)x
-2
¼
1/9
1/16
4
-1
½
1/3
¼
2
0
1
1
1
0
1
2
3
4
½
2
4
9
16
¼
3
8
27
64
1/8
y
1
 
2
x
9
8
4x 3x
2x
4
x
Defn.
The exponential function for which the base is the irrational number
-2 -1
2 3
2.71828..., denoted by e, that
is,
1
2
x
f(x) = e or exp(x)
is called the (natural) exponential function.
Remarks:
1. The graph of ex lies between the curves yx = 2x and y = 3x.
2. Any exponential function can be expressed in terms of the natural
exponential function, e.g. 2t = ekt for k  0.69
Example: Simplifying expressions involving exponents
a.
x1 / 2 (8x ) 2 / 3
x 3 / 4
 2x 1 / 2 

b. 
 y 
4
1
x
  (3x10 / 3 )
y
Solution:
x1 / 9 (8x ) 2 / 3
x1.3 2 / 3 3 / 4 x 5 / 12
1 / 3 2 / 3 2 / 3 3 / 4
a.
x 8
x x 

x 3 / 4
82 / 3
4
 2x 1 / 2 

b. 
y


4
1
 16x  2  y 
x
  (3x10 / 3 )   4  (3x10 / 3 )
y
 y  x 
48x 2 110 / 3 48x1 / 3
=

y 4 1
y3
8.1.1 Exercise Set 8.1
A.
exponents
1. (3a1/2)(-2a3/2)
8. (2x3/4)(5x-3/4)
2. (21/2x-2/3)6
9. ( 3 x-1/4y3/4)4
3. (xy-2/3)3(x1/2y)2
10. (a2b-1/4)2(a-1/3b1/2)3
4.
 a1 / 2 b1 / 3 
11.  5 / 6 
 c

(2x 1y 2 / 3 ) 2
x 2 y2 / 3
 x 2 y3 / 4 
5.  1 / 2 
 x

12.
x1 / 3 y 3 / 4
x  2 / 3 y1 / 2
6. y2/3(2y4/3 - y-5/3)
2 

13. x-3/4   x 7 / 4  4 
x

7. (x1/2 + y1/2)2
14. (a3/2 + )2
B.
Evaluate the following exponential functions:
1. f(2), given f(x) = 24x4x-3
2. g(-3), given g(x) = 22x - 2x+1
3. P(3), given P(t) = 580 e.09t
4. A(5), given A(t) = 4(10)9(1.03)t
 .08 
5. A(15), given A(t) = 1000 1 

4 

4t
6. P(80,000), given P(m) = 100 (2.71)-.000025 m
8.2
Logarithmic Function
Recall:
Defn.
A function g with domain Y and range X is the inverse of a function f
with domain X and range Y if xX, g[f(x)] = x and yY, f[g(y)] = y. g
is denoted by g(x) = f-1(x) or g = f-1.
Remark: If g is the inverse of f, then the following are also true:
a.
b.
c.
f(x) = y  g(y) = f-1(y) = x
f is also the inverse of g, that is
f = g-1 and if x = g(y)  g-1(x) = f(x) = y XX, y  Y.
The graph of a function and its inverse are reflections of each other in the
line y = x.
Examples:
1.
2.
f(x) = x2  g(x) = f-1(x) = x
because f[g(x)] = f( x ) = ( x )2 = x and g[f(x)] = g(x2) =
x2=x
f(x) = 3x  g(x) = f-1(x) = x/3
because f[g(x)] = f(x/3) = 3(x/3) = x and g[f(x)] = g(3x) = (3x)/3 = x
Defn.
The logarithmic function with base b, where b > 0 and b  1, is
denoted by logb and is defined by
y = logbx if by = x
Remarks:
1.
2.
3.
4.
The domain of logb is the set of all positive numbers and the range is the
set of all real numbers.
The logarithmic and exponential functions are inverse functions. Thus, we
can easily transform expressions from exponential to logarithmic form and
vice versa. Furthermore, blogb(x) = x and logbbx = x, and the graph of y =
logbx is the reflection of y = bx, b > 1 in the line y = x.
Logarithm to the base 10 is called common logarithm.
Logarithm to the base e is called natural logarithm and is denoted by ln.
y
Log
x Ln x
Log x
x
Examples:
1. Express 52 = 25 in logarithmic form. Ans. log5 25 = 2
2. Given: log2 32 = 5. Express it in exponential form. Ans. 25 = 32
3. Given: log3 N = 4, find the value of N.
Solution: We transform log3 N = 4 in exponential form.
N = 34
N=3.3.3.3
N = 81 Ans.
4. Given: logb 125 = 3, find the value of b.
Solution: We transform logb 125 = 3 in exponential form
b3 = 125
b3 = 5 3
 b = 5 Ans.
5. Given: log8 32 = x, find the value of x.
Solution: Expressing log8 32 = x in exponential form, we get
8x = 32
= 25
23x = 25
 3x = 5
x = 5/3 Ans.
(23)x
6.
From remark #2, we have blogb(x) = x and logbbx = x. This property of
exponential and logarithmic functions are illustrated in the following
examples.
log10 1000 = 3
bec. 103 = 1000
 log10 103 = 3,
10log10 3 = 3
log2 16
bec. 24 = 16
 log2 24 = 4,
2log2
=4
4
=4
log2 1/8 = -3 bec. 2-3 = 1/23 = 1/5
7.
From remark #3, common logarithms are logarithms to the base 10. This
is assumed whenever the base is not written. Hence,
log 100 = 2
log 1,000,000 = 6
log 104 = 4.
Furthermore, given any number N, this can be written in scientific notation
of the form
N = p x 10k,
where 1 < p < 10. Note that the decimal point of N can be anywhere while
the decimal point of p is immediately to the right of the first non-zero digit in
the number, e.g.
753,800 = 7.538 x 105
0.00059012 = 5.9012 x 10-4
Such a decimal point is called the reference position of the decimal point.
If we let k be the number of digits between the decimal point and the
reference position, then k is positive or negative depending on whether the
decimal point is to the right or to the left of the reference position.
Also, since log N = log p x 10k
= log (10k x p)
= k log 10 + log p
 log N = k + log p
Here, the number k is called the characteristic, and log p is called the
mantissa, the decimal part less than 1. Note that the mantissa is
independent of the decimal point. That is, the mantissa of log 7.538 is the
same as that of log 7538, log 753.8, log 0.07538, and so on, wherever the
decimal point is placed. It is the characteristic which determines the
position of the decimal point. Using the examples above, we can then say
log 753,800 = 5 + log 7.538  5 + 0.8773  5.8773
log 0.00059012 = -4 + log 5.9012  -4 + 0.7709  -3.2291.
Here, the characteristic and mantissa, respectively, are 5 and 0.8773, for
the first, and, -4 and 0.7709 for the second.
8.
From remark #4, we have ln x = logex. Hence,
ln 2  .69315 because e.69315 = 2
This example can also be viewed as follows:
From remark #2 on the definition of exponential function (Section 8.1), we
have
2t = ekt, how do we solve for k?
Solution:
By taking the tth root of both sides of the equation, we obtain,
2 = ek
ln 2 = ln ek
k = ln 2  .69
(remark #2, defn of logarithmic function)
8.2.1 Exercise Set 8.2
A.
Express each of the given equations in logarithmic form.
1. 32 = 9
2. 52 = 25
3. 43 = 64
6. 81/3 = 2
7. 321/5 = 2
1
8. 144 = 12
2
2
1
1
4.   
16
4
9. 3-2 = 1/9
3
1
10.   = 8
3
Express each of the given equations in exponential form.
1. log4 16 = 2
6. log16 8 = 3/4
2. log3 27 = 3
7. log1/27 1/3 = 1/3
3. log5 625 = 4
8. log10 (0.01) = -2
4. log15 1 = 0
9. log0.5 16 = -4
1
5. log25 5 =
10. log7 (1/49) = -2
2
5. (7)o = 1
B.
C.
Determine the value of the unknowns for the following equations.
1. log7 N = 3
2. log6 N = 4
3. log4 N = 3
1
4. log4 N =
2
5. log27 N = 4/3
6. log125 N = 2/3
7. log64 N = 5/3
8. log32 N = -4/5
9. log16/81 N = -3/2
10. log25/36 N = -3/2
11. logb 81 = 2
12. logb 128 = 7
13. logb 1024 = 5
14. logb 64 = 3/4
15. logb 81 = 4/5
16. logb 32 = 5/6
17. logb 125 = 3/4
18. logb 4 = -1/3
19. logb 1/4 = -1/3
20. logb 5/8 = -1
21. log4 16 = x
22. log5 125 = x
23. log16 5/8 = x
24. log1/27 1/3 = x
25. log64/125 16/25 = x
26. log10 0.01 = x
27. log3 27-1 = x
28. log10 100.2 = x
29. log5 51.3 = x
30. log/6 36 = x
8.3
8.3 Properties of Logarithms
Thm. (Laws of Logarithms)
1.
loga xy = loga x + loga y
2.
loga x/y = loga x - loga y
3.
loga xp = p loga x
Proof:

1.
Let x = logb A and y = logb B
A = bx and B = by
Then, AB = bxby

AB = bx+y
 logb AB = logb bx+y
 logb AB = x + y
 logb AB = logb A + logb B
This property is also true for any finite number of factors, that is,
logb A1A2A3 ...Ak = logbA1 + logbA2 + logbA3 + ... logbAk
2.
A bx

B by
A
 bx y
B
logb A/B = logb bx-y
logb A/B = x - y
logb A/B = logb A - logb B
3.
A = bx
(A)k = (bx)k
Ak = bkx
logb Ak = logbbkx
logb Ak = kx
logb Ak = k logb A
Remarks:
1.
logb 1 = 0 because b0 = 1
2.
logb b = 1 because b1 = b
3. Change of base formula
log B a 
log x a
log x b
Examples:
1.
a)
log 100,000 = log (100)(1000) = log 100 + log 1000
=2+3
b) log2 64 = log2(4)(16)
= log2 4 + log2 16
=2+4
243
 log 3 243  log 3 9  log 3 27
9
log2 64 = log2 43 = 3 log2 4
= 3(2)
2. log3
3.
4. Change of base formula
log4 1,078 =
or log4 1,078 =
log 1078 3.0326

 5.0371
log 4
0.6021
ln 1079 6.9829

 5.0371
ln 4
1.3863
8.4
A.
Exercise Set 8.3
Write each in expanded form
1. log5 xy
6.
2. log6abc
7.
3. ln s/t
8.
4. log4 xy/z
9.
ln (a3)
log6(42  53)
log8 4/125
logb (x2yz/ pq )2/3
10. log10 (68)7 ( 147 )
5. log7 s/pt
B.
Write each in contracted form or as the logarithm of a single quantity
1. ln a + ln c
9. -loga x + 6 loga (x - 1) + 3 logax2
2. log2 3 + log2 x
10. 2 ln 8 + 1
3. log5 9 - log5 3
11. ½loga x - 2/3 loga y
4. log8 6 - log8 a
12. ½logb x - 3 logb y - 4 logb x
3
5. log4 3 + log4 9
13. 1/3 (2 ln x - 3 ln y)
6. 2 ln 2 + 3 ln n
14. 2(logb4 x - 1/3 logb y6 + 3 logb z )
7. ½ logb a - 2 logb n
15. 3/2 logb x + 1/3 logb y - 2 logb z)
8. logb 2x + 3(logb x - logb y)
C.
Find the value of the unknown in each of the following
1
4
1. ln x = 2 ln 5
D.
2. 5 log2 2 – ½ log2 16 = x
6. log5 ( x + 2) = log5 4
3
2
7. log2 ( x + 3) = log2 7
3. 2 log10 x = -log10 36
4. log10 (y + 2) = 1
5. ½ ln (x + 3) = 2
6
5
8. log8 ( x + 5) = log8 37
9. ½ log6 (x2 - 4x + 4) = 0
10. ½ log5 (x2 + 2x + 5) = log5 (x+3)
Determine the value of the following expressions using only the given
information:
For numbers 1-3, use log 5 = 0.70 and log 7 = 0.85
 125 
1. log 

 49 
 1
2. log 
o
3
 5 7
3. log 5
710
56




For numbers 4-8, use log10 2 = 0.301030, log10 3 = 0.477121 and log10 5 =
0.6989
4. log10 20
5. log10 50
6. log10 36
8.5
7. log10 9/5
8. log10 240.3
Equations involving Exponential and Logarithmic Expressions
To solve equations involving exponential and logarithmic expressions, we
apply the properties discussed in the previous sections.
Solving unknowns in equations involving logarithms:
a. log x2 + log x = 3
Solution:
log x2 + log x = log (x2  x)
= log x3
 log x3 = 3
10log x3 = 103
x3 = 103
x = 10
b. ln (x + 1) - ln(x - 1) =1
Solution:
 x 1
ln(x + 1) – ln(x – 1) = ln 
 1
 x 1
  x  1 
 exp ln 
  exp(1)
  x  1 

x 1
e
x 1
 x + 1 = xe – e
 x – xe = -1 – e
 x(1 – e) = -1 – e
x=-
e 1
e 1

 2.16
 (e  1) e  1
Solving unknowns in exponents:
a. 10x2 = 73
Solution:
log 10x2 = log 73
x2 = log 73
x =  log 73
=  .365
b. 22x-1 = 10x
Solution:
log 22x-1 = log 10x
(2x - 1)log 2 = x
2x log 2 - x = log 2
x(2 log 2 - 1) = log2
log 2
x = ------------  - 0.756
2 log 2-1
8.5.1 Exercise Set 8.4
Find the unknown for each equation below:
1.
log x = 3 log 2 + 2 log 3
2.
log b = log 2 - log 3 + log 5 - log 7
3.
ln x2 - ln x = ln 18 - ln 6
4.
log (x + 3) - log x + log 2x2 = log 8
5.
log4 (3x + 5) = log4 72 - 3 log4 2
6.
3x = 300
7.
34x = 5x+1
8. 23t = 3t+2
9. 102-3s = 6240
10. 2x2 = 3x
Math 12 - Plane Trigonometry
COURSE OUTLINE
CHAPTER I. INTRODUCTION ..................................................................................................................... 1
1.1 THE CARTESIAN PLANE AND THE DISTANCE FORMULA.......................................................... 1
1.1.1 The Cartesian Coordinate System ............................................................................................... 1
1.1.2 Exercise Set 1.1.1 ......................................................................................................................... 3
1.2
THE DISTANCE FORMULA .................................................................................................................. 4
1.2.1 Exercise Set 1.1.2 ......................................................................................................................... 6
1.3
ANGLES AND THEIR MEASURES ......................................................................................................... 7
1.4
1.4.1 Exercise Set 1.2 .......................................................................................................................... 13
1.5
CENTRAL ANGLES, ARC LENGTHS AND SECTOR AREAS .................................................................. 14
1.6
APPLICATIONS IN ANGULAR AND LINEAR VELOCITIES .................................................................... 16
1.6.1 Exercise Set 1.3 .......................................................................................................................... 19
1.7
FUNCTIONS AND THEIR GRAPHS ....................................................................................................... 23
1.7.1 Exercise Set 1.4: ........................................................................................................................ 25
CHAPTER 2
TRIGONOMETRIC FUNCTIONS - THE RIGHT TRIANGLE ................................... 26
2.1
RIGHT TRIANGLES AND THEIR PROPERTIES ...................................................................................... 26
2.1.1 Exercise 2.1 ................................................................................................................................ 28
2.2
TRIGONOMETRIC FUNCTIONS OF ANGLES ON A RIGHT TRIANGLE ................................................... 29
2.3
FUNDAMENTAL TRIGONOMETRIC IDENTITIES .................................................................................. 30
2.3.1 Exercise Set 2.3: ........................................................................................................................ 34
2.4
SOLUTIONS TO RIGHT TRIANGLES AND APPLICATIONS .................................................................... 35
2.5
APPLICATION PROBLEMS INVOLVING RIGHT TRIANGLES ................................................................. 36
2.5.1 Exercise Set 2.4 .......................................................................................................................... 38
CHAPTER 3
CIRCULAR FUNCTIONS ................................................................................................. 41
3.1
THE UNIT CIRCLE AND CIRCULAR FUNCTIONS ................................................................................ 41
3.2
THE CIRCULAR FUNCTIONS.............................................................................................................. 41
3.3
TRIGONOMETRIC FUNCTIONS OF QUADRANTAL ANGLES ................................................................. 42
3.4
TRIGONOMETRIC FUNCTIONS OF SPECIAL ANGLES: THE CIRCULAR APPROACH .............................. 43
3.4.1 Exercise Set 3.4 .......................................................................................................................... 46
CHAPTER 4
TRIGONOMETRIC FUNCTIONS THE GENERAL ANGLE ...................................... 47
4.1
TRIGONOMETRIC FUNCTIONS OF GENERAL ANGLES ........................................................................ 47
4.1.1 Exercise Set 4.1: ......................................................................................................................... 49
4.2
TRIGONOMETRIC FUNCTIONS OF SPECIAL ANGLES .......................................................................... 50
4.3
THE TRIGONOMETRIC TABLE ........................................................................................................... 53
4.3.1 Exercise Set 4.3: ........................................................................................................................ 57
CHAPTER 5
TRIGONOMETRIC IDENTITIES ................................................................................... 58
5.1
PYTHAGOREAN IDENTITIES .............................................................................................................. 58
5.1.1 Exercise Set 5.1 .......................................................................................................................... 64
5.2
SUM AND DIFFERENCE IDENTITIES ................................................................................................... 65
5.3
SPECIAL REDUCTION FORMULA ....................................................................................................... 66
5.3.1 Exercise Set 3.2 .......................................................................................................................... 70
5.4
REDUCTION FORMULAS ................................................................................................................... 71
5.4.1 Exercise Set 5.3 .......................................................................................................................... 76
5.4
DOUBLE-ANGLE FORMULAS ..................................................................................................... 77
5.4.2 Exercise Set 5.4 .......................................................................................................................... 79
5.5
HALF-ANGLE FORMULAS................................................................................................................. 80
5.5.1 Exercise Set 5.5 .......................................................................................................................... 82
5.6
SUM AND PRODUCT FORMULAS ....................................................................................................... 83
5.6.1 Thm. Product Formulas ............................................................................................................ 83
5.6.2 Thm. Sum formulas .................................................................................................................. 84
5.6.3 Exercise Set 5.6 .......................................................................................................................... 87
5.7
CHAPTER EXERCISE: ........................................................................................................................ 88
CHAPTER 6
GRAPHS OF TRIGONOMETRIC FUNCTIONS AND THEIR INVERSES ............... 91
6.1
ARCCSC ARCSEC AND ARCCOT ........................................................................................................ 96
6.1.1 Exercises 6.2 ............................................................................................................................ 100
CHAPTER 7
SOLUTIONS TO OBLIQUE TRIANGLES AND APPLICATIONS ........................... 101
7.1
LAW OF SINES ................................................................................................................................ 101
7.2
SOLVING TRIANGLES: FOUR CASES .................................................................................... 103
7.2.1 Exercise Set 7.1 ........................................................................................................................ 108
7.3
LAW OF COSINES ........................................................................................................................... 110
7.4
SOLVING A TRIANGLE ............................................................................................................. 114
7.4.1 Exercise Set 7.2 ........................................................................................................................ 115
7.4.2 Exercise Set 7.3 ........................................................................................................................ 122
CHAPTER 8
EXPONENTIAL AND LOGARITHMIC FUNCTIONS ............................................... 123
8.1
EXPONENTIAL FUNCTIONS ............................................................................................................. 123
8.1.1 Exercise Set 8.1 ........................................................................................................................ 127
8.2
LOGARITHMIC FUNCTION ............................................................................................................... 128
8.2.1 Exercise Set 8.2 ........................................................................................................................ 133
8.3
8.3 PROPERTIES OF LOGARITHMS .................................................................................................. 134
8.4
EXERCISE SET 8.3 .......................................................................................................................... 136
8.5
EQUATIONS INVOLVING EXPONENTIAL AND LOGARITHMIC EXPRESSIONS ..................................... 137
8.5.1 Exercise Set 8.4 ........................................................................................................................ 140
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