LHS Trig 8th ed Ch 42 Notes F07 O’Brien
Trigonometry Lecture Notes
Chapter 2
Section 2.1Trigonometric Functions of Acute Angles
I.Right-Triangle Based Definitions of Trigonometric Functions
For any acute angle A in standard position,
sin A = = cos A = = tan A = =
csc A = = sec A = = cot A = =
Example 1Find the values of sin A, cos A, and tan A in the right triangle shown.
II.Cofunction Identities
For any acute angle A,
sin A = cos (90° - A)cos A = sin (90° - A)tan A = cot (90° - A)
csc A = sec (90° - A)sec A = csc (90° - A)cot A = tan (90° - A)
Example 2Write each function in terms of its cofunction.
a) cos 38°b) sec 78°
Example 3Find one solution for the equation:cot (4θ + 8°) = tan (2θ + 4°)
Assume all angles are acute angles.
III.Special Triangles
Example 4Give the exact value.
a) cos 30°b) cot 45°
IV.Function Values of Special Angles
Example 5Find the exact value of each part labeled with a variable.
Section 2.1 Homework: 1, 3, 5, 9, 17, 23, 37, 41, 43, 63
Section 2.2Trigonometric Functions of Non-Acute Angles
I.Reference Angles
A reference angle for an angle θ is the positive acute angle made by the terminal side of angle θ and the x-axis.
Caution: The reference angle is always found with reference to the x-axis, never the y-axis.
Example 1Find the reference angle for each angle.
a) 218°b) 1387°
II.Finding Trigonometric Function Values for any Nonquadrantal Angle θ
1.If θ > 360°, or if θ < 0°, then find a coterminal angle by adding or subtracting 360° as many times as needed to get an angle greater than 0° but less than 360°.
2.Find the reference angle θ′.
3.Find the trigonometric function values for reference angle θ′.
4.Determine the correct signs for the values found in Step 3. This gives the values of the trigonometric functions for angle θ.
Example 2Find exact values of the six trigonometric functions for 210°.
Example 3Find exact values of the six trigonometric functions for -1020°.
III.Finding the Angle Given the Trigonometric Function Value
Example 4Find all values of θ in the interval [0°, 360°) that has the given function value.
Section 2.2 Homework: 1, 4, 6, 19, 22, 30, 62, 64, 65
Section 2.3Finding Trigonometric Function Values Using a Calculator
I.Function Values Using a Calculator
A.When evaluating trigonometric functions of angles given in degrees, remember that the calculator must be set in degree mode.
B.Remember that most calculator values of trigonometric functions are approximations.
Example 1Use a calculator to give a decimal approximation for each value. Give as many digits as your calculator displays.
a) sin 38° 24′b) cot 68.4832°
Example 2Use a calculator to evaluate the expression:cos 75° cos 14° - sin 75° sin 14°
II.Finding Angle Measures Using a Calculator
A.Graphing calculators have three inverse trigonometric functions.
B.If sin θ = x, then θ = sin -1 x for θ in the interval [0°, 90°].
Example 3Find a value of θ in the interval [0°, 90°] that satisfies each statement.
a) sin θ = .8535508b) sec θ = 2.486879
Section 2.3 Homework: 5, 8, 13, 20, 23, 26, 34, 35
Section 2.4Solving Right Triangles
I.Significant Digits for Angles
A.A significant digit is a digit obtained by actual measurement.
B.When performing calculations, your answer is no more accurate than the least accurate number in your calculation.
C.Start by determining the least number of significant digits in the given numbers and round your final answer to the same number of significant digits as this number.
Number ofSignificant Digits / Angle Measure to Nearest: / Example
2 / Degree / 52˚
3 / Ten minutes, or nearest tenth of a degree / 52˚ 30′ = 52.5˚
4 / Minute, or nearest hundredth of a degree / 52˚ 45′ = 52.75˚
5 / Tenth of a minute, or nearest thousandth of a degree / 52˚ 40.5′ = 52.675˚
II.Solving Triangles
A.To solve a triangle means to find the measures of all the angles and sides of the triangle.
B.Denote the angles of a triangle by capital letters. Then use the corresponding lower case letters to denote the respective opposite sides.
C.If the letters A, B, and C, are used to denote the angles of a right triangle, then it is usually assumed that C is the right angle.
Example 1Solve the following right triangle.
Example 2Solve right triangle ABC if b = 219 m and c = 647 m and C = 90˚.
(When two sides are given, give angles in degrees and minutes.)
III.Solving Applied Trigonometry Problems
A.Draw a sketch, and label it with the given information. Label the quantity to be found with a variable.
B.Use the sketch to write an equation relating the given quantities to the variable.
C.Solve the equation, and check that your answer makes sense.
Example 3Find the altitude of an isosceles triangle having base 184.2 cm if the angle opposite the base is 68˚ 44′.
IV.Angles of Elevation and Depression
When a horizontal line of sight is used as a reference line, the angle measured above the line of
sight is called an angle of elevation, while the angle measured below the line of sight is called
an angle of depression.
Example 4An airplane is flying 10,500 feet above the level ground. The angle of depression from the plane to the base of a tree is 13˚ 50′. How far horizontally must the plane fly to be directly over the tree?
Section 2.4 Homework: 9, 10, 13, 14, 21, 23, 35, 39, 43, 44
Section 2.5Further Applications of Right Triangles
I.Bearing
A.When a single angle is given as a bearing, it is understood that the bearing is measured in a clockwise direction from the north.
bearing of 32°bearing of 229°
B.The second method for expressing bearing starts with a north-south line and uses an acute angle to show the direction, either east or west, from this line.
Example 1A ship travels 50 km on a bearing of 27°, then travels on a bearing of 117° for 140 km. Find the distance traveled from the starting point to the ending point.
Example 2The bearing from Atlanta to Macon is S 27° E, and the bearing from Macon to Augusta is N 63° E. An automobile traveling at 60 mph needs 1 ¼ hour to go from Atlanta to Macon and 1 ¾ hour to go from Macon to Augusta. Find the distance from Atlanta to Augusta.
II.Problems Involving Angles of Elevation/Depression
Example 3Sean wants to know the height of a Ferris wheel. From a given point on the ground, he finds the angle of elevation to the top of the Ferris wheel is 42.3°. He then moves back 75 ft. From the second point, the angle of elevation to the top of the Ferris wheel is 25.4°. Find the height of the Ferris wheel.
Section 2.5 Homework: 5, 11, 15, 23, 25, 26, 32
Trigonometry
Chapter 4 Lecture Notes
Section 4.1Graphs of the Sine and Cosine Functions
I.Periodic Functions
A periodic function is one whose function values repeat in a regular pattern. They are used
to describe cyclical phenomena. All six trigonometric functions are periodic.
The smallest increment containing one full repetition of the pattern is called the period of the
function.
II.The Graph of the “Plain Vanilla” Sine Function
The graph of the “plain vanilla” sine function, y = sin x, has the following characteristics:
a.Domain: (-∞, ∞) Range: [-1, 1]
b.Amplitude: 1
Note: The amplitude is the height above or the depth below the graph’s horizontal axis.
It is half the distance between the maximum and minimum y-values.
c.Period: 2π; x-increment: Note: The x-increment is usually of the period.
d.5 key points: (0, 0) .
Note that its y-values start in the “neutral position” with y1 = 0.
e.Sine is an odd function with origin symmetry. For odd functions, f(–x) = –f(x), so for
sine, sin (–x) = –sin x. If the point (a, b) is on the graph of an odd function, the point (–a, –b)
is also on the graph.
III.The Graph of the “Plain Vanilla” Cosine Function
The graph of the “plain vanilla” cosine function, y = cos x, has the following characteristics:
a.Domain: (-∞, ∞) Range: [-1, 1]
b.Amplitude: 1
c.Period: 2π; x-increment:
d.5 key points: (0, 1) .
Note that its y-values start in the “up position” with y1 = 1.
e.Cosine is an even function with y-axis symmetry. For even functions, f(–x) = f(x), so for
cosine, cos (–x) = cos (x). Thus, if the point (a, b) is on the graph of an even function, the
point (–a, b) is also on the graph.
IV.Transformations of the “Plain Vanilla” Sine and Cosine Functions
A.Amplitude and x-axis reflection
In the functions y = a sin x and y = a cos x, the a-value controls two things – the amplitude of
the graph and whether there has been an x-axis reflection.
The amplitude of the graph is .
If a < 0, the graph has been reflected over the x-axis.
B.Period and x-increment
In the functions y = sin bx and y = cos bx, the b-value controls the period, which in turn
controls the x-increment.
The period is and the x-increment is .
Example 1Find the amplitude, period, and x-increment for each of the given functions and
identify whether there has been an x-axis reflection.
a.b.y = –5 sin πx
a.a = amp = ; there is no x-axis reflection.
b = 3 period = ; x-increment =
b.a = –5 amp = 5; there is an x-axis reflection
b = π period = ; x-increment =
Example 2Graph two full periods of y = 3 sin 2x. Give a, b, amplitude, whether there is an
x-axis reflection, period, x-increment, and 5 key points.(#24)
a = 3amp = 3no x-axis reflection
b = 2period = x-increment =
(0, 0)
Example 3Graph two full periods of . Give a, b, amplitude, whether there is an
x-axis reflection, period, x-increment, and 5 key points.(#24)
a = –2amp = 2There is an x-axis reflection
b = period = x-increment =
(0, –2) (2, 0) (4, 2) (6, 0) (8, –2)
Section 4.2Translations of the Graphs of the Sine and Cosine Functions
I.Phase Shifts (a.k.a. Horizontal Translations)
To find the phase shift of a function in the form y = a sin (bx – c) or y = a cos (bx – c),
set the argument of the function (what is inside parentheses, after the function name)
equal to zero and solve for x: .
If is positive (i.e. you see subtraction in the parentheses), the phase shift is to the right.
If is negative (i.e. you see addition in the parentheses), the phase shift is to the left.
II.Vertical Shifts (a.k.a. Vertical Translations)
In the functions y = a sin (bx – c) + d and y = a cos (bx – c) + d, the d controls the vertical shift.
If d is positive, vertical the shift is up. If d is negative, the vertical shift is down.
Example 1Find the amplitude, period, x-increment, phase shift, and vertical shift for each of the
given functions and identify whether there has been an x-axis reflection.
a.y = cos (2x – 3π) – 1 (#22)b.y = 2 – 3sin
a.a = amp = ; there is no x-axis reflection.
b = 2 period = ; x-increment =
c = 3π 2x – 3π = 0 2x = 3π x = phase shift: right
d = –1 vertical shift: 1 down
b.a = –3 amp = 3; there is an x-axis reflection
b = period = ; x-increment =
c = = 0 phase shift: left
d = 2 vertical shift: 2 up
III.How to Graph y = a sin (bx – c) + d and y = a cos (bx – c) + d
1.Identify a, b, c, and d.
2a.Find the amplitude: amp = |a|.
b.Determine if there has been an x-axis reflection: if a is negative, there has been an x-axis
reflection.
3a.Find the period: period = .
b.Find the x-increment: x-increment = .
4.Find the phase shift by setting the argument, bx – c, equal to zero and solving for x:
P.S. = right (if we see bx – c and is positive) or
P.S. = left (if we see bx + c and is negative).
5.Find the vertical shift: |d| up if d is positive or |d| down if d is negative.
6.Identify the five key points:
for sine: (x1 + x-inc, ) (x2 + x-inc, d) (x3 + x-inc, d – a) (x4 + x-inc, d)
for cosine: (x1 + x-inc, d) (x2 + x-inc, d – a) (x3 + x-inc, d) (x4 + x-inc, d + a)
7.Plot the five key points and connect them with a smooth, continuous curve, then extend the graph
to two full periods.
Example 2: y = –3 sin (2x – π) + 1
1.a = –3 b = 2 c = π d = 1
2.amp = |–3| = 3 There is an x-axis reflection.
3.period = x-increment =
4.phase shift = or right
5.vertical shift = 1 up
6.5 key points:
7.
Example 3: y = cos (4x + π) – 2
1.a = b = 4 c = –π d = –2 = –
2.amp = = There is no x-axis reflection.
3.period = x-increment =
4.phase shift = or left
5.vertical shift = 2 or down
6.5 key points:
7.
Section 4.3Graphs of the Other Circular Functions
Note:The concept of amplitude does not apply to cosecant, secant, tangent, or cotangent. We can only
discuss the amplitude of a sine or cosine function.
I.The Basic Cosecant Function
The graph of y = csc x has a period of 2π and an x- increment of . It is an odd function with
origin symmetry. Because , the cosecant function is undefined whenever sin x = 0.
Thus the domain of csc x is all real numbers except integer multiples of π. At x = nπ, where n is
an integer, the graph of y = sin x crosses the horizontal axis and the graph of y = csc x has a
vertical asymptote. The range of csc x is (-∞, 1] U [1, ∞).
The general procedure for graphing y = a csc (bx – c) + d is to graph y = a sin (bx – c) + d with a
dotted line and then to use that graph as a guide for graphing the cosecant. The graph of
y = csc x has vertical asymptotes at x = 0, x = π, and x = 2π. It has a minimum (low point) at
and a maximum (high point) at .
II.The Basic Secant Function
The graph of y = sec x has a period of 2π and an x- increment of . It is an even function with
y-axis symmetry. Because , the secant function is undefined whenever cos x = 0.
Thus the domain of sec x is all real numbers except odd multiples of . At , where n
is an integer, the graph of y = cos x crosses the horizontal axis and the graph of y = sec x has a
vertical asymptote. The range of sec x is (-∞, 1] U [1, ∞).
The general procedure for graphing y = a sec (bx – c) + d is to graph y = a cos (bx – c) + d with a
dotted line and then to use that graph as a guide for graphing the secant. The graph of y = sec x
has vertical asymptotes at and . It has minima (low points) at (0, 1) and (2π, 1).
It has a maximum (high point) at .
III.How to Graph y = a csc (bx – c) + d and y = a sec (bx – c) + d
1.Write the guide function. For the guide function is and
for the guide function is .
2.Identify a, b, c, and d.
3a.Find the amplitude of the guide function: amp = |a|.
Note: Cosecant and secant do not have amplitude.
b.Determine if there has been an x-axis reflection: if a is negative, there has been an x-axis
reflection.
4a.Find the period: period = .
b.Find the x-increment: x-increment = .
5.Find the phase shift by setting the argument, bx – c, equal to zero and solving for x:
P.S. = right (if we see bx – c and is positive) or
P.S. = left (if we see bx + c and is negative).
6.Find the vertical shift: |d| up if d is positive or |d| down if d is negative.
7.Identify the five key points of the guide function:
for sine: (x1 + x-inc, d+a) (x2 + x-inc, d) (x3 + x-inc, d – a) (x4 + x-inc, d)
for cosine: (x1 + x-inc, d) (x2 + x-inc, d – a) (x3 + x-inc, d) (x4 + x-inc, d + a)
8.Plot the five key points of the guide function and connect them with a smooth, continuous,
dotted curve, then extend the graph to two full periods.
9.Draw the given function using these guidelines:
Anywhere the guide function crosses its horizontal axis (i.e., when y = d), the given function has
a vertical asymptote. Anywhere the guide function has a minimum, the given function has a
maximum and we draw a parabola shaped curve opening down. Anywhere the guide function
has a maximum, the given function has a minimum, and we draw a parabola shaped curve
opening up.
Example 1: y = 2 csc (x – 3π) – 1
1.Guide function: y = 2 sin (x – 3π) – 1
2.a = 2 b = 1 c = 3π = d = –1
3.amp of sine = |2| = 2 There is no x-axis reflection.
4.period = x-increment =
5.phase shift = 3π or right
6.vertical shift = 1 down
7.5 key points of the sine:
8.9.
Example 2: y = – sec + 3
1.Guide function: y = – cos + 3
2.a = – b = 1 c = – d = 3 =
3.amp of cosine = = There is an x-axis reflection.
4.period = x-increment =
5.phase shift = left
6.vertical shift = 3 or up
7.5 key points of the cosine:
8.9.
IV.The Basic Tangent Function
The graph of y = tan x has a period of π and an x- increment of . It is an odd function with
origin symmetry. Because , the tangent function is undefined whenever cos x = 0.
Thus the domain of tan x is all real numbers except odd multiples of . At , where n
is an integer, the graph of y = tan x has a vertical asymptote. The range of tan x is (-∞, ∞).
Notice that the graph of y = tan x is increasing along its entire length. Its left “hand” points
down and its right hand points up. The 5 key features of the graph of y = tan x are:
left asymptote: 3 key points: (0, 0) right asymptote:
V.The Basic Cotangent Function
The graph of y = cot x has a period of π and an x- increment of . It is an odd function with
origin symmetry. Because , the cotangent function is undefined whenever sin x = 0.
Thus the domain of cot x is all real numbers except integer multiples of π. At , where n is
an integer, the graph of y = cot x has a vertical asymptote. The range of cot x is (-∞,∞).
Notice that the graph of y = cot x is decreasing along its entire length. Its left “hand” points up
and its right hand points down. The 5 key features of the graph of y = cot x are:
left asymptote: x = 0 3 key points: right asymptote:
VI.How to Graph y = a tan (bx – c) + d and y = a cot (bx – c) + d
1.Identify a, b, c, and d.
2.Determine if there has been an x-axis reflection: if a is negative, there has been an x-axis
reflection.
3a.Find the period: period = .
b.Find the x-increment: x-increment = .
4.Find the phase shift by setting the argument, bx – c, equal to zero and solving for x:
P.S. = right (if we see bx – c and is positive) or
P.S. = left (if we see bx + c and is negative).
5.Find the vertical shift: |d| up if d is positive or |d| down if d is negative.
6.Identify the left asymptote, the three key points, and the right asymptote.
cotangent: To find the left asymptote, let x = phase shift, .
LA: (x of LA + x-increment, d + a) (x1 + x-inc, d) (x2 + x-inc, d – a) RA: x = x3 + x-inc
Note: The left asymptote of a tangent function is NOT x = phase shift.
tangent: To find the left asymptote, set the argument, bx – c, equal to and solve for x.
LA: (x of LA + x-inc, d – a) (x1 + x-inc, d) (x2 + x-inc, d + a) RA: x = x3 + x-inc
7.Plot the asymptotes and the three key points. Connect the points with a smooth, continuous curve.
Extend the graph to two full periods.
Example 3: y = 2 tan – 3
1.a = 2 b = c = π d = –3
2.There is no x-axis reflection.
3.period = x-increment =
4.phase shift = 4π right
5.vertical shift = 3 down
6.LA: (3π, –5) (4π, –3) (5π, –1) RA: x = 6π
7.
Example 4: y = –3 cot (2x + π) +1
1.a = –3 b = 2 c = –π d = 1
2.There is an x-axis reflection.
3.period = x-increment =
4.phase shift = left
5.vertical shift = 1 up
6.LA: RA: x = 0
7.
Section 4.4Harmonic Motion
I.Definition
Simple harmonic motion is a periodic, oscillating movement which can be described by a
sinusoidal function. Examples of physical phenomena which can be described as simple
harmonic motion include radio and television waves; light waves; sound waves; water waves;
the swinging of a pendulum; the vibrations of a tuning fork; and the bobbing of a weight attached
to a coiled spring.
II.The Movement of a Mass on a Spring
Consider a mass on a spring. The system is said to be in equilibrium when the mass is at rest.
The point of rest is called the origin of the system. We consider the distance above equilibrium
as positive and the distance below equilibrium as negative.
If the mass is lifted a distance a and released, it will oscillate up and down in periodic motion.
If there is no friction, the motion repeats itself in a certain period of time. The distance a is
called the displacement from the origin. The number of times the mass oscillates in 1 unit of
time is called the frequency f of the motion, and the time it takes for the mass to complete one
oscillation is called its period. The frequency and period are related by the formulas: and . The maximum displacement from equilibrium is called the amplitude of the
motion.
III.Equations Modeling Simple Harmonic Motion
If the displacement from the origin is at a maximum at time t = 0, simple harmonic motion can
be modeled by y = a cos ωt.
If the displacement from the origin is zero at time t = 0, simple harmonic motion can be modeled