unit 12 trigonometry homework 1 answer key pdf

Trigonometry (Algebra 2 Curriculum - Unit 12) | All Things Algebra®

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Due to the length of this Trigonometry Unit Bundle , it is divided into two parts with two unit tests. In addition to the unit tests, each part includes guided notes, homework assignments, quizzes, and study guides to cover the following topics:

Unit 12 Part I:

• Pythagorean Theorem

• Special Right Triangles

• Trigonometric Functions (sin, cos, tan, csc, sec, cot)

• Finding Side and Angle Measures

• Applications: Angle of Elevation and Depression

• Angles in Standard Position

• Converting between Degrees and Radians

• Coterminal and Reference Angles

• Trigonometric Functions in the Coordinate Plane

• The Unit Circle

• Law of Sines

• Law of Cosines

• Area of Triangles

• Applications of Law of Sines, Law of Cosines, and Area

Unit 12 Part II:

• Graphing Trigonometric Functions

• Trigonometric Identities

• Sum and Difference of Angle Identities

• Double-Angle and Half-Angle Identities

• Solving Trigonometric Equations

ADDITIONAL COMPONENTS INCLUDED:

(1) Links to Instructional Videos: Links to videos of each lesson in the unit are included. Videos were created by fellow teachers for their students using the guided notes and shared in March 2020 when schools closed with no notice.  Please watch through first before sharing with your students. Many teachers still use these in emergency substitute situations. (2) Editable Assessments: Editable versions of each quiz and the unit test are included. PowerPoint is required to edit these files. Individual problems can be changed to create multiple versions of the assessment. The layout of the assessment itself is not editable. If your Equation Editor is incompatible with mine (I use MathType), simply delete my equation and insert your own.

(3) Google Slides Version of the PDF: The second page of the Video links document contains a link to a Google Slides version of the PDF. Each page is set to the background in Google Slides. There are no text boxes;  this is the PDF in Google Slides.  I am unable to do text boxes at this time but hope this saves you a step if you wish to use it in Slides instead! 

This resource is included in the following bundle(s):

Algebra 2 Curriculum

More Algebra 2 Units:

Unit 1 – Equations and Inequalities

Unit 2 – Linear Functions and Systems

Unit 3 – Parent Functions and Transformations

Unit 4 – Solving Quadratics and Complex Numbers

Unit 5 – Polynomial Functions

Unit 6 – Radical Functions

Unit 7 – Exponential and Logarithmic Functions

Unit 8 – Rational Functions

Unit 9 – Conic Sections

Unit 10 – Sequences and Series

Unit 11 – Probability and Statistics

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COPYRIGHT TERMS: This resource may not be uploaded to the internet in any form, including classroom/personal websites or network drives, unless the site is password protected and can only be accessed by students.

© All Things Algebra (Gina Wilson), 2012-present

Questions & Answers

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1.1 Functions and Function Notation

• ⓑ yes. (Note: If two players had been tied for, say, 4th place, then the name would not have been a function of rank.)

w = f ( d ) w = f ( d )

g ( 5 ) = 1 g ( 5 ) = 1

m = 8 m = 8

y = f ( x ) = x 3 2 y = f ( x ) = x 3 2

g ( 1 ) = 8 g ( 1 ) = 8

x = 0 x = 0 or x = 2 x = 2

• ⓐ yes, because each bank account has a single balance at any given time
• ⓑ no, because several bank account numbers may have the same balance
• ⓒ no, because the same output may correspond to more than one input.
• ⓐ Yes, letter grade is a function of percent grade;
• ⓑ No, it is not one-to-one. There are 100 different percent numbers we could get but only about five possible letter grades, so there cannot be only one percent number that corresponds to each letter grade.

No, because it does not pass the horizontal line test.

1.2 Domain and Range

{ − 5 , 0 , 5 , 10 , 15 } { − 5 , 0 , 5 , 10 , 15 }

( − ∞ , ∞ ) ( − ∞ , ∞ )

( − ∞ , 1 2 ) ∪ ( 1 2 , ∞ ) ( − ∞ , 1 2 ) ∪ ( 1 2 , ∞ )

[ − 5 2 , ∞ ) [ − 5 2 , ∞ )

• ⓐ values that are less than or equal to –2, or values that are greater than or equal to –1 and less than 3;
• ⓑ { x | x ≤ − 2 or − 1 ≤ x < 3 } { x | x ≤ − 2 or − 1 ≤ x < 3 } ;
• ⓒ ( − ∞ , − 2 ] ∪ [ − 1 , 3 ) ( − ∞ , − 2 ] ∪ [ − 1 , 3 )

domain =[1950,2002] range = [47,000,000,89,000,000]

domain: ( − ∞ , 2 ] ; ( − ∞ , 2 ] ; range: ( − ∞ , 0 ] ( − ∞ , 0 ]

1.3 Rates of Change and Behavior of Graphs

$ 2.84 − $ 2.31 5  years = $ 0.53 5  years = $ 0.106 $ 2.84 − $ 2.31 5  years = $ 0.53 5  years = $ 0.106 per year.

a + 7 a + 7

The local maximum appears to occur at ( − 1 , 28 ) , ( − 1 , 28 ) , and the local minimum occurs at ( 5 , − 80 ) . ( 5 , − 80 ) . The function is increasing on ( − ∞ , − 1 ) ∪ ( 5 , ∞ ) ( − ∞ , − 1 ) ∪ ( 5 , ∞ ) and decreasing on ( − 1 , 5 ) . ( − 1 , 5 ) .

1.4 Composition of Functions

( f g ) ( x ) = f ( x ) g ( x ) = ( x − 1 ) ( x 2 − 1 ) = x 3 − x 2 − x + 1 ( f − g ) ( x ) = f ( x ) − g ( x ) = ( x − 1 ) − ( x 2 − 1 ) = x − x 2 ( f g ) ( x ) = f ( x ) g ( x ) = ( x − 1 ) ( x 2 − 1 ) = x 3 − x 2 − x + 1 ( f − g ) ( x ) = f ( x ) − g ( x ) = ( x − 1 ) − ( x 2 − 1 ) = x − x 2

No, the functions are not the same.

A gravitational force is still a force, so a ( G ( r ) ) a ( G ( r ) ) makes sense as the acceleration of a planet at a distance r from the Sun (due to gravity), but G ( a ( F ) ) G ( a ( F ) ) does not make sense.

f ( g ( 1 ) ) = f ( 3 ) = 3 f ( g ( 1 ) ) = f ( 3 ) = 3 and g ( f ( 4 ) ) = g ( 1 ) = 3 g ( f ( 4 ) ) = g ( 1 ) = 3

g ( f ( 2 ) ) = g ( 5 ) = 3 g ( f ( 2 ) ) = g ( 5 ) = 3

[ − 4 , 0 ) ∪ ( 0 , ∞ ) [ − 4 , 0 ) ∪ ( 0 , ∞ )

Possible answer:

g ( x ) = 4 + x 2 g ( x ) = 4 + x 2 h ( x ) = 4 3 − x h ( x ) = 4 3 − x f = h ∘ g f = h ∘ g

1.5 Transformation of Functions

The graphs of f ( x ) f ( x ) and g ( x ) g ( x ) are shown below. The transformation is a horizontal shift. The function is shifted to the left by 2 units.

g ( x ) = 1 x - 1 + 1 g ( x ) = 1 x - 1 + 1

g ( x ) = − f ( x ) g ( x ) = − f ( x )

h ( x ) = f ( − x ) h ( x ) = f ( − x )

Notice: g ( x ) = f ( − x ) g ( x ) = f ( − x ) looks the same as f ( x ) f ( x ) .

g ( x ) = 3 x - 2 g ( x ) = 3 x - 2

g ( x ) = f ( 1 3 x ) g ( x ) = f ( 1 3 x ) so using the square root function we get g ( x ) = 1 3 x g ( x ) = 1 3 x

1.6 Absolute Value Functions

| x − 2 | ≤ 3 | x − 2 | ≤ 3

using the variable p p for passing, | p − 80 | ≤ 20 | p − 80 | ≤ 20

f ( x ) = − | x + 2 | + 3 f ( x ) = − | x + 2 | + 3

x = − 1 x = − 1 or x = 2 x = 2

f ( 0 ) = 1 , f ( 0 ) = 1 , so the graph intersects the vertical axis at ( 0 , 1 ) . ( 0 , 1 ) . f ( x ) = 0 f ( x ) = 0 when x = − 5 x = − 5 and x = 1 x = 1 so the graph intersects the horizontal axis at ( − 5 , 0 ) ( − 5 , 0 ) and ( 1 , 0 ) . ( 1 , 0 ) .

- 8 ≤ x ≤ 4 - 8 ≤ x ≤ 4

k ≤ 1 k ≤ 1 or k ≥ 7 ; k ≥ 7 ; in interval notation, this would be ( − ∞ , 1 ] ∪ [ 7 , ∞ ) ( − ∞ , 1 ] ∪ [ 7 , ∞ )

1.7 Inverse Functions

h ( 2 ) = 6 h ( 2 ) = 6

The domain of function f − 1 f − 1 is ( − ∞ , − 2 ) ( − ∞ , − 2 ) and the range of function f − 1 f − 1 is ( 1 , ∞ ) . ( 1 , ∞ ) .

• f ( 60 ) = 50. f ( 60 ) = 50. In 60 minutes, 50 miles are traveled.
• f − 1 ( 60 ) = 70. f − 1 ( 60 ) = 70. To travel 60 miles, it will take 70 minutes.

a. 3; b. 5.6

x = 3 y + 5 x = 3 y + 5

f − 1 ( x ) = ( 2 − x ) 2 ; domain of f : [ 0 , ∞ ) ; domain of f − 1 : ( − ∞ , 2 ] f − 1 ( x ) = ( 2 − x ) 2 ; domain of f : [ 0 , ∞ ) ; domain of f − 1 : ( − ∞ , 2 ]

1.1 Section Exercises

A relation is a set of ordered pairs. A function is a special kind of relation in which no two ordered pairs have the same first coordinate.

When a vertical line intersects the graph of a relation more than once, that indicates that for that input there is more than one output. At any particular input value, there can be only one output if the relation is to be a function.

When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to only one input.

not a function

f ( − 3 ) = − 11 ; f ( − 3 ) = − 11 ; f ( 2 ) = − 1 ; f ( 2 ) = − 1 ; f ( − a ) = − 2 a − 5 ; f ( − a ) = − 2 a − 5 ; − f ( a ) = − 2 a + 5 ; − f ( a ) = − 2 a + 5 ; f ( a + h ) = 2 a + 2 h − 5 f ( a + h ) = 2 a + 2 h − 5

f ( − 3 ) = 5 + 5 ; f ( − 3 ) = 5 + 5 ; f ( 2 ) = 5 ; f ( 2 ) = 5 ; f ( − a ) = 2 + a + 5 ; f ( − a ) = 2 + a + 5 ; − f ( a ) = − 2 − a − 5 ; − f ( a ) = − 2 − a − 5 ; f ( a + h ) = 2 − a − h + 5 f ( a + h ) = 2 − a − h + 5

f ( − 3 ) = 2 ; f ( − 3 ) = 2 ; f ( 2 ) = 1 − 3 = − 2 ; f ( 2 ) = 1 − 3 = − 2 ; f ( − a ) = | − a − 1 | − | − a + 1 | ; f ( − a ) = | − a − 1 | − | − a + 1 | ; − f ( a ) = − | a − 1 | + | a + 1 | ; − f ( a ) = − | a − 1 | + | a + 1 | ; f ( a + h ) = | a + h − 1 | − | a + h + 1 | f ( a + h ) = | a + h − 1 | − | a + h + 1 |

g ( x ) − g ( a ) x − a = x + a + 2 , x ≠ a g ( x ) − g ( a ) x − a = x + a + 2 , x ≠ a

• ⓐ f ( − 2 ) = 14 ; f ( − 2 ) = 14 ;
• ⓑ x = 3 x = 3
• ⓐ f ( 5 ) = 10 ; f ( 5 ) = 10 ;
• ⓑ x = − 1 x = − 1 or x = 4 x = 4
• ⓐ f ( t ) = 6 − 2 3 t ; f ( t ) = 6 − 2 3 t ;
• ⓑ f ( − 3 ) = 8 ; f ( − 3 ) = 8 ;
• ⓒ t = 6 t = 6
• ⓐ f ( 0 ) = 1 ; f ( 0 ) = 1 ;
• ⓑ f ( x ) = − 3 , x = − 2 f ( x ) = − 3 , x = − 2 or x = 2 x = 2

not a function so it is also not a one-to-one function

one-to-one function

function, but not one-to-one

f ( x ) = 1 , x = 2 f ( x ) = 1 , x = 2

f ( − 2 ) = 14 ; f ( − 1 ) = 11 ; f ( 0 ) = 8 ; f ( 1 ) = 5 ; f ( 2 ) = 2 f ( − 2 ) = 14 ; f ( − 1 ) = 11 ; f ( 0 ) = 8 ; f ( 1 ) = 5 ; f ( 2 ) = 2

f ( − 2 ) = 4 ;    f ( − 1 ) = 4.414 ; f ( 0 ) = 4.732 ; f ( 1 ) = 5 ; f ( 2 ) = 5.236 f ( − 2 ) = 4 ;    f ( − 1 ) = 4.414 ; f ( 0 ) = 4.732 ; f ( 1 ) = 5 ; f ( 2 ) = 5.236

f ( − 2 ) = 1 9 ; f ( − 1 ) = 1 3 ; f ( 0 ) = 1 ; f ( 1 ) = 3 ; f ( 2 ) = 9 f ( − 2 ) = 1 9 ; f ( − 1 ) = 1 3 ; f ( 0 ) = 1 ; f ( 1 ) = 3 ; f ( 2 ) = 9

[ 0 ,  100 ] [ 0 ,  100 ]

[ − 0.001 ,  0 .001 ] [ − 0.001 ,  0 .001 ]

[ − 1 , 000 , 000 ,  1,000,000 ] [ − 1 , 000 , 000 ,  1,000,000 ]

[ 0 ,  10 ] [ 0 ,  10 ]

[ −0.1 , 0.1 ] [ −0.1 , 0.1 ]

[ − 100 ,  100 ] [ − 100 ,  100 ]

• ⓐ g ( 5000 ) = 50 ; g ( 5000 ) = 50 ;
• ⓑ The number of cubic yards of dirt required for a garden of 100 square feet is 1.
• ⓐ The height of a rocket above ground after 1 second is 200 ft.
• ⓑ the height of a rocket above ground after 2 seconds is 350 ft.

1.2 Section Exercises

The domain of a function depends upon what values of the independent variable make the function undefined or imaginary.

There is no restriction on x x for f ( x ) = x 3 f ( x ) = x 3 because you can take the cube root of any real number. So the domain is all real numbers, ( − ∞ , ∞ ) . ( − ∞ , ∞ ) . When dealing with the set of real numbers, you cannot take the square root of negative numbers. So x x -values are restricted for f ( x ) = x f ( x ) = x to nonnegative numbers and the domain is [ 0 , ∞ ) . [ 0 , ∞ ) .

Graph each formula of the piecewise function over its corresponding domain. Use the same scale for the x x -axis and y y -axis for each graph. Indicate inclusive endpoints with a solid circle and exclusive endpoints with an open circle. Use an arrow to indicate − ∞ − ∞ or ∞ . ∞ . Combine the graphs to find the graph of the piecewise function.

( − ∞ , 3 ] ( − ∞ , 3 ]

( − ∞ , − 1 2 ) ∪ ( − 1 2 , ∞ ) ( − ∞ , − 1 2 ) ∪ ( − 1 2 , ∞ )

( − ∞ , − 11 ) ∪ ( − 11 , 2 ) ∪ ( 2 , ∞ ) ( − ∞ , − 11 ) ∪ ( − 11 , 2 ) ∪ ( 2 , ∞ )

( − ∞ , − 3 ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ ) ( − ∞ , − 3 ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ )

( − ∞ , 5 ) ( − ∞ , 5 )

[ 6 , ∞ ) [ 6 , ∞ )

( − ∞ , − 9 ) ∪ ( − 9 , 9 ) ∪ ( 9 , ∞ ) ( − ∞ , − 9 ) ∪ ( − 9 , 9 ) ∪ ( 9 , ∞ )

domain: ( 2 , 8 ] , ( 2 , 8 ] , range [ 6 , 8 ) [ 6 , 8 )

domain: [ − 4 ,  4], [ − 4 ,  4], range: [ 0 ,  2] [ 0 ,  2]

domain: [ − 5 , 3 ) , [ − 5 , 3 ) , range: [ 0 , 2 ] [ 0 , 2 ]

domain: ( − ∞ , 1 ] , ( − ∞ , 1 ] , range: [ 0 , ∞ ) [ 0 , ∞ )

domain: [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] ; [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] ; range: [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ]

domain: [ − 3 , ∞ ) ; [ − 3 , ∞ ) ; range: [ 0 , ∞ ) [ 0 , ∞ )

domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

f ( − 3 ) = 1 ; f ( − 2 ) = 0 ; f ( − 1 ) = 0 ; f ( 0 ) = 0 f ( − 3 ) = 1 ; f ( − 2 ) = 0 ; f ( − 1 ) = 0 ; f ( 0 ) = 0

f ( − 1 ) = − 4 ; f ( 0 ) = 6 ; f ( 2 ) = 20 ; f ( 4 ) = 34 f ( − 1 ) = − 4 ; f ( 0 ) = 6 ; f ( 2 ) = 20 ; f ( 4 ) = 34

f ( − 1 ) = − 5 ; f ( 0 ) = 3 ; f ( 2 ) = 3 ; f ( 4 ) = 16 f ( − 1 ) = − 5 ; f ( 0 ) = 3 ; f ( 2 ) = 3 ; f ( 4 ) = 16

domain: ( − ∞ , 1 ) ∪ ( 1 , ∞ ) ( − ∞ , 1 ) ∪ ( 1 , ∞ )

window: [ − 0.5 , − 0.1 ] ; [ − 0.5 , − 0.1 ] ; range: [ 4 , 100 ] [ 4 , 100 ]

window: [ 0.1 , 0.5 ] ; [ 0.1 , 0.5 ] ; range: [ 4 , 100 ] [ 4 , 100 ]

[ 0 , 8 ] [ 0 , 8 ]

Many answers. One function is f ( x ) = 1 x − 2 . f ( x ) = 1 x − 2 .

1.3 Section Exercises

Yes, the average rate of change of all linear functions is constant.

The absolute maximum and minimum relate to the entire graph, whereas the local extrema relate only to a specific region around an open interval.

4 ( b + 1 ) 4 ( b + 1 )

4 x + 2 h 4 x + 2 h

− 1 13 ( 13 + h ) − 1 13 ( 13 + h )

3 h 2 + 9 h + 9 3 h 2 + 9 h + 9

4 x + 2 h − 3 4 x + 2 h − 3

increasing on ( − ∞ , − 2.5 ) ∪ ( 1 , ∞ ) , ( − ∞ , − 2.5 ) ∪ ( 1 , ∞ ) , decreasing on ( − 2.5 , 1 ) ( − 2.5 , 1 )

increasing on ( − ∞ , 1 ) ∪ ( 3 , 4 ) , ( − ∞ , 1 ) ∪ ( 3 , 4 ) , decreasing on ( 1 , 3 ) ∪ ( 4 , ∞ ) ( 1 , 3 ) ∪ ( 4 , ∞ )

local maximum: ( − 3 , 50 ) , ( − 3 , 50 ) , local minimum: ( 3 , − 50 ) ( 3 , − 50 )

absolute maximum at approximately ( 7 , 150 ) , ( 7 , 150 ) , absolute minimum at approximately ( −7.5 , −220 ) ( −7.5 , −220 )

a. –3000; b. –1250

Local minimum at ( 3 , − 22 ) , ( 3 , − 22 ) , decreasing on ( − ∞ , 3 ) , ( − ∞ , 3 ) , increasing on ( 3 , ∞ ) ( 3 , ∞ )

Local minimum at ( − 2 , − 2 ) , ( − 2 , − 2 ) , decreasing on ( − 3 , − 2 ) , ( − 3 , − 2 ) , increasing on ( − 2 , ∞ ) ( − 2 , ∞ )

Local maximum at ( − 0.5 , 6 ) , ( − 0.5 , 6 ) , local minima at ( − 3.25 , − 47 ) ( − 3.25 , − 47 ) and ( 2.1 , − 32 ) , ( 2.1 , − 32 ) , decreasing on ( − ∞ , − 3.25 ) ( − ∞ , − 3.25 ) and ( − 0.5 , 2.1 ) , ( − 0.5 , 2.1 ) , increasing on ( − 3.25 , − 0.5 ) ( − 3.25 , − 0.5 ) and ( 2.1 , ∞ ) ( 2.1 , ∞ )

b = 5 b = 5

2.7 gallons per minute

approximately –0.6 milligrams per day

1.4 Section Exercises

Find the numbers that make the function in the denominator g g equal to zero, and check for any other domain restrictions on f f and g , g , such as an even-indexed root or zeros in the denominator.

Yes. Sample answer: Let f ( x ) = x + 1  and  g ( x ) = x − 1. f ( x ) = x + 1  and  g ( x ) = x − 1. Then f ( g ( x ) ) = f ( x − 1 ) = ( x − 1 ) + 1 = x f ( g ( x ) ) = f ( x − 1 ) = ( x − 1 ) + 1 = x and g ( f ( x ) ) = g ( x + 1 ) = ( x + 1 ) − 1 = x . g ( f ( x ) ) = g ( x + 1 ) = ( x + 1 ) − 1 = x . So f ∘ g = g ∘ f . f ∘ g = g ∘ f .

( f + g ) ( x ) = 2 x + 6 , ( f + g ) ( x ) = 2 x + 6 , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f − g ) ( x ) = 2 x 2 + 2 x − 6 , ( f − g ) ( x ) = 2 x 2 + 2 x − 6 , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f g ) ( x ) = − x 4 − 2 x 3 + 6 x 2 + 12 x , ( f g ) ( x ) = − x 4 − 2 x 3 + 6 x 2 + 12 x , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f g ) ( x ) = x 2 + 2 x 6 − x 2 , ( f g ) ( x ) = x 2 + 2 x 6 − x 2 , domain: ( − ∞ , − 6 ) ∪ ( − 6 , 6 ) ∪ ( 6 , ∞ ) ( − ∞ , − 6 ) ∪ ( − 6 , 6 ) ∪ ( 6 , ∞ )

( f + g ) ( x ) = 4 x 3 + 8 x 2 + 1 2 x , ( f + g ) ( x ) = 4 x 3 + 8 x 2 + 1 2 x , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f − g ) ( x ) = 4 x 3 + 8 x 2 − 1 2 x , ( f − g ) ( x ) = 4 x 3 + 8 x 2 − 1 2 x , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f g ) ( x ) = x + 2 , ( f g ) ( x ) = x + 2 , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f g ) ( x ) = 4 x 3 + 8 x 2 , ( f g ) ( x ) = 4 x 3 + 8 x 2 , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f + g ) ( x ) = 3 x 2 + x − 5 , ( f + g ) ( x ) = 3 x 2 + x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f − g ) ( x ) = 3 x 2 − x − 5 , ( f − g ) ( x ) = 3 x 2 − x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f g ) ( x ) = 3 x 2 x − 5 , ( f g ) ( x ) = 3 x 2 x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f g ) ( x ) = 3 x 2 x − 5 , ( f g ) ( x ) = 3 x 2 x − 5 , domain: ( 5 , ∞ ) ( 5 , ∞ )

• ⓑ f ( g ( x ) ) = 2 ( 3 x − 5 ) 2 + 1 ; f ( g ( x ) ) = 2 ( 3 x − 5 ) 2 + 1 ;
• ⓒ g ( f ) ( x ) ) = 6 x 2 − 2 ; g ( f ) ( x ) ) = 6 x 2 − 2 ;
• ⓓ ( g ∘ g ) ( x ) = 3 ( 3 x − 5 ) − 5 = 9 x − 20 ; ( g ∘ g ) ( x ) = 3 ( 3 x − 5 ) − 5 = 9 x − 20 ;
• ⓔ ( f ∘ f ) ( − 2 ) = 163 ( f ∘ f ) ( − 2 ) = 163

f ( g ( x ) ) = x 2 + 3 + 2 , g ( f ( x ) ) = x + 4 x + 7 f ( g ( x ) ) = x 2 + 3 + 2 , g ( f ( x ) ) = x + 4 x + 7

f ( g ( x ) ) = x + 1 x 3 3 = x + 1 3 x , g ( f ( x ) ) = x 3 + 1 x f ( g ( x ) ) = x + 1 x 3 3 = x + 1 3 x , g ( f ( x ) ) = x 3 + 1 x

( f ∘ g ) ( x ) = 1 2 x + 4 − 4 = x 2 , ( g ∘ f ) ( x ) = 2 x − 4 ( f ∘ g ) ( x ) = 1 2 x + 4 − 4 = x 2 , ( g ∘ f ) ( x ) = 2 x − 4

f ( g ( h ( x ) ) ) = ( 1 x + 3 ) 2 + 1 f ( g ( h ( x ) ) ) = ( 1 x + 3 ) 2 + 1

• ⓐ Text ( g ∘ f ) ( x ) = − 3 2 − 4 x ; ( g ∘ f ) ( x ) = − 3 2 − 4 x ;
• ⓑ ( − ∞ , 1 2 ) ( − ∞ , 1 2 )
• ⓐ ( 0 , 2 ) ∪ ( 2 , ∞ ) ; ( 0 , 2 ) ∪ ( 2 , ∞ ) ;
• ⓑ ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) ; ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) ; c. ( 0 , ∞ ) ( 0 , ∞ )

( 1 , ∞ ) ( 1 , ∞ )

sample: f ( x ) = x 3 g ( x ) = x − 5 f ( x ) = x 3 g ( x ) = x − 5

sample: f ( x ) = 4 x g ( x ) = ( x + 2 ) 2 f ( x ) = 4 x g ( x ) = ( x + 2 ) 2

sample: f ( x ) = x 3 g ( x ) = 1 2 x − 3 f ( x ) = x 3 g ( x ) = 1 2 x − 3

sample: f ( x ) = x 4 g ( x ) = 3 x − 2 x + 5 f ( x ) = x 4 g ( x ) = 3 x − 2 x + 5

sample: f ( x ) = x f ( x ) = x g ( x ) = 2 x + 6 g ( x ) = 2 x + 6

sample: f ( x ) = x 3 f ( x ) = x 3 g ( x ) = ( x − 1 ) g ( x ) = ( x − 1 )

sample: f ( x ) = x 3 f ( x ) = x 3 g ( x ) = 1 x − 2 g ( x ) = 1 x − 2

sample: f ( x ) = x f ( x ) = x g ( x ) = 2 x − 1 3 x + 4 g ( x ) = 2 x − 1 3 x + 4

f ( g ( 0 ) ) = 27 , g ( f ( 0 ) ) = − 94 f ( g ( 0 ) ) = 27 , g ( f ( 0 ) ) = − 94

f ( g ( 0 ) ) = 1 5 , g ( f ( 0 ) ) = 5 f ( g ( 0 ) ) = 1 5 , g ( f ( 0 ) ) = 5

18 x 2 + 60 x + 51 18 x 2 + 60 x + 51

g ∘ g ( x ) = 9 x + 20 g ∘ g ( x ) = 9 x + 20

( f ∘ g ) ( 6 ) = 6 ( f ∘ g ) ( 6 ) = 6 ; ( g ∘ f ) ( 6 ) = 6 ( g ∘ f ) ( 6 ) = 6

( f ∘ g ) ( 11 ) = 11 , ( g ∘ f ) ( 11 ) = 11 ( f ∘ g ) ( 11 ) = 11 , ( g ∘ f ) ( 11 ) = 11

A ( t ) = π ( 25 t + 2 ) 2 A ( t ) = π ( 25 t + 2 ) 2 and A ( 2 ) = π ( 25 4 ) 2 = 2500 π A ( 2 ) = π ( 25 4 ) 2 = 2500 π square inches

A ( 5 ) = π ( 2 ( 5 ) + 1 ) 2 = 121 π A ( 5 ) = π ( 2 ( 5 ) + 1 ) 2 = 121 π square units

• ⓐ N ( T ( t ) ) = 23 ( 5 t + 1.5 ) 2 − 56 ( 5 t + 1.5 ) + 1 ; N ( T ( t ) ) = 23 ( 5 t + 1.5 ) 2 − 56 ( 5 t + 1.5 ) + 1 ;
• ⓑ 3.38 hours

1.5 Section Exercises

A horizontal shift results when a constant is added to or subtracted from the input. A vertical shifts results when a constant is added to or subtracted from the output.

A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output.

For a function f , f , substitute ( − x ) ( − x ) for ( x ) ( x ) in f ( x ) . f ( x ) . Simplify. If the resulting function is the same as the original function, f ( − x ) = f ( x ) , f ( − x ) = f ( x ) , then the function is even. If the resulting function is the opposite of the original function, f ( − x ) = − f ( x ) , f ( − x ) = − f ( x ) , then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even.

g ( x ) = | x - 1 | − 3 g ( x ) = | x - 1 | − 3

g ( x ) = 1 ( x + 4 ) 2 + 2 g ( x ) = 1 ( x + 4 ) 2 + 2

The graph of f ( x + 43 ) f ( x + 43 ) is a horizontal shift to the left 43 units of the graph of f . f .

The graph of f ( x - 4 ) f ( x - 4 ) is a horizontal shift to the right 4 units of the graph of f . f .

The graph of f ( x ) + 8 f ( x ) + 8 is a vertical shift up 8 units of the graph of f . f .

The graph of f ( x ) − 7 f ( x ) − 7 is a vertical shift down 7 units of the graph of f . f .

The graph of f ( x + 4 ) − 1 f ( x + 4 ) − 1 is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of f . f .

decreasing on ( − ∞ , − 3 ) ( − ∞ , − 3 ) and increasing on ( − 3 , ∞ ) ( − 3 , ∞ )

decreasing on [ 0 , ∞ ) [ 0 , ∞ )

g ( x ) = f ( x - 1 ) , h ( x ) = f ( x ) + 1 g ( x ) = f ( x - 1 ) , h ( x ) = f ( x ) + 1

f ( x ) = | x - 3 | − 2 f ( x ) = | x - 3 | − 2

f ( x ) = x + 3 − 1 f ( x ) = x + 3 − 1

f ( x ) = ( x - 2 ) 2 f ( x ) = ( x - 2 ) 2

f ( x ) = | x + 3 | − 2 f ( x ) = | x + 3 | − 2

f ( x ) = − x f ( x ) = − x

f ( x ) = − ( x + 1 ) 2 + 2 f ( x ) = − ( x + 1 ) 2 + 2

f ( x ) = − x + 1 f ( x ) = − x + 1

The graph of g g is a vertical reflection (across the x x -axis) of the graph of f . f .

The graph of g g is a vertical stretch by a factor of 4 of the graph of f . f .

The graph of g g is a horizontal compression by a factor of 1 5 1 5 of the graph of f . f .

The graph of g g is a horizontal stretch by a factor of 3 of the graph of f . f .

The graph of g g is a horizontal reflection across the y y -axis and a vertical stretch by a factor of 3 of the graph of f . f .

g ( x ) = | − 4 x | g ( x ) = | − 4 x |

g ( x ) = 1 3 ( x + 2 ) 2 − 3 g ( x ) = 1 3 ( x + 2 ) 2 − 3

g ( x ) = 1 2 ( x - 5 ) 2 + 1 g ( x ) = 1 2 ( x - 5 ) 2 + 1

The graph of the function f ( x ) = x 2 f ( x ) = x 2 is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units.

The graph of f ( x ) = | x | f ( x ) = | x | is stretched vertically by a factor of 2, shifted horizontally 4 units to the right, reflected across the horizontal axis, and then shifted vertically 3 units up.

The graph of the function f ( x ) = x 3 f ( x ) = x 3 is compressed vertically by a factor of 1 2 . 1 2 .

The graph of the function is stretched horizontally by a factor of 3 and then shifted vertically downward by 3 units.

The graph of f ( x ) = x f ( x ) = x is shifted right 4 units and then reflected across the vertical line x = 4. x = 4.

1.6 Section Exercises

Isolate the absolute value term so that the equation is of the form | A | = B . | A | = B . Form one equation by setting the expression inside the absolute value symbol, A , A , equal to the expression on the other side of the equation, B . B . Form a second equation by setting A A equal to the opposite of the expression on the other side of the equation, − B . − B . Solve each equation for the variable.

The graph of the absolute value function does not cross the x x -axis, so the graph is either completely above or completely below the x x -axis.

First determine the boundary points by finding the solution(s) of the equation. Use the boundary points to form possible solution intervals. Choose a test value in each interval to determine which values satisfy the inequality.

| x + 4 | = 1 2 | x + 4 | = 1 2

| f ( x ) − 8 | < 0.03 | f ( x ) − 8 | < 0.03

{ 1 , 11 } { 1 , 11 }

{ - 9 4 , 13 4 } { - 9 4 , 13 4 }

{ 10 3 , 20 3 } { 10 3 , 20 3 }

{ 11 5 , 29 5 } { 11 5 , 29 5 }

{ 5 2 , 7 2 } { 5 2 , 7 2 }

No solution

{ − 57 , 27 } { − 57 , 27 }

( 0 , − 8 ) ; ( − 6 , 0 ) , ( 4 , 0 ) ( 0 , − 8 ) ; ( − 6 , 0 ) , ( 4 , 0 )

( 0 , − 7 ) ; ( 0 , − 7 ) ; no x x -intercepts

( − ∞ , − 8 ) ∪ ( 12 , ∞ ) ( − ∞ , − 8 ) ∪ ( 12 , ∞ )

− 4 3 ≤ x ≤ 4 − 4 3 ≤ x ≤ 4

( − ∞ , − 8 3 ] ∪ [ 6 , ∞ ) ( − ∞ , − 8 3 ] ∪ [ 6 , ∞ )

( − ∞ , − 8 3 ] ∪ [ 16 , ∞ ) ( − ∞ , − 8 3 ] ∪ [ 16 , ∞ )

range: [ 0 , 20 ] [ 0 , 20 ]

x - x - intercepts:

There is no solution for a a that will keep the function from having a y y -intercept. The absolute value function always crosses the y y -intercept when x = 0. x = 0.

| p − 0.08 | ≤ 0.015 | p − 0.08 | ≤ 0.015

| x − 5.0 | ≤ 0.01 | x − 5.0 | ≤ 0.01

1.7 Section Exercises

Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that y y -values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no y y -values repeat and the function is one-to-one.

Yes. For example, f ( x ) = 1 x f ( x ) = 1 x is its own inverse.

Given a function y = f ( x ) , y = f ( x ) , solve for x x in terms of y . y . Interchange the x x and y . y . Solve the new equation for y . y . The expression for y y is the inverse, y = f − 1 ( x ) . y = f − 1 ( x ) .

f − 1 ( x ) = x − 3 f − 1 ( x ) = x − 3

f − 1 ( x ) = 2 − x f − 1 ( x ) = 2 − x

f − 1 ( x ) = − 2 x x − 1 f − 1 ( x ) = − 2 x x − 1

domain of f ( x ) : [ − 7 , ∞ ) ; f − 1 ( x ) = x − 7 f ( x ) : [ − 7 , ∞ ) ; f − 1 ( x ) = x − 7

domain of f ( x ) : [ 0 , ∞ ) ; f − 1 ( x ) = x + 5 f ( x ) : [ 0 , ∞ ) ; f − 1 ( x ) = x + 5

• ⓐ f ( g ( x ) ) = x f ( g ( x ) ) = x and g ( f ( x ) ) = x . g ( f ( x ) ) = x .
• ⓑ This tells us that f f and g g are inverse functions

f ( g ( x ) ) = x , g ( f ( x ) ) = x f ( g ( x ) ) = x , g ( f ( x ) ) = x

not one-to-one

[ 2 , 10 ] [ 2 , 10 ]

f − 1 ( x ) = ( 1 + x ) 1 / 3 f − 1 ( x ) = ( 1 + x ) 1 / 3

f − 1 ( x ) = 5 9 ( x − 32 ) . f − 1 ( x ) = 5 9 ( x − 32 ) . Given the Fahrenheit temperature, x , x , this formula allows you to calculate the Celsius temperature.

t ( d ) = d 50 , t ( d ) = d 50 , t ( 180 ) = 180 50 . t ( 180 ) = 180 50 . The time for the car to travel 180 miles is 3.6 hours.

Review Exercises

f ( − 3 ) = − 27 ; f ( − 3 ) = − 27 ; f ( 2 ) = − 2 ; f ( 2 ) = − 2 ; f ( − a ) = − 2 a 2 − 3 a ; f ( − a ) = − 2 a 2 − 3 a ; − f ( a ) = 2 a 2 − 3 a ; − f ( a ) = 2 a 2 − 3 a ; f ( a + h ) = − 2 a 2 + 3 a − 4 a h + 3 h − 2 h 2 f ( a + h ) = − 2 a 2 + 3 a − 4 a h + 3 h − 2 h 2

x = − 1.8 x = − 1.8 or  or  x = 1.8  or  x = 1.8

− 64 + 80 a − 16 a 2 − 1 + a = − 16 a + 64 − 64 + 80 a − 16 a 2 − 1 + a = − 16 a + 64

( − ∞ , − 2 ) ∪ ( − 2 , 6 ) ∪ ( 6 , ∞ ) ( − ∞ , − 2 ) ∪ ( − 2 , 6 ) ∪ ( 6 , ∞ )

increasing ( 2 , ∞ ) ; ( 2 , ∞ ) ; decreasing ( − ∞ , 2 ) ( − ∞ , 2 )

increasing ( − 3 , 1 ) ; ( − 3 , 1 ) ; constant ( − ∞ , − 3 ) ∪ ( 1 , ∞ ) ( − ∞ , − 3 ) ∪ ( 1 , ∞ )

local minimum ( − 2 , − 3 ) ; ( − 2 , − 3 ) ; local maximum ( 1 , 3 ) ( 1 , 3 )

Absolute Maximum: 10

( f ∘ g ) ( x ) = 17 − 18 x ; ( g ∘ f ) ( x ) = − 7 − 18 x ( f ∘ g ) ( x ) = 17 − 18 x ; ( g ∘ f ) ( x ) = − 7 − 18 x

( f ∘ g ) ( x ) = 1 x + 2 ; ( f ∘ g ) ( x ) = 1 x + 2 ; ( g ∘ f ) ( x ) = 1 x + 2 ( g ∘ f ) ( x ) = 1 x + 2

( f ∘ g ) ( x ) = 1 + x 1 + 4 x , x ≠ 0 , x ≠ − 1 4 ( f ∘ g ) ( x ) = 1 + x 1 + 4 x , x ≠ 0 , x ≠ − 1 4

( f ∘ g ) ( x ) = 1 x , x > 0 ( f ∘ g ) ( x ) = 1 x , x > 0

sample: g ( x ) = 2 x − 1 3 x + 4 ; f ( x ) = x g ( x ) = 2 x − 1 3 x + 4 ; f ( x ) = x

f ( x ) = | x − 3 | f ( x ) = | x − 3 |

f ( x ) = 1 2 | x + 2 | + 1 f ( x ) = 1 2 | x + 2 | + 1

f ( x ) = − 3 | x − 3 | + 3 f ( x ) = − 3 | x − 3 | + 3

x = − 22 , x = 14 x = − 22 , x = 14

( − 5 3 , 3 ) ( − 5 3 , 3 )

f − 1 ( x ) = x - 1 f − 1 ( x ) = x - 1

The function is one-to-one.

The function is not one-to-one.

Practice Test

The relation is a function.

The graph is a parabola and the graph fails the horizontal line test.

2 a 2 − a 2 a 2 − a

− 2 ( a + b ) + 1 − 2 ( a + b ) + 1

x = − 7 x = − 7 and x = 10 x = 10

f − 1 ( x ) = x + 5 3 f − 1 ( x ) = x + 5 3

( − ∞ , − 1.1 )  and  ( 1.1 , ∞ ) ( − ∞ , − 1.1 )  and  ( 1.1 , ∞ )

( 1.1 , − 0.9 ) ( 1.1 , − 0.9 )

f ( 2 ) = 2 f ( 2 ) = 2

f ( x ) = { | x | if x ≤ 2 3 if x > 2 f ( x ) = { | x | if x ≤ 2 3 if x > 2

x = 2 x = 2

f − 1 ( x ) = − x − 11 2 f − 1 ( x ) = − x − 11 2

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• Book title: Precalculus
• Publication date: Oct 23, 2014
• Location: Houston, Texas
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5.4: Right Triangle Trigonometry

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Learning Objectives

• Use right triangles to evaluate trigonometric functions.
• Find function values for 30°(\(\dfrac{\pi}{6}\)),45°(\(\dfrac{\pi}{4}\)),and 60°(\(\dfrac{\pi}{3}\)).
• Use equal cofunctions of complementary angles.
• Use the definitions of trigonometric functions of any angle.
• Use right-triangle trigonometry to solve applied problems.

Mt. Everest, which straddles the border between China and Nepal, is the tallest mountain in the world. Measuring its height is no easy task and, in fact, the actual measurement has been a source of controversy for hundreds of years. The measurement process involves the use of triangles and a branch of mathematics known as trigonometry. In this section, we will define a new group of functions known as trigonometric functions, and find out how they can be used to measure heights, such as those of the tallest mountains.

We have previously defined the sine and cosine of an angle in terms of the coordinates of a point on the unit circle intersected by the terminal side of the angle:

\[ \begin{align*} \cos t &= x \\ \sin t &=y \end{align*} \]

In this section, we will see another way to define trigonometric functions using properties of right triangles .

Using Right Triangles to Evaluate Trigonometric Functions

In earlier sections, we used a unit circle to define the trigonometric functions . In this section, we will extend those definitions so that we can apply them to right triangles. The value of the sine or cosine function of \(t\) is its value at \(t\) radians. First, we need to create our right triangle. Figure \(\PageIndex{1}\) shows a point on a unit circle of radius 1. If we drop a vertical line segment from the point \((x,y)\) to the x -axis, we have a right triangle whose vertical side has length \(y\) and whose horizontal side has length \(x\). We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle.

\[ \cos t= \frac{x}{1}=x \]

Likewise, we know

\[ \sin t= \frac{y}{1}=y \]

These ratios still apply to the sides of a right triangle when no unit circle is involved and when the triangle is not in standard position and is not being graphed using \((x,y)\) coordinates. To be able to use these ratios freely, we will give the sides more general names: Instead of \(x\),we will call the side between the given angle and the right angle the adjacent side to angle \(t\). (Adjacent means “next to.”) Instead of \(y\),we will call the side most distant from the given angle the opposite side from angle \(t\). And instead of \(1\),we will call the side of a right triangle opposite the right angle the hypotenuse . These sides are labeled in Figure \(\PageIndex{2}\).

Understanding Right Triangle Relationships

Given a right triangle with an acute angle of \(t\),

\[\begin{align} \sin (t) &= \dfrac{\text{opposite}}{\text{hypotenuse}} \label{sindef}\\ \cos (t) &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \label{cosdef}\\ \tan (t) &= \dfrac{\text{opposite}}{\text{adjacent}} \label{tandef}\end{align}\]

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “ S ine is o pposite over h ypotenuse, C osine is a djacent over h ypotenuse, T angent is o pposite over a djacent.”

how to: Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle

• Find the sine as the ratio of the opposite side to the hypotenuse.
• Find the cosine as the ratio of the adjacent side to the hypotenuse.
• Find the tangent is the ratio of the opposite side to the adjacent side.

Example \(\PageIndex{1}\): Evaluating a Trigonometric Function of a Right Triangle

Given the triangle shown in Figure \(\PageIndex{3}\), find the value of \(\cos α\).

The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so via Equation \ref{cosdef}:

\[\begin{align*} \cos (α) &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \\[4pt] &= \dfrac{15}{17} \end{align*}\]

Exercise \(\PageIndex{1}\)

Given the triangle shown in Figure \(\PageIndex{4}\), find the value of \(\sin t\).

\(\frac{7}{25}\)

Relating Angles and Their Functions

When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure \(\PageIndex{5}\). The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

We will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.

how to: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles

• If needed, draw the right triangle and label the angle provided.
• Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
• sine as the ratio of the opposite side to the hypotenuse
• cosine as the ratio of the adjacent side to the hypotenuse
• tangent as the ratio of the opposite side to the adjacent side
• secant as the ratio of the hypotenuse to the adjacent side
• cosecant as the ratio of the hypotenuse to the opposite side
• cotangent as the ratio of the adjacent side to the opposite side

Example \(\PageIndex{2}\): Evaluating Trigonometric Functions of Angles Not in Standard Position

Using the triangle shown in Figure \(\PageIndex{6}\), evaluate \( \sin α, \cos α, \tan α, \sec α, \csc α,\) and \( \cot α\).

\[ \begin{align*} \sin α &= \dfrac{\text{opposite } α}{\text{hypotenuse}} = \dfrac{4}{5} \\ \cos α &= \dfrac{\text{adjacent to }α}{\text{hypotenuse}}=\dfrac{3}{5} \\ \tan α &= \dfrac{\text{opposite }α}{\text{adjacent to }α}=\dfrac{4}{3} \\ \sec α &= \dfrac{\text{hypotenuse}}{\text{adjacent to }α}= \dfrac{5}{3} \\ \csc α &= \dfrac{\text{hypotenuse}}{\text{opposite }α}=\dfrac{5}{4} \\ \cot α &= \dfrac{\text{adjacent to }α}{\text{opposite }α}=\dfrac{3}{4} \end{align*}\]

Exercise \(\PageIndex{2}\)

Using the triangle shown in Figure \(\PageIndex{7}\), evaluate \( \sin t, \cos t,\tan t, \sec t, \csc t,\) and \(\cot t\).

\[\begin{align*} \sin t &= \frac{33}{65}, \cos t= \frac{56}{65},\tan t= \frac{33}{56}, \\ \\ \sec t &= \frac{65}{56},\csc t= \frac{65}{33},\cot t= \frac{56}{33} \end{align*}\]

Finding Trigonometric Functions of Special Angles Using Side Lengths

We have already discussed the trigonometric functions as they relate to the special angles on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of \(30°, 60°,\) and \(45°\), however, remember that when dealing with right triangles, we are limited to angles between \(0° \text{ and } 90°\).

Suppose we have a \(30°,60°,90°\) triangle, which can also be described as a \(\frac{π}{6}, \frac{π}{3},\frac{π}{2}\) triangle. The sides have lengths in the relation \(s,\sqrt{3}s,2s.\) The sides of a \(45°,45°,90° \)triangle, which can also be described as a \(\frac{π}{4},\frac{π}{4},\frac{π}{2}\) triangle, have lengths in the relation \(s,s,\sqrt{2}s.\) These relations are shown in Figure \(\PageIndex{8}\).

We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.

Given trigonometric functions of a special angle, evaluate using side lengths.

• Use the side lengths shown in Figure \(\PageIndex{8}\) for the special angle you wish to evaluate.
• Use the ratio of side lengths appropriate to the function you wish to evaluate.

Example \(\PageIndex{3}\): Evaluating Trigonometric Functions of Special Angles Using Side Lengths

Find the exact value of the trigonometric functions of \(\frac{π}{3}\), using side lengths.

\[\begin{align*} \sin (\dfrac{π}{3}) &= \dfrac{\text{opp}}{\text{hyp}}=\dfrac{\sqrt{3}s}{2s}=\dfrac{\sqrt{3}}{2} \\ \cos (\dfrac{π}{3}) &= \dfrac{\text{adj}}{\text{hyp}}=\dfrac{s}{2s}=\dfrac{1}{2} \\ \tan (\dfrac{π}{3}) &= \dfrac{\text{opp}}{\text{adj}} =\dfrac{\sqrt{3}s}{s}=\sqrt{3} \\ \sec (\dfrac{π}{3}) &= \dfrac{\text{hyp}}{\text{adj}} = \dfrac{2s}{s}=2 \\ \csc (\dfrac{π}{3}) &= \dfrac{\text{hyp}}{\text{opp}} =\dfrac{2s}{\sqrt{3}s}=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3} \\ \cot (\dfrac{π}{3}) &= \dfrac{\text{adj}}{\text{opp}}=\dfrac{s}{\sqrt{3}s}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3} \end{align*}\]

Exercise \(\PageIndex{3}\)

Find the exact value of the trigonometric functions of \(\frac{π}{4}\) using side lengths.

\( \sin (\frac{π}{4})=\frac{\sqrt{2}}{2}, \cos (\frac{π}{4})=\frac{\sqrt{2}}{2}, \tan (\frac{π}{4})=1,\)

\( \sec (\frac{π}{4})=\sqrt{2}, \csc (\frac{π}{4})=\sqrt{2}, \cot (\frac{π}{4}) =1 \)

Using Equal Cofunction of Complements

If we look more closely at the relationship between the sine and cosine of the special angles relative to the unit circle, we will notice a pattern. In a right triangle with angles of \(\frac{π}{6}\) and \(\frac{π}{3}\), we see that the sine of \(\frac{π}{3}\), namely \(\frac{\sqrt{3}}{2}\), is also the cosine of \(\frac{π}{6}\), while the sine of \(\frac{π}{6}\), namely \(\frac{1}{2},\) is also the cosine of \(\frac{π}{3}\) (Figure \(\PageIndex{9}\)).

\[\begin{align*} \sin \frac{π}{3} &= \cos \frac{π}{6}=\frac{\sqrt{3}s}{2s}=\frac{\sqrt{3}}{2} \\ \sin \frac{π}{6} &= \cos \frac{π}{3}=\frac{s}{2s}=\frac{1}{2} \end{align*}\]

This result should not be surprising because, as we see from Figure \(\PageIndex{9}\), the side opposite the angle of \(\frac{π}{3}\) is also the side adjacent to \(\frac{π}{6}\), so \(\sin (\frac{π}{3})\) and \(\cos (\frac{π}{6})\) are exactly the same ratio of the same two sides, \(\sqrt{3} s\) and \(2s.\) Similarly, \( \cos (\frac{π}{3})\) and \( \sin (\frac{π}{6})\) are also the same ratio using the same two sides, \(s\) and \(2s\).

The interrelationship between the sines and cosines of \(\frac{π}{6}\) and \(\frac{π}{3}\) also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to π, and the right angle is \(\frac{π}{2}\), the remaining two angles must also add up to \(\frac{π}{2}\). That means that a right triangle can be formed with any two angles that add to \(\frac{π}{2}\)—in other words, any two complementary angles. So we may state a cofunction identity : If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure \(\PageIndex{10}\).

Using this identity, we can state without calculating, for instance, that the sine of \(\frac{π}{12}\) equals the cosine of \(\frac{5π}{12}\), and that the sine of \(\frac{5π}{12}\) equals the cosine of \(\frac{π}{12}\). We can also state that if, for a certain angle \(t, \cos t= \frac{5}{13},\) then \( \sin (\frac{π}{2}−t)=\frac{5}{13}\) as well.

COFUNCTION IDENTITIES

The cofunction identities in radians are listed in Table \(\PageIndex{1}\).

how to: Given the sine and cosine of an angle, find the sine or cosine of its complement.

• To find the sine of the complementary angle, find the cosine of the original angle.
• To find the cosine of the complementary angle, find the sine of the original angle.

Example \(\PageIndex{4}\): Using Cofunction Identities

If \( \sin t = \frac{5}{12},\) find \(( \cos \frac{π}{2}−t)\).

According to the cofunction identities for sine and cosine,

\[ \sin t= \cos (\dfrac{π}{2}−t). \nonumber\]

\[ \cos (\dfrac{π}{2}−t)= \dfrac{5}{12}. \nonumber\]

Exercise \(\PageIndex{4}\)

If \(\csc (\frac{π}{6})=2,\) find \( \sec (\frac{π}{3}).\)

Using Trigonometric Functions

In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.

how to: Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides

• For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.
• Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
• Using the value of the trigonometric function and the known side length, solve for the missing side length.

Example \(\PageIndex{5}\): Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in Figure \(\PageIndex{11}\).

We know the angle and the opposite side, so we can use the tangent to find the adjacent side.

\[ \tan (30°)= \dfrac{7}{a} \nonumber\]

We rearrange to solve for \(a\).

\[\begin{align} a &=\dfrac{7}{ \tan (30°)} \\ & =12.1 \end{align} \nonumber\]

We can use the sine to find the hypotenuse.

\[ \sin (30°)= \dfrac{7}{c} \nonumber\]

Again, we rearrange to solve for \(c\).

\[\begin{align*} c &= \dfrac{7}{\sin (30°)} =14 \end{align*}\]

Exercise \(\PageIndex{5}\):

A right triangle has one angle of \(\frac{π}{3}\) and a hypotenuse of 20. Find the unknown sides and angle of the triangle.

\(\mathrm{adjacent=10; opposite=10 \sqrt{3}; }\) missing angle is \(\frac{π}{6}\)

Using Right Triangle Trigonometry to Solve Applied Problems

Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. See Figure \(\PageIndex{12}\).

how to: Given a tall object, measure its height indirectly

• Make a sketch of the problem situation to keep track of known and unknown information.
• Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.
• At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.
• Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.
• Solve the equation for the unknown height.

Example \(\PageIndex{6}\): Measuring a Distance Indirectly

To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of 57° 57° between a line of sight to the top of the tree and the ground, as shown in Figure \(\PageIndex{13}\). Find the height of the tree.

We know that the angle of elevation is \(57°\) and the adjacent side is 30 ft long. The opposite side is the unknown height.

The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of \(57°\), letting \(h\) be the unknown height.

\[\begin{array}{cl} \tan θ = \dfrac{\text{opposite}}{\text{adjacent}} & \text{} \\ \tan (57°) = \dfrac{h}{30} & \text{Solve for }h. \\ h=30 \tan (57°) & \text{Multiply.} \\ h≈46.2 & \text{Use a calculator.} \end{array} \]

The tree is approximately 46 feet tall.

Exercise \(\PageIndex{6}\):

How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of \(\frac{5π}{12}\) with the ground? Round to the nearest foot.

About 52 ft

Access these online resources for additional instruction and practice with right triangle trigonometry.

• Finding Trig Functions on Calculator
• Finding Trig Functions Using a Right Triangle
• Relate Trig Functions to Sides of a Right Triangle
• Determine Six Trig Functions from a Triangle
• Determine Length of Right Triangle Side

Visit this website for additional practice questions from Learningpod.

Key Equations

Cofunction Identities

\[\begin{align*} \cos t &= \sin ( \frac{π}{2}−t) \\ \sin t &= \cos (\frac{π}{2}−t) \\ \tan t &= \cot (\frac{π}{2}−t) \\ \cot t &= \tan (\frac{π}{2}−t) \\ \sec t &= \csc (\frac{π}{2}−t) \\ \csc t &= \sec (\frac{π}{2}−t) \end{align*}\]

Key Concepts

• We can define trigonometric functions as ratios of the side lengths of a right triangle. See Example .
• The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle. See Example .
• We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur. See Example .
• Any two complementary angles could be the two acute angles of a right triangle.
• If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa. See Example .
• We can use trigonometric functions of an angle to find unknown side lengths.
• Select the trigonometric function representing the ratio of the unknown side to the known side. See Example .
• Right-triangle trigonometry permits the measurement of inaccessible heights and distances.
• The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known. See Example .