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How can the graph of y = cos x be used to construct the graph of y = sec x ?

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Explain why the period of tan x is equal to π .

Answers will vary. Using the unit circle, one can show that tan ( x + π ) = tan x .

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Why are there no intercepts on the graph of y = csc x ?

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How does the period of y = csc x compare with the period of y = sin x ?

The period is the same: 2 π .

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Algebraic

For the following exercises, match each trigonometric function with one of the following graphs.

Trigonometric graph of tangent of x. Trigonometric graph of secant of x. Trigonometric graph of cosecant of x. Trigonometric graph of cotangent of x.

For the following exercises, find the period and horizontal shift of each of the functions.

f ( x ) = 2 tan ( 4 x 32 )

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h ( x ) = 2 sec ( π 4 ( x + 1 ) )

period: 8; horizontal shift: 1 unit to left

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m ( x ) = 6 csc ( π 3 x + π )

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If tan x = 1.5 , find tan ( x ) .

1.5

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If sec x = 2 , find sec ( x ) .

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If csc x = 5 , find csc ( x ) .

5

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If x sin x = 2 , find ( x ) sin ( x ) .

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For the following exercises, rewrite each expression such that the argument x is positive.

cot ( x ) cos ( x ) + sin ( x )

cot x cos x sin x

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cos ( x ) + tan ( x ) sin ( x )

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Graphical

For the following exercises, sketch two periods of the graph for each of the following functions. Identify the stretching factor, period, and asymptotes.

f ( x ) = 2 tan ( 4 x 32 )

A graph of two periods of a modified tangent function. There are two vertical asymptotes.

stretching factor: 2; period:   π 4 ;   asymptotes:   x = 1 4 ( π 2 + π k ) + 8 ,  where  k  is an integer

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h ( x ) = 2 sec ( π 4 ( x + 1 ) )

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m ( x ) = 6 csc ( π 3 x + π )

A graph of two periods of a modified cosecant function. Vertical Asymptotes at x= -6, -3, 0, 3, and 6.

stretching factor: 6; period: 6; asymptotes:   x = 3 k ,  where  k  is an integer

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j ( x ) = tan ( π 2 x )

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p ( x ) = tan ( x π 2 )

A graph of two periods of a modified tangent function. Vertical asymptotes at multiples of pi.

stretching factor: 1; period:   π ;   asymptotes:   x = π k ,  where  k  is an integer

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f ( x ) = tan ( x + π 4 )

A graph of two periods of a modified tangent function. Three vertical asymptiotes shown.

Stretching factor: 1; period:   π ;   asymptotes:   x = π 4 + π k ,  where  k  is an integer

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f ( x ) = π tan ( π x π ) π

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f ( x ) = 2 csc ( x )

A graph of two periods of a modified cosecant function. Vertical asymptotes at multiples of pi.

stretching factor: 2; period:   2 π ;   asymptotes:   x = π k ,  where  k  is an integer

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f ( x ) = 1 4 csc ( x )

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f ( x ) = 4 sec ( 3 x )

A graph of two periods of a modified secant function. Vertical asymptotes at x=-pi/2, -pi/6, pi/6, and pi/2.

stretching factor: 4; period:   2 π 3 ;   asymptotes:   x = π 6 k ,  where  k  is an odd integer

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f ( x ) = 3 cot ( 2 x )

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f ( x ) = 7 sec ( 5 x )

A graph of two periods of a modified secant function. There are four vertical asymptotes all pi/5 apart.

stretching factor: 7; period:   2 π 5 ;   asymptotes:   x = π 10 k ,  where  k  is an odd integer

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f ( x ) = 9 10 csc ( π x )

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f ( x ) = 2 csc ( x + π 4 ) 1

A graph of two periods of a modified cosecant function. Three vertical asymptotes, each pi apart.

stretching factor: 2; period:   2 π ;   asymptotes:   x = π 4 + π k ,  where  k  is an integer

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f ( x ) = sec ( x π 3 ) 2

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f ( x ) = 7 5 csc ( x π 4 )

A graph of a modified cosecant function. Four vertical asymptotes.

stretching factor:   7 5 ;   period:   2 π ;   asymptotes:   x = π 4 + π k ,  where  k  is an integer

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f ( x ) = 5 ( cot ( x + π 2 ) 3 )

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For the following exercises, find and graph two periods of the periodic function with the given stretching factor, | A | , period, and phase shift.

A tangent curve, A = 1 , period of π 3 ; and phase shift ( h , k ) = ( π 4 , 2 )

y = tan ( 3 ( x π 4 ) ) + 2

A graph of two periods of a modified tangent function. Vertical asymptotes at x=-pi/4 and pi/12.
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A tangent curve, A = −2 , period of π 4 , and phase shift ( h , k ) = ( π 4 , −2 )

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For the following exercises, find an equation for the graph of each function.

graph of two periods of a modified tangent function. Vertical asymptotes at x=-0.005 and x=0.005.

f ( x ) = 1 2 tan ( 100 π x )

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Technology

For the following exercises, use a graphing calculator to graph two periods of the given function. Note: most graphing calculators do not have a cosecant button; therefore, you will need to input csc x as 1 sin x .

f ( x ) = csc ( x ) sec ( x )

A graph of tangent of x.
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Graph f ( x ) = 1 + sec 2 ( x ) tan 2 ( x ) . What is the function shown in the graph?

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f ( x ) = sec ( 0.001 x )

A graph of two periods of a modified secant function. Vertical asymptotes at multiples of 500pi.
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f ( x ) = cot ( 100 π x )

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f ( x ) = sin 2 x + cos 2 x

A graph of y=1.
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Real-world applications

The function f ( x ) = 20 tan ( π 10 x ) marks the distance in the movement of a light beam from a police car across a wall for time x , in seconds, and distance f ( x ) , in feet.

  1. Graph on the interval [ 0 , 5 ] .
  2. Find and interpret the stretching factor, period, and asymptote.
  3. Evaluate f ( 1 ) and f ( 2.5 ) and discuss the function’s values at those inputs.
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Standing on the shore of a lake, a fisherman sights a boat far in the distance to his left. Let x , measured in radians, be the angle formed by the line of sight to the ship and a line due north from his position. Assume due north is 0 and x is measured negative to the left and positive to the right. (See [link] .) The boat travels from due west to due east and, ignoring the curvature of the Earth, the distance d ( x ) , in kilometers, from the fisherman to the boat is given by the function d ( x ) = 1.5 sec ( x ) .

  1. What is a reasonable domain for d ( x ) ?
  2. Graph d ( x ) on this domain.
  3. Find and discuss the meaning of any vertical asymptotes on the graph of d ( x ) .
  4. Calculate and interpret d ( π 3 ) . Round to the second decimal place.
  5. Calculate and interpret d ( π 6 ) . Round to the second decimal place.
  6. What is the minimum distance between the fisherman and the boat? When does this occur?
An illustration of a man and the distance he is away from a boat.
  1. ( π 2 , π 2 ) ;
  2. A graph of a half period of a secant function. Vertical asymptotes at x=-pi/2 and pi/2.
  3. x = π 2 and x = π 2 ; the distance grows without bound as | x | approaches π 2 —i.e., at right angles to the line representing due north, the boat would be so far away, the fisherman could not see it;
  4. 3; when x = π 3 , the boat is 3 km away;
  5. 1.73; when x = π 6 , the boat is about 1.73 km away;
  6. 1.5 km; when x = 0
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A laser rangefinder is locked on a comet approaching Earth. The distance g ( x ) , in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g ( x ) = 250,000 csc ( π 30 x ) .

  1. Graph g ( x ) on the interval [ 0 , 35 ] .
  2. Evaluate g ( 5 ) and interpret the information.
  3. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond?
  4. Find and discuss the meaning of any vertical asymptotes.
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A video camera is focused on a rocket on a launching pad 2 miles from the camera. The angle of elevation from the ground to the rocket after x seconds is π 120 x .

  1. Write a function expressing the altitude h ( x ) , in miles, of the rocket above the ground after x seconds. Ignore the curvature of the Earth.
  2. Graph h ( x ) on the interval ( 0 , 60 ) .
  3. Evaluate and interpret the values h ( 0 ) and h ( 30 ) .
  4. What happens to the values of h ( x ) as x approaches 60 seconds? Interpret the meaning of this in terms of the problem.
  1. h ( x ) = 2 tan ( π 120 x ) ;
  2. An exponentially increasing function with a vertical asymptote at x=60.
  3. h ( 0 ) = 0 : after 0 seconds, the rocket is 0 mi above the ground; h ( 30 ) = 2 : after 30 seconds, the rockets is 2 mi high;
  4. As x approaches 60 seconds, the values of h ( x ) grow increasingly large. The distance to the rocket is growing so large that the camera can no longer track it.
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Questions & Answers

f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
Ken Reply
proof
AUSTINE
sebd me some questions about anything ill solve for yall
Manifoldee Reply
how to solve x²=2x+8 factorization?
Kristof Reply
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
SO THE ANSWER IS X=-8
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
1KI POWER 1/3 PLEASE SOLUTIONS
Prashant Reply
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
Reuben Reply
which of these functions is not uniformly cintinuous on (0, 1)? sinx
Pooja Reply
which of these functions is not uniformly continuous on 0,1
Basant Reply
solve this equation by completing the square 3x-4x-7=0
Jamiz Reply
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
mantu
what's the formula
Modress
-x=7
Modress
new member
siame
what is trigonometry
Jean Reply
deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
Thomas
solve for me this equational y=2-x
Rubben Reply
what are you solving for
Alex
solve x
Rubben
you would move everything to the other side leaving x by itself. subtract 2 and divide -1.
Nikki
then I got x=-2
Rubben
it will b -y+2=x
Alex
goodness. I'm sorry. I will let Alex take the wheel.
Nikki
ouky thanks braa
Rubben
I think he drive me safe
Rubben
how to get 8 trigonometric function of tanA=0.5, given SinA=5/13? Can you help me?m
Pab Reply
More example of algebra and trigo
Stephen Reply
What is Indices
Yashim Reply
If one side only of a triangle is given is it possible to solve for the unkown two sides?
Felix Reply
cool
Rubben
kya
Khushnama
please I need help in maths
Dayo Reply
Okey tell me, what's your problem is?
Navin

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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