# 8.2 Graphs of the other trigonometric functions  (Page 9/9)

 Page 9 / 9

How can the graph of $\text{\hspace{0.17em}}y=\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ be used to construct the graph of $\text{\hspace{0.17em}}y=\mathrm{sec}\text{\hspace{0.17em}}x?$

Explain why the period of $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is equal to $\text{\hspace{0.17em}}\pi .$

Answers will vary. Using the unit circle, one can show that $\text{\hspace{0.17em}}\mathrm{tan}\left(x+\pi \right)=\mathrm{tan}\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$

Why are there no intercepts on the graph of $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x?$

How does the period of $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ compare with the period of $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x?$

The period is the same: $\text{\hspace{0.17em}}2\pi .$

## Algebraic

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

$f\left(x\right)=\mathrm{tan}\text{\hspace{0.17em}}x$

$f\left(x\right)=\mathrm{sec}\text{\hspace{0.17em}}x$

IV

$f\left(x\right)=\mathrm{csc}\text{\hspace{0.17em}}x$

$f\left(x\right)=\mathrm{cot}\text{\hspace{0.17em}}x$

III

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

$f\left(x\right)=2\mathrm{tan}\left(4x-32\right)$

$h\left(x\right)=2\mathrm{sec}\left(\frac{\pi }{4}\left(x+1\right)\right)$

period: 8; horizontal shift: 1 unit to left

$m\left(x\right)=6\mathrm{csc}\left(\frac{\pi }{3}x+\pi \right)$

If $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x=-1.5,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}\mathrm{tan}\left(-x\right).$

1.5

If $\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}\mathrm{sec}\left(-x\right).$

If $\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x=-5,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}\mathrm{csc}\left(-x\right).$

5

If $\text{\hspace{0.17em}}x\mathrm{sin}\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}\left(-x\right)\mathrm{sin}\left(-x\right).$

For the following exercises, rewrite each expression such that the argument $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is positive.

$\mathrm{cot}\left(-x\right)\mathrm{cos}\left(-x\right)+\mathrm{sin}\left(-x\right)$

$-\mathrm{cot}x\mathrm{cos}x-\mathrm{sin}x$

$\mathrm{cos}\left(-x\right)+\mathrm{tan}\left(-x\right)\mathrm{sin}\left(-x\right)$

## Graphical

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

$f\left(x\right)=2\mathrm{tan}\left(4x-32\right)$

stretching factor: 2; period: asymptotes:

$\text{\hspace{0.17em}}h\left(x\right)=2\mathrm{sec}\left(\frac{\pi }{4}\left(x+1\right)\right)\text{\hspace{0.17em}}$

$m\left(x\right)=6\mathrm{csc}\left(\frac{\pi }{3}x+\pi \right)$

stretching factor: 6; period: 6; asymptotes:

$j\left(x\right)=\mathrm{tan}\left(\frac{\pi }{2}x\right)$

$p\left(x\right)=\mathrm{tan}\left(x-\frac{\pi }{2}\right)$

stretching factor: 1; period: asymptotes:

$f\left(x\right)=4\mathrm{tan}\left(x\right)$

$f\left(x\right)=\mathrm{tan}\left(x+\frac{\pi }{4}\right)$

Stretching factor: 1; period: asymptotes:

$f\left(x\right)=\pi \mathrm{tan}\left(\pi x-\pi \right)-\pi$

$f\left(x\right)=2\mathrm{csc}\left(x\right)$

stretching factor: 2; period: asymptotes:

$f\left(x\right)=-\frac{1}{4}\mathrm{csc}\left(x\right)$

$f\left(x\right)=4\mathrm{sec}\left(3x\right)$

stretching factor: 4; period: asymptotes:

$f\left(x\right)=-3\mathrm{cot}\left(2x\right)$

$f\left(x\right)=7\mathrm{sec}\left(5x\right)$

stretching factor: 7; period: asymptotes:

$f\left(x\right)=\frac{9}{10}\mathrm{csc}\left(\pi x\right)$

$f\left(x\right)=2\mathrm{csc}\left(x+\frac{\pi }{4}\right)-1$

stretching factor: 2; period: asymptotes:

$f\left(x\right)=-\mathrm{sec}\left(x-\frac{\pi }{3}\right)-2$

$f\left(x\right)=\frac{7}{5}\mathrm{csc}\left(x-\frac{\pi }{4}\right)$

stretching factor: period: asymptotes:

$f\left(x\right)=5\left(\mathrm{cot}\left(x+\frac{\pi }{2}\right)-3\right)$

For the following exercises, find and graph two periods of the periodic function with the given stretching factor, $\text{\hspace{0.17em}}|A|,\text{\hspace{0.17em}}$ period, and phase shift.

A tangent curve, $\text{\hspace{0.17em}}A=1,\text{\hspace{0.17em}}$ period of $\text{\hspace{0.17em}}\frac{\pi }{3};\text{\hspace{0.17em}}$ and phase shift $\text{\hspace{0.17em}}\left(h,\text{\hspace{0.17em}}k\right)=\left(\frac{\pi }{4},2\right)$

$y=\mathrm{tan}\left(3\left(x-\frac{\pi }{4}\right)\right)+2$

A tangent curve, $\text{\hspace{0.17em}}A=-2,\text{\hspace{0.17em}}$ period of $\text{\hspace{0.17em}}\frac{\pi }{4},\text{\hspace{0.17em}}$ and phase shift $\text{\hspace{0.17em}}\left(h,\text{\hspace{0.17em}}k\right)=\left(-\frac{\pi }{4},\text{\hspace{0.17em}}-2\right)$

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

$f\left(x\right)=\mathrm{csc}\left(2x\right)$

$f\left(x\right)=\mathrm{csc}\left(4x\right)$

$f\left(x\right)=2\mathrm{csc}x$

$f\left(x\right)=\frac{1}{2}\mathrm{tan}\left(100\pi x\right)$

## 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 $\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}x}.$

$f\left(x\right)=|\mathrm{csc}\left(x\right)|$

$f\left(x\right)=|\mathrm{cot}\left(x\right)|$

$f\left(x\right)={2}^{\mathrm{csc}\left(x\right)}$

$f\left(x\right)=\frac{\mathrm{csc}\left(x\right)}{\mathrm{sec}\left(x\right)}$

Graph $\text{\hspace{0.17em}}f\left(x\right)=1+{\mathrm{sec}}^{2}\left(x\right)-{\mathrm{tan}}^{2}\left(x\right).\text{\hspace{0.17em}}$ What is the function shown in the graph?

$f\left(x\right)=\mathrm{sec}\left(0.001x\right)$

$f\left(x\right)=\mathrm{cot}\left(100\pi x\right)$

$f\left(x\right)={\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x$

## Real-world applications

The function $\text{\hspace{0.17em}}f\left(x\right)=20\mathrm{tan}\left(\frac{\pi }{10}x\right)\text{\hspace{0.17em}}$ marks the distance in the movement of a light beam from a police car across a wall for time $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ in seconds, and distance $\text{\hspace{0.17em}}f\left(x\right),$ in feet.

1. Graph on the interval $\text{\hspace{0.17em}}\left[0,\text{\hspace{0.17em}}5\right].$
2. Find and interpret the stretching factor, period, and asymptote.
3. Evaluate $\text{\hspace{0.17em}}f\left(1\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(2.5\right)\text{\hspace{0.17em}}$ and discuss the function’s values at those inputs.

Standing on the shore of a lake, a fisherman sights a boat far in the distance to his left. Let $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}d\left(x\right),\text{\hspace{0.17em}}$ in kilometers, from the fisherman to the boat is given by the function $\text{\hspace{0.17em}}d\left(x\right)=1.5\mathrm{sec}\left(x\right).$

1. What is a reasonable domain for $\text{\hspace{0.17em}}d\left(x\right)?$
2. Graph $\text{\hspace{0.17em}}d\left(x\right)\text{\hspace{0.17em}}$ on this domain.
3. Find and discuss the meaning of any vertical asymptotes on the graph of $\text{\hspace{0.17em}}d\left(x\right).$
4. Calculate and interpret $\text{\hspace{0.17em}}d\left(-\frac{\pi }{3}\right).\text{\hspace{0.17em}}$ Round to the second decimal place.
5. Calculate and interpret $\text{\hspace{0.17em}}d\left(\frac{\pi }{6}\right).\text{\hspace{0.17em}}$ Round to the second decimal place.
6. What is the minimum distance between the fisherman and the boat? When does this occur?
1. $\text{\hspace{0.17em}}\left(-\frac{\pi }{2},\text{\hspace{0.17em}}\frac{\pi }{2}\right);\text{\hspace{0.17em}}$
2. $\text{\hspace{0.17em}}x=-\frac{\pi }{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=\frac{\pi }{2};\text{\hspace{0.17em}}$ the distance grows without bound as $\text{\hspace{0.17em}}|x|$ approaches $\text{\hspace{0.17em}}\frac{\pi }{2}\text{\hspace{0.17em}}$ —i.e., at right angles to the line representing due north, the boat would be so far away, the fisherman could not see it;
3. 3; when $\text{\hspace{0.17em}}x=-\frac{\pi }{3},\text{\hspace{0.17em}}$ the boat is 3 km away;
4. 1.73; when $\text{\hspace{0.17em}}x=\frac{\pi }{6},\text{\hspace{0.17em}}$ the boat is about 1.73 km away;
5. 1.5 km; when $\text{\hspace{0.17em}}x=0\text{\hspace{0.17em}}$

A laser rangefinder is locked on a comet approaching Earth. The distance $\text{\hspace{0.17em}}g\left(x\right),\text{\hspace{0.17em}}$ in kilometers, of the comet after $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ days, for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in the interval 0 to 30 days, is given by $\text{\hspace{0.17em}}g\left(x\right)=250,000\mathrm{csc}\left(\frac{\pi }{30}x\right).$

1. Graph $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ on the interval $\text{\hspace{0.17em}}\left[0,\text{\hspace{0.17em}}35\right].$
2. Evaluate $\text{\hspace{0.17em}}g\left(5\right)\text{\hspace{0.17em}}$ 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.

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 $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ seconds is $\text{\hspace{0.17em}}\frac{\pi }{120}x.$

1. Write a function expressing the altitude $\text{\hspace{0.17em}}h\left(x\right),\text{\hspace{0.17em}}$ in miles, of the rocket above the ground after $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ seconds. Ignore the curvature of the Earth.
2. Graph $\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$ on the interval $\text{\hspace{0.17em}}\left(0,\text{\hspace{0.17em}}60\right).$
3. Evaluate and interpret the values $\text{\hspace{0.17em}}h\left(0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h\left(30\right).$
4. What happens to the values of $\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ approaches 60 seconds? Interpret the meaning of this in terms of the problem.
1. $h\left(x\right)=2\mathrm{tan}\left(\frac{\pi }{120}x\right);$
2. $h\left(0\right)=0:\text{\hspace{0.17em}}$ after 0 seconds, the rocket is 0 mi above the ground; $h\left(30\right)=2:\text{\hspace{0.17em}}$ after 30 seconds, the rockets is 2 mi high;
3. As $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ approaches 60 seconds, the values of $\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$ grow increasingly large. The distance to the rocket is growing so large that the camera can no longer track it.

f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
how to solve x²=2x+8 factorization?
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
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
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
which of these functions is not uniformly cintinuous on (0, 1)? sinx
which of these functions is not uniformly continuous on 0,1
solve this equation by completing the square 3x-4x-7=0
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
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
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
More example of algebra and trigo
What is Indices
If one side only of a triangle is given is it possible to solve for the unkown two sides?
cool
Rubben
kya
Khushnama
please I need help in maths
Okey tell me, what's your problem is?
Navin