# 6.2 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.

is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
Given
co ordinates for x x=0,(-2,0) x=1,(1,1) x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?