Question

[Algebra 2]

Can anyone help explain how to do this as easy as possible? I know barely anything on this and I have a quiz on this Monday. Any help is greatly appreciated, and feel free to use any example of a question.

Answer 1

More pictures on questions. Also could someone explain the formula 'period'? Not exactly sure what it does.

Answer 2

Final picture. Thanks again for any help!

Answer 3

Forgot to add the instructions:

"Using radians, find the amplitude and period of each function. Then graph."

Answer 4

@dan815

Answer 5

most of this just requires memorization ....

Answer 6

cos is the flip of sec .. for a visual

cos <=> sec, rides along the humps of cos

sin <=> csc, rides along the hums of sin

tan <=> cot, is the mirror of tan and shifted.

Answer 7

Could you explain how I would solve a problem such as #12 in the final picture?
@amistre64

Answer 8

whats our amp and period?

Answer 9

Unsure what those are... can you briefly explain?

Answer 10

youve already worked some problems out, its on your sheets.

Answer 11

how did you determine them?

Answer 12

They're notes from my teacher. He didn't really explain what they actually meant.

Answer 13

amp is best viewed on cos and sin function; its represents how tall they are from the midline

period is the shortest distance for which the function repeats itself

Answer 14

now here is where memorization comes into play

sin and cos have a normal period of 2pi
tan has a normal period of pi

Answer 15

Makes sense, so to find the amp of #12 wouldn't it be the number beside cos, which is 4?

Answer 16

the amp of cot ... yes, its the multiplier to the function.

4 cot(u) has an amp of 4

Answer 17

Okay, now to solve the rest of that problem what else would I need to do?

Answer 18

if you need to determine the period ...

tan(u) has a normal period of pi; its complement has a normal period of pi

in other words: tan(0) = tan(pi)

so lets take the \(\theta\) variable; when is it equal to pi?

Answer 19

O/3 = pi, when O=??

Answer 20

Wouldn't 0 be equal to the amp? Really unsure.

Answer 21

@amistre64
Unsure if you're notified of my responses sorry if spamming

Answer 22

4cot(u) 4 is the amp

now u = x/3, simply becasue its a pain to type theta

cot is periodic in pi, so when does x/3 = pi?

Answer 23

Whenever theta equals pi?

Answer 24

pi/3 is not equal to pi

Answer 25

x/3 = pi ... its a pretty basic algebra process to solve for x

Answer 26

Oh... sorry wasn't thinking so 3 cancels out itself, multiplies pi by 3 and that's it?

Answer 27

so when x=3pi, therefore the period of this function is 3pi

Answer 28

Okay, what would have to be done next? Would we have to move onto graphing now once the period and amp is found?

Answer 29

graphing is next yes, and this takes some memorization as well

this is what a normal tan(x) function looks like

cot(x) = 1/tan(x)

Answer 30

I know there's something called the 'unit circle', and I will have that for the test. Would that be needed?

Answer 31

Could you also break down the graphing please

Answer 32

imnot sure how useful the unit circle is unless your trying to find special values

this is comparing tan(x) to cot(x)

Answer 33

notice that at pi/4 they are normally equal to 1, our amplitude changes that now to 4

Answer 34

Just to make sure, amp is for the x-coordinates and period is for y?

Answer 35

no, amplitude is a measure of height ... y moves up and down

x is a measure of width, x moves left and right

Answer 36

So amp would have to be for Y and period would need to be for X coordinates, right?

Answer 37

yep

Answer 38

period means: shortest interval of X, in which the function repeats itself

Answer 39

Ok and could you explain why its at 1 (y coord) due to tan? And also, the asymptote is doubled by the period correct!

Answer 40

If the asymptote isn't doubled by the period, then what is? Or am I wrong and there is no asymtote

Answer 41

vertical asymps are periodic as well yes

tan(x) = sin(x)/cos(x) = 1 ; when sin(x) = cos(x), this is only true for x=pi/4

cos(x) = cos(x)/sin(x) = 1 .... for the same reason

Answer 42

the tan (and cot) amplitude can be defined by its relevance to pi/4

Answer 43

Okay... That seems to make sense. Now final question, how would I do the math for something such as #3 when trying to find the period?

Answer 44

lets change the variable

cos(u) let u = 0 to t, when is cos(0) = cos(t) for the first time?

Answer 45

what value of t, greater than 0, is cos(0) = cos(t) ??

Answer 46

Sorry one quick question what determines tan/sine/cos when finding the period?

Answer 47

memorization ..... trig is 98% memorization which is why its such a pain

Answer 48

So what would you suggest to finding it the easiest way?

Answer 49

memorize when cos(0) comes around again.

cos(0) equals what?

Answer 50

Unsure how to determine it with the 3 relationships sine tan and cos

Answer 51

what is the function define in #3?

Answer 52

Cos = sin?

Answer 53

\[y=2+4cos(u)\]
what function is being used in the definition of y?

Answer 54

Cos?

Answer 55

yeah, cos is used in the definition so we apply it to the graph

Answer 56

we could just as well solve:

x1 /3 + pi/2 = 0

x2 /3 + pi/2 = 2pi

and then draw the cosine along the interval form x1 to x2; for a period of x2-x1

Answer 57

Oh.. The cosine part's answer was right there... One thing though when simplifying did you have 2 times itself and times the entire equation? How would you have gotten x2-x1 more specifically?

Answer 58

Also what did you do with the period to turn it into x6 according to the graph?

Answer 59

the argument for the function, the stuff thats inside the (....)

it modifies the x axis.

factor out the coefficient of x

u = x/3 + pi/2 = 1/3 (x + 3pi/2)

does this make sense?

Answer 60

2 +- Amp is the range of the graphs height for #3

amp = 4 soo

2 +- 4 = -2 to 6

Answer 61

Somewhat, but doesn't fully explain how you got x6 on the graph as your period

Answer 62

edit ****

the amplitude ISNT x6, its 4
^^^^^^^^^^^

the normal midline of cos is y=0

when we add 2 to it its shifts the graph upwards by ... well 2

y = 2 + 4cos(u); cos(u) = -1 to 1 soo

max y = 2+4 = 6
min y = 2-4 = -2

there is no x6 amplitude.

Answer 63

I get what you mean as for amp now, it can go as low as -2 to as high as 6 correct?

Answer 64

Sorry I meant 6pi for period

Answer 65

the range of y is therefore -2 to 6, a distance of 8; and 8/2 = 4 ... our amp

so your question is how to determine the period

which is why i asked the question earlier:

cos(0) = cos(t) for what value of t?

Answer 66

When you were solving for period it didn't really say how you got 6pi, its the only thing I don't get at this point

Answer 67

x1 /3 + pi/2 = 3(0-pi/2); when x1 =-3pi/2

x2 /3 + pi/2 = 3(2pi - pi/2); when x2 = 9pi/2

the interval of X that gives us ONE copy of this function is therefore from: -3pi/2 to 9pi/2

the distance between x1 and x2 is

9pi/2 - -3pi/2 = 12pi/2

the period is therefore 12pi/2, or 6pi

Answer 68

Thanks for the help.... Your latest response really cleared that up. Will you be on tomorrow in case I were to run into any problems with graphing?

Answer 69

the shortcut to period ... is to take the coefficient of x, and divide 2pi by it

x/3 has a coeff of 1/3

2pi
--- = 3*2pi = 6pi
1/3

Answer 70

i cant guarentee what im doing in 3 seconds much less 6 hours :)

Answer 71

and i seem to have let my brain wander .. that calculations are:

x1 /3 + pi/2 = 0
x1 /3 = 0-pi/2
x1 =-3pi/2

x2 /3 + pi/2 = 2pi
x2 /3 = 2pi - pi/2
x2 = 3(2pi - pi/2)
x2 = 9pi/2

Answer 72

Great help, I have a decent chance of doing well now. Thanks!

Answer 73

good luck

Answer 74

@amistre64 looking at the way you did the period, could you break this down in the easiest way you know how and explain where the values came from? I want to make sure I understand it completely and one more thing could you explain how you would find the period if you had fractions in the equation adding/subtracting with theta involved?

Answer 75

when does a circle, complete itself and start over again? trig is all about circle function ... and relating them to right triangles.

the unit circle is a measure of how many radians?

Answer 76

and lets call theta t instead

Answer 77

A circle completes itself when going all the way around itself, which is 360 degrees; the unit circle would be 360 in radians?

Answer 78

yes, now lets place the unit circle so that 0 degree is the positivie x axis, and work it up and over; let x=cos(t) and y=sin(t)



the period of sin(t) and cos(t) is one complete circle; they start their values over again after one complete turn.

memorize; period of sin and cos = 2pi, or 360 degrees