Really annoying trigonometric graphing, help!

- anonymous

Really annoying trigonometric graphing, help!

- Stacey Warren - Expert brainly.com

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- schrodinger

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- anonymous

- anonymous

I need 1,2,5, and 10-24

- anonymous

I'm just not grasping the concept.

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## More answers

- amistre64

f(x) = tan πx/4
g(x) =1/2 sec πx/4
Approximate the interval where f < g.
lets set this up then, we know f and g
tan (πx/4) < 1/2 sec(πx/4)

- amistre64

one thing that might help is to rewrite sec, or maybe even tan. into equivalent terms

- anonymous

how so?

- amistre64

well, tan = sin/cos, right?
and sec = 1/cos .... that might help us see the resemblence

- amistre64

\[\frac{sin (πx/4)}{cos (πx/4)} < \frac{1}{2}\frac{1}{cos(πx/4)}\]
we can get rid of that 1/2 by multiplying by 2
\[\frac{(2)sin (πx/4)}{cos (πx/4)} < \frac{1(2)}{2}\frac{1}{cos(πx/4)}\]
\[\frac{2sin (πx/4)}{cos (πx/4)} < \frac{1}{cos(πx/4)}\]
now it looks better

- amistre64

since the denominators are the same, lets equate numberators

- amistre64

2sin(pi x/4) < 1

- anonymous

so now we graph the equation to find the point?

- amistre64

id stick to the analysis rather than a graph; the graph can be used to dbl chk the results; but i doubt it will give a definitive result

- amistre64

sin (t) < 1/2; well, when does sin(t) = 1/2?

- anonymous

how would we go about the analysis?

- amistre64

divide each side by 2 to get: sin(t) < 1/2
im using "y" to help clean up the argument

- amistre64

"t" that is ... cant type ;)

- anonymous

that's not one of the choices though :/ ,

- amistre64

why would it be? we are only in the middle of it

- anonymous

Hang in there and understand the steps.

- anonymous

OH! Gotcha. I just have a terrible teacher who goes about everything explaining really fast, and then expecting us to know all this. I'm sorry.

- amistre64

it would be good to remember the basic angles of trig; i believe this is one of them

- amistre64

|dw:1323627720794:dw|

- amistre64

sin(t) = 1/2, when t = 30 degrees, or pi/6

- anonymous

is that the special right triangle?

- amistre64

it is

- amistre64

i have to remember the interval:
-pi/4 to pi/4, that is between 45 and -45 degrees if i recall it correctly

- amistre64

now i would see about equating pi x/4 and t
pi x/4 = pi/6
divide off the pis
x/4 = 1/6
and multiply off the 1/4
x = 4/6 , and simplify
x = 2/3
so, if i did it right, it should be (-1, 2/3)
but if you have questions as to why, it would be good to ask. The numbers tend to be unimportant and it is the process that matters.

- anonymous

why the -1?

- amistre64

its part of our initial interval; and sin(pi *-1/4) is less than 1/2

- anonymous

i see. so for number 2 i would just multiply our original interval?

- amistre64

it would be nice if we could, but no.
trig functions are periodic, they do not act like linear functions. so its best to retrace the steps to make sure

- anonymous

or would it not change?

- amistre64

2f just changes how high or low the graph would go; so id guess that its the same intervals. But i would still have to dbl chk the results

- anonymous

its not it.

- anonymous

like, number 1 was right, but 2f changed it in some way, because now it's not the same

- amistre64

\[2*\frac{sin (πx/4)}{cos (πx/4)} < 2*\frac{1}{2}\frac{1}{cos(πx/4)}\]
\[\frac{2sin (πx/4)}{cos (πx/4)} < \frac{1}{cos(πx/4)}\]
\[2sin (πx/4) < 1\]
\[sin (πx/4) < 1/2\]
it looks to be the same

- anonymous

must be a problem with the software. now for number 5,i cant tell the graphs apart

- amistre64

which one is number 5? this thing is hard to follow

- anonymous

graph for y=csc (x)

- amistre64

well, I know csc is the humpbacks of sine|dw:1323628829369:dw|

- amistre64

the closest one I see is A

- amistre64

we can draw the sin(x) in A and see that csc rides the humpbacks

- anonymous

yeah, me too! thanks so much

- anonymous

number 10?

- amistre64

id ditch the graph...
see that sqrt(3)? that comes about in the 30-60-90 triangle

- amistre64

and since cot = cos/sin; this is negative when either cos is negative OR sin is negative; but not both. that occurs in q2 and q4

- anonymous

so we would relate the degrees with a sin fuction?

- amistre64

|dw:1323629102467:dw|

- amistre64

almost; lets draw the 30-60-90 tri again and see what cot comes from

- amistre64

|dw:1323629160526:dw|
the basic angle is then cot(30) do you agree?
30 = pi/6

- anonymous

yeah, i see where you're coming from

- amistre64

q2: 180-30 = 150 degrees
q4: 360 - 30 = 330 degrees

- amistre64

hmmm, 5pi/6 and 11pi/6 seem to fit in there somewhere. but i do get these mixed aroung at times :)

- amistre64

i see the interval now; it one full revolution forward and one full revolution backwards

- anonymous

ignore that

- anonymous

it sent a previous message

- anonymous

would the pi/6 be negative? since were dealing with cot?

- amistre64

|dw:1323629494595:dw|

- amistre64

-pi/6, 5pi/6, -7pi/6, 11pi/6 is my best assessment

- amistre64

thats option 4 right?

- anonymous

for 11, is the period 16/

- anonymous

YES, YOU WERE RIGHT ONCE AGAIN! :)

- amistre64

tan and cot have a normal period of pi; this speeds that period up; by a factor of 8.; so id say its a period of 1/8

- amistre64

that doesnt work out right ....

- anonymous

if we divide it by 2pi, you get 16

- amistre64

that does sound familiar :)

- amistre64

but the 2pi is wrong

- amistre64

divide by pi; since this is a pi period
2pi is for sin and cos

- anonymous

so it would be 1/8?

- amistre64

pi/(pi/8) = pi * 8/pi = 8

- amistre64

normal period/w

- anonymous

and the normal period is 8??

- amistre64

look again

- amistre64

tan/cot has a normal pf pi
everything else is 2pi

- amistre64

pi/(w) = periodicity

- amistre64

since w = pi/8 .... we plug it in

- anonymous

we end up with 1/8.

- amistre64

\[\frac{pi}{pi/8}=\frac{pi*8}{pi}=8\]

- anonymous

oh, i was solving it incorrectly. can you explain 12 to me please? i know it's similar to 10, but i want to grasp the concept.

- amistre64

since csc rides the humpbacks of sin; lets equate this to an angle of sin.
csc = 1/sin is a good thing to remember

- amistre64

\[csc(x)=\frac{2\sqrt{3}}{3}\]
\[\frac{1}{sin(x)}=\frac{2\sqrt{3}}{3}\]
\[\frac{sin(x)}{1}=\frac{3}{2\sqrt{3}}\]
and draw a triangle to help visualise it

- amistre64

|dw:1323630627950:dw|

- amistre64

this doesnt seem to match any of our basic triangles does it?

- amistre64

maybe .... divide off sqrt(3)
2sqrt(3)/sqrt(3) = 2
3/sqrt(3) = 3sqrt(3)/3 = sqrt(3)
its our 30-60-90 in disguise

- amistre64

|dw:1323630843700:dw|

- anonymous

'disguise' lol

- anonymous

so we're just rearranging the equation so the triangle makes sense, in a way

- amistre64

\[csc(60) = \frac{2}{\sqrt{3}}\]
\[csc(60) = (\frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}})\]
\[csc(60) = \frac{2\sqrt{3}}{3}\]
so we are good :)

- amistre64

60 = pi/3 .... so we need to adapt that to the interval

- amistre64

|dw:1323631158475:dw|

- amistre64

|dw:1323631248664:dw|

- amistre64

thatss what, option 7?

- anonymous

seven is incorrect :-(nt and so is 3

- amistre64

trig takes alot of mental gymnastics :)

- amistre64

hmmm

- amistre64

i think I know where I went astray at

- amistre64

csc is only positive in q1 and q2, i confused it with a tan ....

- amistre64

|dw:1323631597639:dw|

- amistre64

-5,-4,1,2
---------, might be better
3

- anonymous

what choice is that?

- amistre64

option 4, since that aint in the choices

- amistre64

youre right tho, these ARE really annoying ;)

- anonymous

finally correct! this thing penalizes me for every wrong answer too... now for the period of number 13, is 7?

- amistre64

the value attached to x is out "w"
and sec is a 2pi normal
\[\frac{2pi}{-7pi/2}\]
\[\frac{2pi*2}{-7pi}\]
\[\frac{2*2}{-7}=-4/7\]
id go with option 1

- anonymous

ow do we find the frequency?

- amistre64

frequency is how fast it moves along .... id have to look it up :)

- amistre64

http://regentsprep.org/Regents/math/algtrig/ATT7/graphvocab.htm

- amistre64

freq = 1/period

- anonymous

initial phase?

- amistre64

i dont recall ever doing an initial phase ...

- amistre64

is that a phase shift perhaps?

- anonymous

it says initial phase, it might be referring to the same thing

- amistre64

i cant make any sense of the online stuff. I doubt that it is the same thing then

- amistre64

initial phase is when t=0 ....

- amistre64

but i got no idea at the moment what the means

- anonymous

what about the period of 1
7? i got 10/9, but that's apparently wrong

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