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anonymous
 5 years ago
Really annoying trigonometric graphing, help!
anonymous
 5 years ago
Really annoying trigonometric graphing, help!

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I need 1,2,5, and 1024

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm just not grasping the concept.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1f(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
 5 years ago
Best ResponseYou've already chosen the best response.1one thing that might help is to rewrite sec, or maybe even tan. into equivalent terms

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1well, tan = sin/cos, right? and sec = 1/cos .... that might help us see the resemblence

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[\frac{sin (πx/4)}{cos (πx/4)} < \frac{1}{2}\frac{1}{cos(πx/4)}\] 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
 5 years ago
Best ResponseYou've already chosen the best response.1since the denominators are the same, lets equate numberators

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so now we graph the equation to find the point?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1id 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
 5 years ago
Best ResponseYou've already chosen the best response.1sin (t) < 1/2; well, when does sin(t) = 1/2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how would we go about the analysis?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1divide each side by 2 to get: sin(t) < 1/2 im using "y" to help clean up the argument

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1"t" that is ... cant type ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's not one of the choices though :/ ,

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1why would it be? we are only in the middle of it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hang in there and understand the steps.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OH! 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
 5 years ago
Best ResponseYou've already chosen the best response.1it would be good to remember the basic angles of trig; i believe this is one of them

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1dw:1323627720794:dw

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1sin(t) = 1/2, when t = 30 degrees, or pi/6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that the special right triangle?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i have to remember the interval: pi/4 to pi/4, that is between 45 and 45 degrees if i recall it correctly

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1now 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.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1its part of our initial interval; and sin(pi *1/4) is less than 1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i see. so for number 2 i would just multiply our original interval?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1it 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
 5 years ago
Best ResponseYou've already chosen the best response.0or would it not change?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.12f 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
 5 years ago
Best ResponseYou've already chosen the best response.0like, number 1 was right, but 2f changed it in some way, because now it's not the same

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[2*\frac{sin (πx/4)}{cos (πx/4)} < 2*\frac{1}{2}\frac{1}{cos(πx/4)}\] \[\frac{2sin (πx/4)}{cos (πx/4)} < \frac{1}{cos(πx/4)}\] \[2sin (πx/4) < 1\] \[sin (πx/4) < 1/2\] it looks to be the same

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0must be a problem with the software. now for number 5,i cant tell the graphs apart

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1which one is number 5? this thing is hard to follow

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1well, I know csc is the humpbacks of sinedw:1323628829369:dw

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the closest one I see is A

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1we can draw the sin(x) in A and see that csc rides the humpbacks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, me too! thanks so much

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1id ditch the graph... see that sqrt(3)? that comes about in the 306090 triangle

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1and 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
 5 years ago
Best ResponseYou've already chosen the best response.0so we would relate the degrees with a sin fuction?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1dw:1323629102467:dw

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1almost; lets draw the 306090 tri again and see what cot comes from

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1dw:1323629160526:dw the basic angle is then cot(30) do you agree? 30 = pi/6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, i see where you're coming from

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1q2: 18030 = 150 degrees q4: 360  30 = 330 degrees

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1hmmm, 5pi/6 and 11pi/6 seem to fit in there somewhere. but i do get these mixed aroung at times :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i see the interval now; it one full revolution forward and one full revolution backwards

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it sent a previous message

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would the pi/6 be negative? since were dealing with cot?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1dw:1323629494595:dw

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1pi/6, 5pi/6, 7pi/6, 11pi/6 is my best assessment

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1thats option 4 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for 11, is the period 16/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0YES, YOU WERE RIGHT ONCE AGAIN! :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1tan 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
 5 years ago
Best ResponseYou've already chosen the best response.1that doesnt work out right ....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if we divide it by 2pi, you get 16

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1that does sound familiar :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1divide by pi; since this is a pi period 2pi is for sin and cos

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1pi/(pi/8) = pi * 8/pi = 8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and the normal period is 8??

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1tan/cot has a normal pf pi everything else is 2pi

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1since w = pi/8 .... we plug it in

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[\frac{pi}{pi/8}=\frac{pi*8}{pi}=8\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, 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
 5 years ago
Best ResponseYou've already chosen the best response.1since csc rides the humpbacks of sin; lets equate this to an angle of sin. csc = 1/sin is a good thing to remember

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[csc(x)=\frac{2\sqrt{3}}{3}\] \[\frac{1}{sin(x)}=\frac{2\sqrt{3}}{3}\] \[\frac{sin(x)}{1}=\frac{3}{2\sqrt{3}}\] and draw a triangle to help visualise it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1dw:1323630627950:dw

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1this doesnt seem to match any of our basic triangles does it?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1maybe .... divide off sqrt(3) 2sqrt(3)/sqrt(3) = 2 3/sqrt(3) = 3sqrt(3)/3 = sqrt(3) its our 306090 in disguise

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1dw:1323630843700:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we're just rearranging the equation so the triangle makes sense, in a way

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[csc(60) = \frac{2}{\sqrt{3}}\] \[csc(60) = (\frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}})\] \[csc(60) = \frac{2\sqrt{3}}{3}\] so we are good :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.160 = pi/3 .... so we need to adapt that to the interval

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1dw:1323631158475:dw

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1dw:1323631248664:dw

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1thatss what, option 7?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0seven is incorrect :(nt and so is 3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1trig takes alot of mental gymnastics :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i think I know where I went astray at

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1csc is only positive in q1 and q2, i confused it with a tan ....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1dw:1323631597639:dw

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.15,4,1,2 , might be better 3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1option 4, since that aint in the choices

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1youre right tho, these ARE really annoying ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0finally correct! this thing penalizes me for every wrong answer too... now for the period of number 13, is 7?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the 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
 5 years ago
Best ResponseYou've already chosen the best response.0ow do we find the frequency?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1frequency is how fast it moves along .... id have to look it up :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1http://regentsprep.org/Regents/math/algtrig/ATT7/graphvocab.htm

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i dont recall ever doing an initial phase ...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1is that a phase shift perhaps?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it says initial phase, it might be referring to the same thing

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i cant make any sense of the online stuff. I doubt that it is the same thing then

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1initial phase is when t=0 ....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1but i got no idea at the moment what the means

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what about the period of 1 7? i got 10/9, but that's apparently wrong
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