pablobegins
Really annoying trigonometric graphing, help!
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pablobegins
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pablobegins
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I need 1,2,5, and 10-24
pablobegins
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I'm just not grasping the concept.
amistre64
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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
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one thing that might help is to rewrite sec, or maybe even tan. into equivalent terms
pablobegins
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how so?
amistre64
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well, tan = sin/cos, right?
and sec = 1/cos .... that might help us see the resemblence
amistre64
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\[\frac{sin (πx/4)}{cos (πx/4)} < \frac{1}{2}\frac{1}{cos(πx/4)}\]
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
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since the denominators are the same, lets equate numberators
amistre64
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2sin(pi x/4) < 1
pablobegins
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so now we graph the equation to find the point?
amistre64
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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
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sin (t) < 1/2; well, when does sin(t) = 1/2?
pablobegins
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how would we go about the analysis?
amistre64
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divide each side by 2 to get: sin(t) < 1/2
im using "y" to help clean up the argument
amistre64
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"t" that is ... cant type ;)
pablobegins
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that's not one of the choices though :/ ,
amistre64
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why would it be? we are only in the middle of it
GT
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Hang in there and understand the steps.
pablobegins
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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
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it would be good to remember the basic angles of trig; i believe this is one of them
amistre64
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|dw:1323627720794:dw|
amistre64
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sin(t) = 1/2, when t = 30 degrees, or pi/6
pablobegins
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is that the special right triangle?
amistre64
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it is
amistre64
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i have to remember the interval:
-pi/4 to pi/4, that is between 45 and -45 degrees if i recall it correctly
amistre64
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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.
pablobegins
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why the -1?
amistre64
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its part of our initial interval; and sin(pi *-1/4) is less than 1/2
pablobegins
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i see. so for number 2 i would just multiply our original interval?
amistre64
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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
pablobegins
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or would it not change?
amistre64
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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
pablobegins
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its not it.
pablobegins
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like, number 1 was right, but 2f changed it in some way, because now it's not the same
amistre64
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\[2*\frac{sin (πx/4)}{cos (πx/4)} < 2*\frac{1}{2}\frac{1}{cos(πx/4)}\]
\[\frac{2sin (πx/4)}{cos (πx/4)} < \frac{1}{cos(πx/4)}\]
\[2sin (πx/4) < 1\]
\[sin (πx/4) < 1/2\]
it looks to be the same
pablobegins
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must be a problem with the software. now for number 5,i cant tell the graphs apart
amistre64
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which one is number 5? this thing is hard to follow
pablobegins
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graph for y=csc (x)
amistre64
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well, I know csc is the humpbacks of sine|dw:1323628829369:dw|
amistre64
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the closest one I see is A
amistre64
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we can draw the sin(x) in A and see that csc rides the humpbacks
pablobegins
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yeah, me too! thanks so much
pablobegins
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number 10?
amistre64
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id ditch the graph...
see that sqrt(3)? that comes about in the 30-60-90 triangle
amistre64
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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
pablobegins
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so we would relate the degrees with a sin fuction?
amistre64
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|dw:1323629102467:dw|
amistre64
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almost; lets draw the 30-60-90 tri again and see what cot comes from
amistre64
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|dw:1323629160526:dw|
the basic angle is then cot(30) do you agree?
30 = pi/6
pablobegins
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yeah, i see where you're coming from
amistre64
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q2: 180-30 = 150 degrees
q4: 360 - 30 = 330 degrees
amistre64
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hmmm, 5pi/6 and 11pi/6 seem to fit in there somewhere. but i do get these mixed aroung at times :)
amistre64
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i see the interval now; it one full revolution forward and one full revolution backwards
pablobegins
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ignore that
pablobegins
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it sent a previous message
pablobegins
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would the pi/6 be negative? since were dealing with cot?
amistre64
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|dw:1323629494595:dw|
amistre64
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-pi/6, 5pi/6, -7pi/6, 11pi/6 is my best assessment
amistre64
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thats option 4 right?
pablobegins
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for 11, is the period 16/
pablobegins
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YES, YOU WERE RIGHT ONCE AGAIN! :)
amistre64
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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
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that doesnt work out right ....
pablobegins
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if we divide it by 2pi, you get 16
amistre64
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that does sound familiar :)
amistre64
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but the 2pi is wrong
amistre64
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divide by pi; since this is a pi period
2pi is for sin and cos
pablobegins
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so it would be 1/8?
amistre64
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pi/(pi/8) = pi * 8/pi = 8
amistre64
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normal period/w
pablobegins
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and the normal period is 8??
amistre64
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look again
amistre64
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tan/cot has a normal pf pi
everything else is 2pi
amistre64
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pi/(w) = periodicity
amistre64
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since w = pi/8 .... we plug it in
pablobegins
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we end up with 1/8.
amistre64
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\[\frac{pi}{pi/8}=\frac{pi*8}{pi}=8\]
pablobegins
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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
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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
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\[csc(x)=\frac{2\sqrt{3}}{3}\]
\[\frac{1}{sin(x)}=\frac{2\sqrt{3}}{3}\]
\[\frac{sin(x)}{1}=\frac{3}{2\sqrt{3}}\]
and draw a triangle to help visualise it
amistre64
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|dw:1323630627950:dw|
amistre64
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this doesnt seem to match any of our basic triangles does it?
amistre64
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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
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|dw:1323630843700:dw|
pablobegins
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'disguise' lol
pablobegins
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so we're just rearranging the equation so the triangle makes sense, in a way
amistre64
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\[csc(60) = \frac{2}{\sqrt{3}}\]
\[csc(60) = (\frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}})\]
\[csc(60) = \frac{2\sqrt{3}}{3}\]
so we are good :)
amistre64
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60 = pi/3 .... so we need to adapt that to the interval
amistre64
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|dw:1323631158475:dw|
amistre64
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|dw:1323631248664:dw|
amistre64
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thatss what, option 7?
pablobegins
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seven is incorrect :-(nt and so is 3
amistre64
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trig takes alot of mental gymnastics :)
amistre64
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hmmm
amistre64
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i think I know where I went astray at
amistre64
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csc is only positive in q1 and q2, i confused it with a tan ....
amistre64
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|dw:1323631597639:dw|
amistre64
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-5,-4,1,2
---------, might be better
3
pablobegins
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what choice is that?
amistre64
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option 4, since that aint in the choices
amistre64
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youre right tho, these ARE really annoying ;)
pablobegins
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finally correct! this thing penalizes me for every wrong answer too... now for the period of number 13, is 7?
amistre64
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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
pablobegins
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ow do we find the frequency?
amistre64
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frequency is how fast it moves along .... id have to look it up :)
amistre64
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freq = 1/period
pablobegins
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initial phase?
amistre64
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i dont recall ever doing an initial phase ...
amistre64
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is that a phase shift perhaps?
pablobegins
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it says initial phase, it might be referring to the same thing
amistre64
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i cant make any sense of the online stuff. I doubt that it is the same thing then
amistre64
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initial phase is when t=0 ....
amistre64
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but i got no idea at the moment what the means
pablobegins
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what about the period of 1
7? i got 10/9, but that's apparently wrong