anonymous
  • anonymous
Really annoying trigonometric graphing, help!
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
I need 1,2,5, and 10-24
anonymous
  • anonymous
I'm just not grasping the concept.

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amistre64
  • 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
  • amistre64
one thing that might help is to rewrite sec, or maybe even tan. into equivalent terms
anonymous
  • anonymous
how so?
amistre64
  • amistre64
well, tan = sin/cos, right? and sec = 1/cos .... that might help us see the resemblence
amistre64
  • 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
  • amistre64
since the denominators are the same, lets equate numberators
amistre64
  • amistre64
2sin(pi x/4) < 1
anonymous
  • anonymous
so now we graph the equation to find the point?
amistre64
  • 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
  • amistre64
sin (t) < 1/2; well, when does sin(t) = 1/2?
anonymous
  • anonymous
how would we go about the analysis?
amistre64
  • amistre64
divide each side by 2 to get: sin(t) < 1/2 im using "y" to help clean up the argument
amistre64
  • amistre64
"t" that is ... cant type ;)
anonymous
  • anonymous
that's not one of the choices though :/ ,
amistre64
  • amistre64
why would it be? we are only in the middle of it
anonymous
  • anonymous
Hang in there and understand the steps.
anonymous
  • 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
  • amistre64
it would be good to remember the basic angles of trig; i believe this is one of them
amistre64
  • amistre64
|dw:1323627720794:dw|
amistre64
  • amistre64
sin(t) = 1/2, when t = 30 degrees, or pi/6
anonymous
  • anonymous
is that the special right triangle?
amistre64
  • amistre64
it is
amistre64
  • 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
  • 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
  • anonymous
why the -1?
amistre64
  • amistre64
its part of our initial interval; and sin(pi *-1/4) is less than 1/2
anonymous
  • anonymous
i see. so for number 2 i would just multiply our original interval?
amistre64
  • 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
  • anonymous
or would it not change?
amistre64
  • 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
  • anonymous
its not it.
anonymous
  • anonymous
like, number 1 was right, but 2f changed it in some way, because now it's not the same
amistre64
  • 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
  • anonymous
must be a problem with the software. now for number 5,i cant tell the graphs apart
amistre64
  • amistre64
which one is number 5? this thing is hard to follow
anonymous
  • anonymous
graph for y=csc (x)
amistre64
  • amistre64
well, I know csc is the humpbacks of sine|dw:1323628829369:dw|
amistre64
  • amistre64
the closest one I see is A
amistre64
  • amistre64
we can draw the sin(x) in A and see that csc rides the humpbacks
anonymous
  • anonymous
yeah, me too! thanks so much
anonymous
  • anonymous
number 10?
amistre64
  • amistre64
id ditch the graph... see that sqrt(3)? that comes about in the 30-60-90 triangle
amistre64
  • 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
  • anonymous
so we would relate the degrees with a sin fuction?
amistre64
  • amistre64
|dw:1323629102467:dw|
amistre64
  • amistre64
almost; lets draw the 30-60-90 tri again and see what cot comes from
amistre64
  • amistre64
|dw:1323629160526:dw| the basic angle is then cot(30) do you agree? 30 = pi/6
anonymous
  • anonymous
yeah, i see where you're coming from
amistre64
  • amistre64
q2: 180-30 = 150 degrees q4: 360 - 30 = 330 degrees
amistre64
  • amistre64
hmmm, 5pi/6 and 11pi/6 seem to fit in there somewhere. but i do get these mixed aroung at times :)
amistre64
  • amistre64
i see the interval now; it one full revolution forward and one full revolution backwards
anonymous
  • anonymous
ignore that
anonymous
  • anonymous
it sent a previous message
anonymous
  • anonymous
would the pi/6 be negative? since were dealing with cot?
amistre64
  • amistre64
|dw:1323629494595:dw|
amistre64
  • amistre64
-pi/6, 5pi/6, -7pi/6, 11pi/6 is my best assessment
amistre64
  • amistre64
thats option 4 right?
anonymous
  • anonymous
for 11, is the period 16/
anonymous
  • anonymous
YES, YOU WERE RIGHT ONCE AGAIN! :)
amistre64
  • 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
  • amistre64
that doesnt work out right ....
anonymous
  • anonymous
if we divide it by 2pi, you get 16
amistre64
  • amistre64
that does sound familiar :)
amistre64
  • amistre64
but the 2pi is wrong
amistre64
  • amistre64
divide by pi; since this is a pi period 2pi is for sin and cos
anonymous
  • anonymous
so it would be 1/8?
amistre64
  • amistre64
pi/(pi/8) = pi * 8/pi = 8
amistre64
  • amistre64
normal period/w
anonymous
  • anonymous
and the normal period is 8??
amistre64
  • amistre64
look again
amistre64
  • amistre64
tan/cot has a normal pf pi everything else is 2pi
amistre64
  • amistre64
pi/(w) = periodicity
amistre64
  • amistre64
since w = pi/8 .... we plug it in
anonymous
  • anonymous
we end up with 1/8.
amistre64
  • amistre64
\[\frac{pi}{pi/8}=\frac{pi*8}{pi}=8\]
anonymous
  • 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
  • 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
  • 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
  • amistre64
|dw:1323630627950:dw|
amistre64
  • amistre64
this doesnt seem to match any of our basic triangles does it?
amistre64
  • 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
  • amistre64
|dw:1323630843700:dw|
anonymous
  • anonymous
'disguise' lol
anonymous
  • anonymous
so we're just rearranging the equation so the triangle makes sense, in a way
amistre64
  • 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
  • amistre64
60 = pi/3 .... so we need to adapt that to the interval
amistre64
  • amistre64
|dw:1323631158475:dw|
amistre64
  • amistre64
|dw:1323631248664:dw|
amistre64
  • amistre64
thatss what, option 7?
anonymous
  • anonymous
seven is incorrect :-(nt and so is 3
amistre64
  • amistre64
trig takes alot of mental gymnastics :)
amistre64
  • amistre64
hmmm
amistre64
  • amistre64
i think I know where I went astray at
amistre64
  • amistre64
csc is only positive in q1 and q2, i confused it with a tan ....
amistre64
  • amistre64
|dw:1323631597639:dw|
amistre64
  • amistre64
-5,-4,1,2 ---------, might be better 3
anonymous
  • anonymous
what choice is that?
amistre64
  • amistre64
option 4, since that aint in the choices
amistre64
  • amistre64
youre right tho, these ARE really annoying ;)
anonymous
  • anonymous
finally correct! this thing penalizes me for every wrong answer too... now for the period of number 13, is 7?
amistre64
  • 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
  • anonymous
ow do we find the frequency?
amistre64
  • amistre64
frequency is how fast it moves along .... id have to look it up :)
amistre64
  • amistre64
http://regentsprep.org/Regents/math/algtrig/ATT7/graphvocab.htm
amistre64
  • amistre64
freq = 1/period
anonymous
  • anonymous
initial phase?
amistre64
  • amistre64
i dont recall ever doing an initial phase ...
amistre64
  • amistre64
is that a phase shift perhaps?
anonymous
  • anonymous
it says initial phase, it might be referring to the same thing
amistre64
  • amistre64
i cant make any sense of the online stuff. I doubt that it is the same thing then
amistre64
  • amistre64
initial phase is when t=0 ....
amistre64
  • amistre64
but i got no idea at the moment what the means
anonymous
  • anonymous
what about the period of 1 7? i got 10/9, but that's apparently wrong

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