Really annoying trigonometric graphing, help!

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Really annoying trigonometric graphing, help!

Mathematics
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I need 1,2,5, and 10-24
I'm just not grasping the concept.

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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)
one thing that might help is to rewrite sec, or maybe even tan. into equivalent terms
how so?
well, tan = sin/cos, right? and sec = 1/cos .... that might help us see the resemblence
\[\frac{sin (πx/4)}{cos (πx/4)} < \frac{1}{2}\frac{1}{cos(πx/4)}\] 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
since the denominators are the same, lets equate numberators
2sin(pi x/4) < 1
so now we graph the equation to find the point?
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
sin (t) < 1/2; well, when does sin(t) = 1/2?
how would we go about the analysis?
divide each side by 2 to get: sin(t) < 1/2 im using "y" to help clean up the argument
"t" that is ... cant type ;)
that's not one of the choices though :/ ,
why would it be? we are only in the middle of it
Hang in there and understand the steps.
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.
it would be good to remember the basic angles of trig; i believe this is one of them
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sin(t) = 1/2, when t = 30 degrees, or pi/6
is that the special right triangle?
it is
i have to remember the interval: -pi/4 to pi/4, that is between 45 and -45 degrees if i recall it correctly
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.
why the -1?
its part of our initial interval; and sin(pi *-1/4) is less than 1/2
i see. so for number 2 i would just multiply our original interval?
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
or would it not change?
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
its not it.
like, number 1 was right, but 2f changed it in some way, because now it's not the same
\[2*\frac{sin (πx/4)}{cos (πx/4)} < 2*\frac{1}{2}\frac{1}{cos(πx/4)}\] \[\frac{2sin (πx/4)}{cos (πx/4)} < \frac{1}{cos(πx/4)}\] \[2sin (πx/4) < 1\] \[sin (πx/4) < 1/2\] it looks to be the same
must be a problem with the software. now for number 5,i cant tell the graphs apart
which one is number 5? this thing is hard to follow
graph for y=csc (x)
well, I know csc is the humpbacks of sine|dw:1323628829369:dw|
the closest one I see is A
we can draw the sin(x) in A and see that csc rides the humpbacks
yeah, me too! thanks so much
number 10?
id ditch the graph... see that sqrt(3)? that comes about in the 30-60-90 triangle
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
so we would relate the degrees with a sin fuction?
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almost; lets draw the 30-60-90 tri again and see what cot comes from
|dw:1323629160526:dw| the basic angle is then cot(30) do you agree? 30 = pi/6
yeah, i see where you're coming from
q2: 180-30 = 150 degrees q4: 360 - 30 = 330 degrees
hmmm, 5pi/6 and 11pi/6 seem to fit in there somewhere. but i do get these mixed aroung at times :)
i see the interval now; it one full revolution forward and one full revolution backwards
ignore that
it sent a previous message
would the pi/6 be negative? since were dealing with cot?
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-pi/6, 5pi/6, -7pi/6, 11pi/6 is my best assessment
thats option 4 right?
for 11, is the period 16/
YES, YOU WERE RIGHT ONCE AGAIN! :)
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
that doesnt work out right ....
if we divide it by 2pi, you get 16
that does sound familiar :)
but the 2pi is wrong
divide by pi; since this is a pi period 2pi is for sin and cos
so it would be 1/8?
pi/(pi/8) = pi * 8/pi = 8
normal period/w
and the normal period is 8??
look again
tan/cot has a normal pf pi everything else is 2pi
pi/(w) = periodicity
since w = pi/8 .... we plug it in
we end up with 1/8.
\[\frac{pi}{pi/8}=\frac{pi*8}{pi}=8\]
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.
since csc rides the humpbacks of sin; lets equate this to an angle of sin. csc = 1/sin is a good thing to remember
\[csc(x)=\frac{2\sqrt{3}}{3}\] \[\frac{1}{sin(x)}=\frac{2\sqrt{3}}{3}\] \[\frac{sin(x)}{1}=\frac{3}{2\sqrt{3}}\] and draw a triangle to help visualise it
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this doesnt seem to match any of our basic triangles does it?
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
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'disguise' lol
so we're just rearranging the equation so the triangle makes sense, in a way
\[csc(60) = \frac{2}{\sqrt{3}}\] \[csc(60) = (\frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}})\] \[csc(60) = \frac{2\sqrt{3}}{3}\] so we are good :)
60 = pi/3 .... so we need to adapt that to the interval
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thatss what, option 7?
seven is incorrect :-(nt and so is 3
trig takes alot of mental gymnastics :)
hmmm
i think I know where I went astray at
csc is only positive in q1 and q2, i confused it with a tan ....
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-5,-4,1,2 ---------, might be better 3
what choice is that?
option 4, since that aint in the choices
youre right tho, these ARE really annoying ;)
finally correct! this thing penalizes me for every wrong answer too... now for the period of number 13, is 7?
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
ow do we find the frequency?
frequency is how fast it moves along .... id have to look it up :)
http://regentsprep.org/Regents/math/algtrig/ATT7/graphvocab.htm
freq = 1/period
initial phase?
i dont recall ever doing an initial phase ...
is that a phase shift perhaps?
it says initial phase, it might be referring to the same thing
i cant make any sense of the online stuff. I doubt that it is the same thing then
initial phase is when t=0 ....
but i got no idea at the moment what the means
what about the period of 1 7? i got 10/9, but that's apparently wrong

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