536 trig Help plz

DeathLetum

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Jun 6, 2007
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ok this might be a long post i got a exam comming up and im having probs with some stuff i need to get some help.Called for a tutor cant be here for a other 2days so i turn to ppl on here in meantime :D

I KNOW THEY SAID NOT TO POST ALOT BUT I REALLY NEED HELP SORRY.


here we go:

#4 What are the coordinates of the trigonometric point P(-25pi/4)?
The solution must indicate the Value of P' in [0, 2pi].

i did:

P(-25pi/4) P'[0.2pi] ---> P' = [0/4, 8pi/4]

-25pi/4 + 8pi = -25pi/4 + 32pi/4 = 7pi/4

Awnser would be P' = 7pi/4.

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#4 b) What is the Value of The trigonometric function Sec (11pi/4)?
The solution must indicate the P' in [0,2pi].

i have no idea about this one :? and i got like 2more like this to do so a equation and a solution could help me figure the next 2 out.


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#5 Graph the function f(T) = Sec T in the interval [-2pi, 3pi].
Indicate clearly the graduation of the axes and the asymptopes, if applicable.


i got as far as:

Periodic = 2pi
Amplitude = 1, -1
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#7 Sketch the graph of f(t) = -2cos ( 3x - pi ) for t є [ -4pi/3, 7pi/6].
Determine the amplitude, phase shift and period of this function.

i did:

A= -2 B= 3 H= -pi

Amp: 2

Period: 2pi/3

Phase shift: -pi/3

p/4 = 2pi/3 /4 = 2pi/12

First vertex: -pi/3

y= -2cos(3 ( -pi/3 ) -pi) <----- got stuck here.

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DeathLetum said:
I KNOW THEY SAID NOT TO POST ALOT BUT I REALLY NEED HELP SORRY.
Why does that mean you should violate the premise?

#4 What are the coordinates of the trigonometric point P(-25pi/4)?
What does that mean? Keep adding or subtracting 2\(\displaystyle \pi\) until you are in the desired Domain, I guess. I really don't know what the problem statement wants.

#4 b) What is the Value of The trigonometric function Sec (11pi/4)?
Exactly like the last one.

#5 Graph the function f(T) = Sec T in the interval [-2pi, 3pi].
sec(T) = 1/cos(T). What else can you need?

#7 Sketch the graph of f(t) = -2cos ( 3x - pi ) for t є [ -4pi/3, 7pi/6]. A= -2 B= 3 H= -pi
Amp: 2
Period: 2pi/3
Phase shift: -pi/3
p/4 = 2pi/3 /4 = 2pi/12
First vertex: -pi/3
y= -2cos(3 ( -pi/3 ) -pi) <----- got stuck here.
Vertex? Who asked for that? What even is a vertex on these things.

\(\displaystyle 3x-\pi = 3(x-\frac{\pi}{3})\)
Look more obvious?

If you really need help in all these areas, I sincerely hope your exam is not very soon. Good luck.
 
This is just basic unit circle sort of stuff. You got the first one right.

\(\displaystyle \L\ sec(\frac{11\pi}{4}\) = sec(2\pi\ + \frac{3\pi}{4}\) = sec(\frac{3\pi}{4}\)\)

The assymptotes for the function \(\displaystyle f(x) = sec(x)\) occur when \(\displaystyle cos(x) = 0\), so the asymptotes have equation \(\displaystyle \L\ x = \frac{\pi}{2}\ + k\pi\\), \(\displaystyle k \in\ J\).
 
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