DeathLetum
New member
- Joined
- Jun 6, 2007
- Messages
- 8
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
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.
--------------------------------------------------------------------------------------
#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.
---------------------------------------------------------------------------------------
#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
---------------------------------------------------------------------------------------
#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.
----------------------------------------------------------------------------------------
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.
--------------------------------------------------------------------------------------
#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.
---------------------------------------------------------------------------------------
#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
---------------------------------------------------------------------------------------
#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.
----------------------------------------------------------------------------------------