• Support PF! Buy your school textbooks, materials and every day products Here!

Trigometric functions

  • Thread starter Jess048
  • Start date
  • #1
9
0
State the period and phase shift of the function y= -4 tan (1/2x + 3pie/8).

I know the answer is pie; -3pie/8, but I don't understand the process could someone explain how the period and phase shift are found in these functions.
 

Answers and Replies

  • #2
Integral
Staff Emeritus
Science Advisor
Gold Member
7,198
55
Could you share with us the general forumla given in your text or notes for this type of function?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,806
932
Do you know the definitions of "phase shift" and period? What is the period of tan(x)?
 
  • #4
9
0
The period of function y = tan k0 is pie/k where k>o. I'm not sure about the phase shift.
 
  • #5
286
0
Maybe this'll help:
Consider y=x^2

y=(x-2.5)^2 is the same function, shifted to the right 2.5 units.

y=(2x-6)^2 would have to first be written as
y=(2(x-3))^2
This is the function y=x^2 shifted 3 units to the right. The 2 does something else to the function (stretches it vertically in this case.)
Can you get (x-#) in your problem?
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,806
932
tan(x) has period [itex]\pi[/itex]. In particular, [itex]tan(0)= tan(\pi)[/itex].
One period starts at x= 0 and ends at [itex]x= \pi[/itex].

Okay, one period of [itex]-4tan((1/2)x- 3\pi/8)[/itex] "starts" when [itex](1/2)x- 3\pi/8= 0[/itex] and ends when [itex](1/2)x- 3\pi/8= \pi[/itex]. What is the "starting" value of x (the phase shift) and what is the difference between the two values of x (the period)?
 

Related Threads on Trigometric functions

Replies
5
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
3K
Top