# Trigometric functions

1. Mar 19, 2007

### Jess048

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.

2. Mar 19, 2007

### Integral

Staff Emeritus
Could you share with us the general forumla given in your text or notes for this type of function?

3. Mar 19, 2007

### HallsofIvy

Staff Emeritus
Do you know the definitions of "phase shift" and period? What is the period of tan(x)?

4. Mar 20, 2007

### Jess048

The period of function y = tan k0 is pie/k where k>o. I'm not sure about the phase shift.

5. Mar 20, 2007

### drpizza

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. Mar 20, 2007

### HallsofIvy

Staff Emeritus
tan(x) has period $\pi$. In particular, $tan(0)= tan(\pi)$.
One period starts at x= 0 and ends at $x= \pi$.

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