# Trigometric functions

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.

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

HallsofIvy
Homework Helper
Do you know the definitions of "phase shift" and period? What is the period of tan(x)?

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

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?

HallsofIvy
Homework Helper
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)?