# Finding the period of this trig function

1. Apr 10, 2009

### Dell

f(x)=sin3x*cos3x

i am looking for the period of this function,
what i did, and now i see that it is wrong was compared it to zero and looked for any time that

sin3x=0 or cos3x=0

this really did give an answer for every time the function reached 0, but did not take into account that once i get 0+ and once i get 0-, how do i find the period?

what i did
sin3x=0 or cos3x=0
x=K*pi/3 or x=pi/6 + K*pi/3

then i got
T=pi/6

do i need to use trig identities to bring it to a function of just sin or cos?? which identities?

2. Apr 10, 2009

### Dick

How about sin(2a)=2*sin(a)*cos(a)? More generally you can use product to sum formulas like sin(a)*cos(b)=(1/2)*(sin(a+b)+sin(a-b)).

3. Apr 10, 2009

### Dell

now what do i do to find the period? what do i compare it to
1/2*sin(6x)=1/2 ???
sin(6x)=1
6x=pi/2
x=pi/12

not right,

mathematically how do i ptove that the period is pi/3

4. Apr 10, 2009

### Dick

sin(x) has period 2pi, right? For sin(6x), 6x goes from 0 to 2pi (one period) while x goes from 0 to 2pi/6. 2pi/6 is pi/3.

5. Apr 10, 2009

### Dell

okay i understand that, but is there no way to get to this with equations, for example, i had another question to find period for cos^2(x)

so i said
cos^2(x)=cos^2(0)=1
cos^2(x)=1
cos(x)=+-1
x=0+2pi*K or x=pi + 2pi*K
x=[0 pi 2pi 3pi...]
T=pi

how would you solve this like you solved the sin, using cos(x) has a period of 2pi,

6. Apr 10, 2009

### Dick

I would use product forms to write cos(x)^2 as a sum of sines and cosines, like I said in the first post. cos(x)^2=(cos(2x)+1)/2. The '2x' part is what determines the period.

7. Apr 10, 2009

### Dell

so only the actual trig function makes any difference, the peiod of cosx is the same as, (4*cos(x) +2) for eg??

8. Apr 10, 2009

### Dick

Sure. Isn't that pretty easy to see? Imagine plotting it.

9. Apr 10, 2009

### AUMathTutor

"so only the actual trig function makes any difference, the peiod of cosx is the same as, (4*cos(x) +2) for eg?? "

Yes, a function is periodic with period T iff (d^n)f(x)/(dx^n) = (d^n)f(x + T)/(dx^n) for all n (including zero, where the zeroth derivative is interpreted to mean the original function). At least that's my understanding... someone may please correct me if I'm in error.

10. Apr 10, 2009

### Dick

Being periodic has nothing to do with being differentiable. T is a period of a function f if f(x)=f(x+T) for all x. The period of f is the smallest such value of T.