period of function

**woollybull** · Sep 13th 2008, 01:29 AM

The function cos2x has a period of 2x=2pi, x=pi

The function y=(cos2x)^2 has a period of what? By looking at a graph of the function it is pi/2, but the answer in the book says pi.

Thanks for any help offered.

**Matt Westwood** · Sep 13th 2008, 01:46 AM

?! Sounds to me like the book is wrong.

$(\cos(2x))^2$ turns into an expression with $\cos (4x)$ in it, which indeed has a period $\frac {2 \pi}{4}$ which is, as you say, $\pi / 2$ .

As I say, I don't believe your book. What book is it, so we can all go and check?

**Moo** · Sep 13th 2008, 01:52 AM

Hello,

Originally Posted by woollybull

The function cos2x has a period of 2x=2pi, x=pi

The function y=(cos2x)^2 has a period of what? By looking at a graph of the function it is pi/2, but the answer in the book says pi.

Thanks for any help offered.

Let T be the period of the function $y=\cos^2(2x)$

Thus we have $\cos^2(2(x+T))=\cos^2(2x)$

This means that $|\cos(2(x+T))|=|\cos(2x)|$ or if you don't understand this, this means that :
$\cos^2(2(x+T))-\cos^2(2x)=0$ << difference of two squares :

$(\cos(2(x+T))-\cos(2x))(\cos(2(x+T))+\cos(2x))=0$

So either $\cos(2(x+T))=\cos(2x)$ , either $\cos(2(x+T))=-\cos(2x)$

From the first one, we get $2T=2k\pi \implies T=k \pi$ (k is an integer), from the second one, we get $2T=(2k'+1) \pi \implies T=(2k'+1) \frac \pi 2$ (k' is an integer)
Thus $(2k'+1) \frac \pi 2=k \pi \implies k=2k'+1$

$\boxed{T=(2k'+1) \pi}$

In fact, the period T is the least nonnegative number such that it is periodic. $k'=0 \implies \boxed{T=\frac \pi 2}$

**bkarpuz** · Sep 13th 2008, 02:54 AM

Originally Posted by woollybull

The function cos2x has a period of 2x=2pi, x=pi

The function y=(cos2x)^2 has a period of what? By looking at a graph of the function it is pi/2, but the answer in the book says pi.

Thanks for any help offered.

Let $f(x)=[\cos(2x)]^2$ .
Then, we have $f(x)=\frac{1}{2}[\cos(4x)+1]$ .
Clearly, $\cos(4x)$ is of $\frac{\pi}{2}$ period, and when we add $1$ and then devide by $2$ , the period does not change.
Also see

Therefore, Matt Westwood is right, the book must have a typo.

**ticbol** · Sep 13th 2008, 03:15 AM

Originally Posted by woollybull

The function cos2x has a period of 2x=2pi, x=pi

The function y=(cos2x)^2 has a period of what? By looking at a graph of the function it is pi/2, but the answer in the book says pi.

Thanks for any help offered.

A period here means one cycle. Whhe graph reaches a value on the horizontal axis and then it goes on to repeat another cycle, that value on the horizontal axis is the period of the curve.

The period of cos(x) is 2pi/1 = 2pi.
The graph starts at the maximum at x=0, goes down to zero at pi/2, goes down to minimum at x=pi, goes up to zero at 3pi/2, and goes up to maximum at 2pi. Then at this x-value, pi, the curve goes down again to repeat anorher cylce. That is why 2pi is the period of cos(x).

The period of cos(2x) is 2pi/2 = pi.
The curve starts at maximun at x=0, goes down to zero at pi/4, goes down to minimum at pi/2, goes up to zero at 3pi/4, and goes up again to maximum at pi. Then the curve goes down again to repeat another cycle. That is why the period of cos(2x) is pi.

What about cos^2(2x) or (cos(2x))^2?
Since it is a square, then the function value will always be positive. The curve will not go down below the horizontal axis.
So the curve will start at maximum at x=0, goes down to zero at pi/4, then instead of going down below the x-axis, the curve will go up to maximum at x=pi/2, goes down to zero at x = 3pi/4, and then goes up again to maximum at at x = pi.
The curve has a complete cycle from x = pi/4 to x = 3pi/4. The shape is that of a "dome".
That is 3pi/4 - pi/4 = 2pi/4 = pi/2.
Therefore, the period of (cos(2x))^2 is pi/2. ----------answer.

Or, looking at the curve in another way, the curve has a complete cycle from x = 0 to x = pi/2. The shape is that of a "bird, flying."

**woollybull** · Sep 14th 2008, 12:22 AM

That was a great response.
Thanks for all your help.

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