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.
Thus we have
This means that or if you don't understand this, this means that :
<< difference of two squares :
So either , either
From the first one, we get (k is an integer), from the second one, we get (k' is an integer)
In fact, the period T is the least nonnegative number such that it is periodic.
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."