Determing period in PDE

Question

I am trying to find the period of,

$$f(x) = (\sin2x)^2$$

So I know how to determine the period of $ f(x) = \sin2x $ by the following,

$$f(x) = 2(\sin x)$$

Using the formula $P = \dfrac{2\pi}{|B|}$ here $B$ would be $2$ so, $P = \dfrac{2\pi} 2 = \pi.$

So then how could I solve my original problem? I am not sure how to solve this with the squared function

$2\sin x$ is not the same as $\sin(2x).$ Note that $2\sin x$ oscillates between $-2$ and $+2$ whereas $\sin(2x)$ oscillates between $-1$ and $+1. \qquad$ — Michael Hardy, Oct 15 '17 at 03:58

score 2 · Answer 1 · answered Oct 15 '17 at 02:37

2

Less wordy: $\sin^2{2x}=\frac{1}{2}-\frac{1}{2}\cos4x$. What is the period of $\cos4x$?

answered Oct 15 '17 at 02:37

imranfat

$\frac{\pi}{2}$? so then how do I use that – user104 Oct 15 '17 at 02:39
Hmmm, not sure what you mean by "So then how do I use that?". You answered correctly that the period of your function is $\pi/2$. In your post you asked "I am trying to find the period of $(sin2x)^2$ and I removed the square by using double angle formula and then $2\pi/B$. That's how I would answer.....Another idea to confirm: Graphing. – imranfat Oct 15 '17 at 02:46
so then the period would be $\frac{1}{2}- \frac{\pi}{2}$? – user104 Oct 15 '17 at 02:48
2

No, the $\frac{1}{2}$ is a vertical shift of the sine curve. It has got nothing to do with the period. The period is determined by $cos4x$ and from that you deduce $\pi/2$ – imranfat Oct 15 '17 at 02:50

1 Answers1