anonymous
  • anonymous
how to find the average of sin square theta?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Don't get it :S
anonymous
  • anonymous
me neither...how can u find the average of just one quantity?
apoorvk
  • apoorvk
You need \(<\sin^2\theta>\) right?

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anonymous
  • anonymous
not a quantity.
anonymous
  • anonymous
*just one term
anonymous
  • anonymous
yes @apoorvk
anonymous
  • anonymous
average must be 1/2 ? :P
anonymous
  • anonymous
how you can say that?
apoorvk
  • apoorvk
@Vaidehi09 basically that means average value of \(\sin^2\theta\) in its period - which is \(\pi\). (the graph for rekindling old memories:)|dw:1343221614291:dw|
anonymous
  • anonymous
is this graph applicable for that one? @apoorvk
anonymous
  • anonymous
You can try integration
anonymous
  • anonymous
please contineu how to do so.. ? need to integrate it from zero o pi. but in which period or how?
anonymous
  • anonymous
\[1\div(b-a)\int\limits_{a}^{b}\sin^2\theta d \theta \]
apoorvk
  • apoorvk
For this we need to integrate \(\sin^2\theta\) from 0 to \(\pi\),
anonymous
  • anonymous
take a = 0 and b = 2pi
anonymous
  • anonymous
no bro @apoorvk it should be 2pi
anonymous
  • anonymous
@apoorvk i cant understand why the period is pi.
apoorvk
  • apoorvk
\[\int\limits_0^{\pi}\sin^2\theta.d\theta\] and then I divide this by the difference of the limit, that is \(\pi -0=\pi\). Basically like the sum of the values by the 'domain covered'.
apoorvk
  • apoorvk
@Vaidehi09, do you understand that the period sinx of 2pi?
anonymous
  • anonymous
yes
anonymous
  • anonymous
okay. thanks to all.
apoorvk
  • apoorvk
now, @Vaidehi09 when I square the function - two changes happen to curve - 1. The curve's shape changes. 2. The negative part of the graph is flipped up in the positive axis. So, whereas the period was 2 pi for sinx, the \(\sin^2x\) graph repeats itself after every \(\pi\)-distance. Hence the period is pi.
anonymous
  • anonymous
OmG ! Expert ! xD Keep it up brother.
apoorvk
  • apoorvk
Not really :P
anonymous
  • anonymous
@apoorvk ah yes..of course! i was getting confused with the -ve part and what kind of change will happen to the shape of the graph....but i get it now. thanks a lot! :))
apoorvk
  • apoorvk
No worries :]
anonymous
  • anonymous
I am new to calc.. so I am not sure about my replies :P
anonymous
  • anonymous
@apoorvk good Explaination.

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