shruti
how to find the average of sin square theta?
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vikrantg4
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Don't get it :S
Vaidehi09
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me neither...how can u find the average of just one quantity?
apoorvk
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You need \(<\sin^2\theta>\) right?
shruti
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not a quantity.
Vaidehi09
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*just one term
shruti
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yes @apoorvk
vikrantg4
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average must be 1/2 ? :P
shruti
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how you can say that?
apoorvk
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@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|
shruti
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is this graph applicable for that one? @apoorvk
vikrantg4
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You can try integration
shruti
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please contineu how to do so.. ?
need to integrate it from zero o pi.
but in which period or how?
vikrantg4
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\[1\div(b-a)\int\limits_{a}^{b}\sin^2\theta d \theta \]
apoorvk
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For this we need to integrate \(\sin^2\theta\) from 0 to \(\pi\),
vikrantg4
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take a = 0 and b = 2pi
vikrantg4
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no bro @apoorvk it should be 2pi
Vaidehi09
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@apoorvk i cant understand why the period is pi.
apoorvk
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\[\int\limits_0^{\pi}\sin^2\theta.d\theta\] 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
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@Vaidehi09, do you understand that the period sinx of 2pi?
Vaidehi09
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yes
shruti
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okay.
thanks to all.
apoorvk
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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.
vikrantg4
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OmG ! Expert ! xD Keep it up brother.
apoorvk
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Not really :P
Vaidehi09
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@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
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No worries :]
vikrantg4
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I am new to calc.. so I am not sure about my replies :P
sami-21
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@apoorvk good Explaination.