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shruti Group Title

how to find the average of sin square theta?

  • 2 years ago
  • 2 years ago

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  1. vikrantg4 Group Title
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    Don't get it :S

    • 2 years ago
  2. Vaidehi09 Group Title
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    me neither...how can u find the average of just one quantity?

    • 2 years ago
  3. apoorvk Group Title
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    You need \(<\sin^2\theta>\) right?

    • 2 years ago
  4. shruti Group Title
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    not a quantity.

    • 2 years ago
  5. Vaidehi09 Group Title
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    *just one term

    • 2 years ago
  6. shruti Group Title
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    yes @apoorvk

    • 2 years ago
  7. vikrantg4 Group Title
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    average must be 1/2 ? :P

    • 2 years ago
  8. shruti Group Title
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    how you can say that?

    • 2 years ago
  9. apoorvk Group Title
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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|

    • 2 years ago
  10. shruti Group Title
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    is this graph applicable for that one? @apoorvk

    • 2 years ago
  11. vikrantg4 Group Title
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    You can try integration

    • 2 years ago
  12. shruti Group Title
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    please contineu how to do so.. ? need to integrate it from zero o pi. but in which period or how?

    • 2 years ago
  13. vikrantg4 Group Title
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    \[1\div(b-a)\int\limits_{a}^{b}\sin^2\theta d \theta \]

    • 2 years ago
  14. apoorvk Group Title
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    For this we need to integrate \(\sin^2\theta\) from 0 to \(\pi\),

    • 2 years ago
  15. vikrantg4 Group Title
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    take a = 0 and b = 2pi

    • 2 years ago
  16. vikrantg4 Group Title
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    no bro @apoorvk it should be 2pi

    • 2 years ago
  17. Vaidehi09 Group Title
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    @apoorvk i cant understand why the period is pi.

    • 2 years ago
  18. apoorvk Group Title
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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'.

    • 2 years ago
  19. apoorvk Group Title
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    @Vaidehi09, do you understand that the period sinx of 2pi?

    • 2 years ago
  20. Vaidehi09 Group Title
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    yes

    • 2 years ago
  21. shruti Group Title
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    okay. thanks to all.

    • 2 years ago
  22. apoorvk Group Title
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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.

    • 2 years ago
  23. vikrantg4 Group Title
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    OmG ! Expert ! xD Keep it up brother.

    • 2 years ago
  24. apoorvk Group Title
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    Not really :P

    • 2 years ago
  25. Vaidehi09 Group Title
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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! :))

    • 2 years ago
  26. apoorvk Group Title
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    No worries :]

    • 2 years ago
  27. vikrantg4 Group Title
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    I am new to calc.. so I am not sure about my replies :P

    • 2 years ago
  28. sami-21 Group Title
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    @apoorvk good Explaination.

    • 2 years ago
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