## anonymous 4 years ago how to find the average of sin square theta?

1. anonymous

Don't get it :S

2. anonymous

me neither...how can u find the average of just one quantity?

3. anonymous

You need $$<\sin^2\theta>$$ right?

4. anonymous

not a quantity.

5. anonymous

*just one term

6. anonymous

yes @apoorvk

7. anonymous

average must be 1/2 ? :P

8. anonymous

how you can say that?

9. anonymous

@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|

10. anonymous

is this graph applicable for that one? @apoorvk

11. anonymous

You can try integration

12. anonymous

please contineu how to do so.. ? need to integrate it from zero o pi. but in which period or how?

13. anonymous

$1\div(b-a)\int\limits_{a}^{b}\sin^2\theta d \theta$

14. anonymous

For this we need to integrate $$\sin^2\theta$$ from 0 to $$\pi$$,

15. anonymous

take a = 0 and b = 2pi

16. anonymous

no bro @apoorvk it should be 2pi

17. anonymous

@apoorvk i cant understand why the period is pi.

18. anonymous

$\int\limits_0^{\pi}\sin^2\theta.d\theta$ 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'.

19. anonymous

@Vaidehi09, do you understand that the period sinx of 2pi?

20. anonymous

yes

21. anonymous

okay. thanks to all.

22. anonymous

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.

23. anonymous

OmG ! Expert ! xD Keep it up brother.

24. anonymous

Not really :P

25. 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! :))

26. anonymous

No worries :]

27. anonymous

I am new to calc.. so I am not sure about my replies :P

28. anonymous

@apoorvk good Explaination.