A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
how to find the average of sin square theta?
anonymous
 3 years ago
how to find the average of sin square theta?

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0me neither...how can u find the average of just one quantity?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You need \(<\sin^2\theta>\) right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0average must be 1/2 ? :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how you can say that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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
 3 years ago
Best ResponseYou've already chosen the best response.0is this graph applicable for that one? @apoorvk

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You can try integration

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0please contineu how to do so.. ? need to integrate it from zero o pi. but in which period or how?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[1\div(ba)\int\limits_{a}^{b}\sin^2\theta d \theta \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For this we need to integrate \(\sin^2\theta\) from 0 to \(\pi\),

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0take a = 0 and b = 2pi

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no bro @apoorvk it should be 2pi

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@apoorvk i cant understand why the period is pi.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_0^{\pi}\sin^2\theta.d\theta\] 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'.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Vaidehi09, do you understand that the period sinx of 2pi?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now, @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
 3 years ago
Best ResponseYou've already chosen the best response.0OmG ! Expert ! xD Keep it up brother.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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! :))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am new to calc.. so I am not sure about my replies :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@apoorvk good Explaination.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.