Integrating sin2θ: How to Prove <sin2θ>=1/2 and <cos2θ>=1/2?

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Homework Help Overview

The discussion revolves around proving that the average values of and equal 1/2. Participants are exploring the integration of these functions over specific intervals to establish this proof.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss integrating from -Infinity to +Infinity, with some suggesting that integrating over one cycle may suffice. Questions arise regarding the appropriate interval for averaging and the method of integration used.

Discussion Status

Some participants have offered insights regarding the integration process, noting the need to divide by the interval length to find the average. There appears to be a mix of interpretations and approaches being explored without a clear consensus on the method.

Contextual Notes

There is mention of confusion regarding the application of the average value rule and the specific intervals used for integration, indicating potential gaps in understanding the requirements for the proof.

gennarakis
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How can <sin2θ>=1/2 and <cos2θ>=1/2

How is the proof made?Integrate sin2θ from -Infinity to +Infinity?
 
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gennarakis said:
How can <sin2θ>=1/2 and <cos2θ>=1/2

How is the proof made?Integrate sin2θ from -Infinity to +Infinity?

You only need to integrate over one cycle. Every other cycle will be the same, right?
 
Average value on what interval?
 
I just integrated from 0 to 2Pi changed sin2θ=(1-cos2θ)/2 but the result is Pi and not 1/2...:confused:
 
gennarakis said:
I just integrated from 0 to 2Pi changed sin2θ=(1-cos2θ)/2 but the result is Pi and not 1/2...:confused:

You forgot to divide by 2*PI to get the average over the interval...
 

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