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

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SUMMARY

The discussion focuses on proving that the average values of and equal 1/2 over one complete cycle. Participants clarify that integrating from 0 to 2π yields a total area of π, which must then be divided by the interval length (2π) to find the average value. The correct average value is confirmed to be 1/2 after applying the average value theorem correctly. The integration technique discussed is essential for understanding periodic functions in trigonometry.

PREREQUISITES
  • Understanding of trigonometric functions, specifically and
  • Knowledge of integration techniques in calculus
  • Familiarity with the average value theorem for functions
  • Basic comprehension of periodic functions and their properties
NEXT STEPS
  • Study the average value theorem in calculus
  • Practice integrating trigonometric functions over specified intervals
  • Learn about the properties of periodic functions and their applications
  • Explore advanced integration techniques, such as substitution and integration by parts
USEFUL FOR

Students and educators in mathematics, particularly those focusing on calculus and trigonometry, as well as anyone interested in understanding the behavior of periodic functions and their averages.

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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