Why does the trig expression just disappear?

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SUMMARY

The discussion centers on the disappearance of the trigonometric expression \(1 - \cos(x)\) in the context of time-averaging over one period. The key conclusion is that when averaging the expression over the interval from \(0\) to \(2\pi\), the mean value becomes \(1\). This is established through the integral calculation \(\frac{\int_0^{2\pi}(1-\cos x)dx}{\int_0^{2\pi}dx}=1\), demonstrating that the trigonometric component effectively cancels out in this specific averaging process.

PREREQUISITES
  • Understanding of trigonometric functions, specifically cosine.
  • Knowledge of integral calculus, particularly definite integrals.
  • Familiarity with the concept of time-averaging in periodic functions.
  • Basic principles of mean value calculations in calculus.
NEXT STEPS
  • Study the properties of trigonometric integrals, focusing on \(1 - \cos(x)\).
  • Learn about time-averaging techniques in periodic functions.
  • Explore the concept of mean value in calculus, particularly in relation to integrals.
  • Investigate applications of trigonometric identities in integral calculus.
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Students studying calculus, particularly those focusing on trigonometric functions and integrals, as well as educators looking for examples of time-averaging in mathematical contexts.

adamaero
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Homework Statement


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Homework Equations



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3. Solution
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Then the equations from part (A) are used--but why does the converted trig (1-cos(etc)) go away?
Does this have to do with the mean-squared bit?
 

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Last edited:
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It becomes 1 when you time-average over one period.$$\frac{\int_0^{2\pi}(1-\cos x)dx}{\int_0^{2\pi}dx}=1$$
 
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