Show that the average values of sin(nωt) and cos(nωt) over a period are zero, while the average values of cos² (nωt) and sin² (nωt) are ½.

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Accepted Answer

The average value of sin(nωt) over a period is

[latex] < \sin (nωt)> = (1/T)\int_{t-T/2}^{t+T/2}{dt^\prime \sin(nωt^\prime )} \ \quad \quad \quad \quad \quad = (1/nTω)[\cos(nωt-nωT/2)-\cos(nωt+nωT/2)] \ \quad \quad \quad \quad \quad = (1/2n \pi)[\cos(nωt-n\pi )-\cos(nωt+n\pi)]=0[/latex]

A similar analysis shows that < cos(nωt) > = 0. The interpretation of this result is simple: the functions sin(nωt) and cos(nωt) take opposite values in two consecutive half periods; thus, their average values over a period is zero. As for the squares of these functions, they are always positive. As sin²nωt = ½ - ½ cos(2nωt) and cos² nωt = ½ + ½ cos(2nωt), the average value of cos(2nωt) being zero, we deduce that the average values of sin²nωt and cos²nωt are equal to the constant term ½.

Question 1.4: Show that the average values of sin(nωt) and cos(nωt) over a...

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