[quote author=Fu-Kwun Hwang link=topic=1942.msg7116#msg7116 date=1284391806]
The power loss due to current I(t)=A\, \sin\omega t for resistor R is
P=I^2R=A^2\, \sin^2\omega t \, R=A^2 \frac{1-\cos 2\omega t}{2} \, R
[/quote]

oic, the blue line is to show [color=blue]Power loss in R =A^2 \frac{1-\cos 2\omega t}{2} \, R[/color]
i see now :)