The power loss due to current $I(t)=A\, \sin\omega t$ for resistor R is
$P=I^2R=A^2\, \sin^2\omega t \, R=A^2 \frac{1-\cos 2\omega t}{2} \, R$
oic, the blue line is to show [color=blue]$Power loss in R =A^2 \frac{1-\cos 2\omega t}{2} \, R$[/color]