Yes you are right. This does involve more trig that physics. So let's review the math for a bit.
Cos(2x) = cos^2(x) - sin^2(x) Substitute 1 - sin^2(x) for cos^2(x)
Cos(2x) = 1 - sin^2(x) - sin^2(x)
cos(2x) = 1 - 2*sin^2(x) Now we need to solve for sin^2(x)
2sin^2(x) = 1 - cos(2x)
sin^2(x) = (1 - cos(2x))/2
Now all we need do is make a few substitutions.
x = ωt
Substitute (1 - cos(2ωt))/2 for sin^2(ωt)
P = (Io)^2 *R * (1 - cos(2ωt))/2
P = (Io)^2*R/2 - (Io)^2*R cos(2ωt) / 2
I think if you follow it carefully, you will find that the answer is
B <<<<< Answer.
