Where does the power absorbed of this capacitor come from?
You have

$$p(t)=v(t)i(t)=-100\cdot120\pi\sin(2\pi 60t)\cos(2\pi 60t)=
-50\cdot 120\pi\sin(4\pi 60t)=\\=-6000\pi\sin(240\pi t)=-18850\sin(240\pi t)$$

I think you entered one factor of \$\pi\$ too many in WolframAlpha. Note that I used \$2\sin(x)\cos(x)=2\sin(2x)\$.