The instantaneous current $i$ and the instantaneous voltage $v$ in a pure resistance a.c. circuit is given by:
$$i=i_{\text{max}} \sin \omega t$$ $$ \text{and} \space v= v_{\text{max}} \sin \omega t$$
Since the power follows:
$$P = iv =i^2 R$$
Show that an equation for instantaneous power is;
$$P=i_{\text{max}}^2 R[1-\cos 2(\omega t)]/2$$
I have been revising basic compound angles and I am struggling to understand the following question from the examples I have previously studied on such topic.
I cannot see how the compound angle formulae relates to this question (if I am correct in thinking that it is relevant for a question of this nature).
A first step, or point of direction/suggestion would be brilliant. Thank You.
If you're question is "how can I turn $(\sin(\omega t))^2$ into a double-angle-type form", then you should look at your double angle formulas.