I am working on an applied math assignment and this particular question is bugging me.
The problem states:
The power, $p$, in an electrical circuit is quantified by $p=iv$.
This problem wants me to show the following:
- If $i=Isin(\omega t-\frac{\pi}{2})$ and $v=Vsin(\omega t)$ then $p=-\frac{IV}{2}cos(2\omega t-\frac{\pi}{2})$
Here is what I did so far:
$$Isin(\omega t-\frac{\pi}{2})Vsin(\omega t)$$
Using the identity: $sin(A-B)=sin(A)cos(B)-cos(A)sin(B)$
$$Isin(\omega t-\frac{\pi}{2})=I[sin(\omega t)cos(\frac{\pi}{2})-cos(\omega t)sin(\frac{\pi}{2})]$$
$$I[0-cos(\omega t)(1)]$$
$$\therefore -Icos(\omega t) $$
$$-Icos(\omega t)Vsin(\omega t)$$
$$-IVcos(\omega t)sin(\omega t)$$
I am stuck at this point and I don't really know where to go next. I noticed that I also took out the $\frac{\pi}{2}$ which is supposed to be part of my final answer. Have I selected the correct identity? Are there better ways of getting my way around these type of problems?
Hint: Use the product to sum formula $$\sin A \sin B = \frac {1}{2}[\cos (A - B) - \cos (A + B)]$$ with $A = \omega t - \pi/2$ and $B = \omega t$. What do you notice?