Question

Please help

If v= Vpsinwt
i= Ipsin(wt-Ɵ)

prove: p = 1/2 Vp Ip cos Ɵ - 1/2 Vp Ip cos (2wt-Ɵ)

Answer 1

To do this you need to prove:
\[\sin (wt) \sin(wt - ø) = \frac{ 1 }{ 2 }(\cosø - \cos (2wt - ø) \]

this is nothing more than playing around with the various trig formulae.
i'd start by breaking up the RHS as the cos (2wt - ø) looks somewhat contrived.

RHS becomes:
\[\frac{ 1 }{2 } ( \cos ø - \cos(2wt) \cosø - \sin (2wt) \sin ø ) = \]
\[\frac {1}{2} ( \cosø • 2 \sin^{2}wt - 2 \sin(wt) \cos(wt) \sin(ø) )\]
\[= \sin(wt) (\cos ø \sin(wt) - \sinø \cos(wt) )\]

all that is left is to expand the sin (wt - ø) on the LHS -- I will leave you to do that --- and you will see that LHS = RHS.