prove P=VI for V=5sinwt, I=10sin(wt-x),p=25(cosx-cos(2wt-x))

[vgreen74]

prove that P=VI where V=5sinwt and I=10sin(wt-x) and p=25(cosx-cos(2wt-x))
am I correct so far and where do I go from here?
Let wt=A and x=B
P=(5sinA)(10sin(a-b))
P=50(sinA)(sinAcosB-cosAsinB)
P=50(sin^2AcosB-sinAsinBcosA)

P=50(cosB+(sin^2AcosB-sinAsinBcosA-cos^2AcosB))
P=25(cosB-(2sinAsinBcosA+cosBcos^2A-sin^2AcosB))
P=25(cosB-cos(2A-B))

[stapel_eliz]

How are ''p'' and ''P'' related?

Thank you!

[vgreen74]

Sorry P and p are the samething I just missed the shift key. I have to prove that starting with the equations for V and I I end up with the equation for P.
using the cos(A+B) and cos(A-B) rules

[stapel_eliz]

where do I go from here?

I'm not sure what you mean by this, since you seem to have arrived at the final result...?

P=50(cosB+(sin^2AcosB-sinAsinBcosA-cos^2AcosB))
P=25(cosB-(2sinAsinBcosA+cosBcos^2A-sin^2AcosB))
P=25(cosB-cos(2A-B))

I was able to figure out some of the missing steps in your displayed work, but I'm afraid I'm drawing a blank here. Please reply with the rest of what you did. Thank you!

[vgreen74]

I have tired to work from both ends of the equation as I was drawing a blank on working from the begining however I may not have the thrid line correct when tring to work back from the equation for P

[stapel_eliz]

I could be wrong, but I think the proof isn't working because the equality isn't true.

What did you get when you graphed the left- and right-hand sides of the equations? Did the lines match up? Mine aren't.

[vgreen74]

sorry my first mistake was with my grouping it should have been let A=wt and B=wt-x
so P=(5sinA)(10sinB)
P=50sinAsinB
using product to sum
2sinAsinB=cos(A-B)-cos(A+B)
therefore
P=25(2sinAsinB)
P=25(cos[wt-{wt-x}]-cos[wt+{wt-x}])
P=25(cos[wt-wt+x]-cos[wt+wt-x])
P=25(cosx-cos[2wt-x])
which was what I had to prove