How can I calculate average power for a trigonometric expression?

AI Thread Summary
To calculate the average power for the expression 8sin(200t) - 6cos(200t - pi/4), it is essential to express it in terms of a single frequency and phase angle or as two terms with different frequencies. The user initially attempted to convert the sine term but encountered difficulties with cosine products. Clarification was provided that the focus is on instantaneous voltage rather than power, with a resistance of 4 ohms. A suggested approach is to decompose the cosine term using the identity cos(u-v) = cos(u)cos(v) + sin(u)sin(v) to align the time-dependent functions. This method will help achieve the necessary phase and frequency alignment for further calculations.
jesuslovesu
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Homework Statement




8sin(200t) - 6 cos(200t - pi/4)

Homework Equations





The Attempt at a Solution


I'm trying to calculate the average power for that expression so:
I need to get that expression into either one term with the same frequency and phase angle or two terms with a different frequency.
I've tried a bunch of different methods but can't quite get anything..

I've tried to turn 8sin(200t) into 4cos(100t)sin(100t) = 4cos(100t)cos(100t - pi/2) but then I get stuck with cosines multiplying each other.

Any help?
 
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jesuslovesu said:
8sin(200t) - 6 cos(200t - pi/4)

So is this supposed to be the instantaneous power?
 
Thanks for your reply,

Sorry I kind of misspoke, it's actually instantaneous voltage,
vs = 8sin(200t) - 6cos(200t - 45 deg)
R = 4 ohms

P = 1/2 VI cos(theta - phi)

I think the best way to approach it would be to find it as an expression with two terms of different frequencies (so I can use superposition) but so far that hasn't worked
 
You should decompose the second term using the following identity.

\cos(u-v)=\cos(u)\cos(v)+\sin(u)\sin(v)

Then all of your time-dependent trig functions will have the same phase and frequency.
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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