Engineering Solving 1st Order Circuits: Private Solution & Finding C, D

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In solving a first-order circuit with the input current i_s = Acos(wt + phase), the user is attempting to find the private solution i_p expressed as a combination of cosine and sine functions. They question whether they can equate the transformed functions f[C,D]*cos(wt) + g[C,D]*sin(wt) to Acos(wt) directly. The response clarifies that to find C and D, one must develop the expression Acos(wt + phase) into its sine and cosine components, identifying f[C,D] as A_1 and g[C,D] as A_2. This process is essential for accurately determining the coefficients needed for the solution.
asi123
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Ok, let's say I have a first order circuit with i_s = Acos(wt+phase) (like in the pic).
I'm having problem with the private solution, let's say I pick my private solution like this:
i_p = Bcos(wt + phase) = Ccos(wt) + Dsin(wt) (right?)
After I place this i_p in the equation I come up with this:
f[C,D]*cos(wt) + g[C,D]*sin(wt) = Acos(wt+phase) (right?)
Can I say
f[C,D]*cos(wt) + g[C,D]*sin(wt) = Acos(wt)?
I mean, How do I come up with C and D?

10x in advance.
 

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I'm afraid I'm not understanding your question. What is i_p, f[C,D} and g[C,D]?
 
asi123 said:
Ok, let's say I have a first order circuit with i_s = Acos(wt+phase) (like in the pic).
I'm having problem with the private solution, let's say I pick my private solution like this:
i_p = Bcos(wt + phase) = Ccos(wt) + Dsin(wt) (right?)
After I place this i_p in the equation I come up with this:
f[C,D]*cos(wt) + g[C,D]*sin(wt) = Acos(wt+phase) (right?)
Can I say
f[C,D]*cos(wt) + g[C,D]*sin(wt) = Acos(wt)?
I mean, How do I come up with C and D?

10x in advance.

You must develop Acos(\omega t + \phi) = A_1cos(\omega t) + A_2sin(\omega t)

Now, f[C,D] = A_1 and g[C.D] = A_2
 

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