Solve V = 5 + 12*j in Form v(t) = A*cos(w*t) + B*sin(w*t)

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To solve V = 5 + 12*j in the form v(t) = A*cos(w*t) + B*sin(w*t), the correct approach involves recognizing the relationship between the components and their projections on the unit circle. The initial projection at t=0 gives v(t) = C*cos(phi), leading to the realization that v(t) can be expressed as 5*cos(w*t) - 12*sin(w*t). The confusion arises from the interpretation of the sine and cosine functions on the unit circle, particularly regarding their signs in different quadrants. The solution can be confirmed using the trig identity for the cosine of a sum of angles. Understanding these relationships clarifies the transition from the complex form to the desired trigonometric representation.
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Homework Statement



If V = 5 + 12*j then write v(t) in the form v(t) = A*cos(w*t) + B*sin(w*t)

w = 10000 rad/s

Homework Equations



N/A

The Attempt at a Solution



I can easily put it into the form v(t) = C*cos(w*t + phi) and I can probably get the above form using a trig identity (haven't tried yet).

However, apparently it should be easily solvable by looking at it graphically.

I look at the case when t=0, then v(t) = C*cos(phi), the projection on the x-axis = 5.

Since x = cos(t) and y = sin(t) on the unit circle, then I figured v(t) = 5*cos(w*t) + 12*sin(w*t)

That works for t=0, but the actual answer is 5*cos(w*t) - 12*sin(w*t) (I found this by plotting it and 13*cos(w*t + arctan(12/5), they matched)

I seem to be on the right track but I don't understand the actual answer. The only resource I've found on the net puts +cos(t) on the positive x-axis but +sin(t) on the negative y axis... ?

I just don't understand and no one can help because it's an assignment question. I've got the mark now though because I know the answer, but I don't understand how to get it exactly.
 
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Consider the trig identity for the cosine of the sum of two angles:
cos(\alpha + \beta) = cos(\alpha) cos(\beta) - sin(\alpha) sin(\beta)
and work from your A*cos(ωt + φ) form.
 
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