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Convert the following instantaneous currents to phasors, using cos(of) as the reference. Give your answers in both rectangular and polar form. (a) i(1) = 500\/2 cos(a)t-30) (b) i(r) = 4 sin(c)1 + 30) (c) i(1) 5 cos(a)1-15) + 4V2 sin(a)1 + 30)

I have a question with three parts; a, b, and c. I just don't get the part when to add the negative 90 degree. Also, if there is different cases than the negitive 90 degrees, can anyone please expose it here.  

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