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electrical engineering math about conversion...MEDAL 3cos600t - 5cos(600t + 20deg) find the phasor
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oh, and the answer is 2.41e^(-j134.8 deg) but I don't know how to get it
or (2.41∠34.8 deg) in the old way Here are similar problems: http://www.egr.unlv.edu/~eebag/EE-221-Chap%209.pdf#page=10
OK, so from example 16(c) is broken up into real and imaginary parts then converts from rectangular form to polar...similarly: 3-5e^(j20 deg) is half rectangular form and last part polar...
sob! still not clicking...is it that the Argand diagram--with the one having 90 degs--makes it protrude into the imaginary part of the graph? ...even though it's a visual aid...still...
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