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Conversion to phasor from cos and sine

  1. Apr 2, 2013 #1
    I am given the problem:
    100Sin(ωt+30) +20cos(ωt)

    and the solution is 78.7<38.9degrees

    How do I convert to phasor form?

    I know for this calculation I need the following relationship:
    √2Esin(ωt-∅)=Ee^(-j∅)

    At first I tried making both sines and then splitting up 100Sin(ωt+30) and 20Sin(ωt+90)using the trigonometric identities but I get stuck at that. Please help!
     
  2. jcsd
  3. Apr 2, 2013 #2
    Solved it :D
    Please let me know if anyone would like me to go over it.
     
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