addition of phasors

Discussion in 'Homework Help' started by gunning91, Dec 5, 2009.

  1. gunning91

    gunning91 Thread Starter New Member

    Joined:
    Dec 5, 2009
    Messages:
    1
    hey minor problem, i have two phasors i need to add:
    44.7cos(ωt-161°) + 126.5(ωt + 63°)
    how do i add them when one is a cos and the other is a sin, the answer is supposed to be given as a sin, but i cant find anything similar in my notes.
    the answer is 100sin(ωt +45°).
    all i need to know is how to write the equation with cos in it as a sine.
    thanks for any help, i have an exam on monday so help is much appreciated.
    andy
  2. t_n_k

    t_n_k AAC Fanatic!

    Joined:
    Mar 6, 2009
    Messages:
    5,085
    Use identities such as ...

    cos(a+b)=cos(a).cos(b)-sin(a).sin(b)

    sin(a+b)=sin(a).cos(b)+cos(a).sin(b)

    to expand your given relationship and then collect like terms - in sin(ωt) and cos(ωt)

    You end up with something like

    f(t)=A.cos(ωt)+B.sin(ωt)

    From this you want

    f(t)=C.sin(ωt+θ) [f(t)=C.sin(ωt).cos(θ) +C.cos(ωt).sin(θ)]

    where C=√(A^2+B^2) and θ=arctan(A/B)
  3. steveb

    steveb Senior Member

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    Jul 3, 2008
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    2,431
    Location:
    New England, USA
    Cosine and sine are related by a 90 degree (or pi/2 radian) phase shift. Converting one to the other is an addition or subtraction of this amount in the phase portion of the phasor.

    i.e. sin(x) = cos(x-pi/2) and cos(x)=sin(x+pi/2)
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