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Math Help - Addition of sine & cosine?

  1. #1
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    Unhappy Addition of sine & cosine?

    Hello, can anyone give me some help please!!

    I'm struggling with the following question:

    Output of a circuit current is given by: i1 + i2.
    If i1 = 5sin(50t + pi/3) and i2 = 6cos 50t
    Calculate the amplitude of i and the first time it occurs.

    I have made a start but don't know if i'm going the right way? I have:

    5sin(50t + pi/3) = (5x) sin50tcos pi/3 + cos50tsin pi/3
    = 0.5 x sin50t + 0.866 x cos50t
    Therefore: 6cos50t + 5sin(50t + pi/3) = 6cos50t + 5(0.5sin50t + 0.866cos50t)

    Then I go on to find R and a, using arctan etc, but I don't think i'm going the right way, the inclusion of t is something I haven't come across in these sort of equations and is confusing me?
    Hope this makes some sense and someone can point me in the right direction?

    Thanks in advance.
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  2. #2
    Junior Member lanzailan's Avatar
    Joined
    Mar 2009
    Posts
    27
    Quote Originally Posted by joemc22 View Post
    Hello, can anyone give me some help please!!

    I'm struggling with the following question:

    Output of a circuit current is given by: i1 + i2.
    If i1 = 5sin(50t + pi/3) and i2 = 6cos 50t
    Calculate the amplitude of i and the first time it occurs.

    I have made a start but don't know if i'm going the right way? I have:

    5sin(50t + pi/3) = (5x) sin50tcos pi/3 + cos50tsin pi/3
    = 0.5 x sin50t + 0.866 x cos50t
    Therefore: 6cos50t + 5sin(50t + pi/3) = 6cos50t + 5(0.5sin50t + 0.866cos50t)

    Then I go on to find R and a, using arctan etc, but I don't think i'm going the right way, the inclusion of t is something I haven't come across in these sort of equations and is confusing me?
    Hope this makes some sense and someone can point me in the right direction?

    Thanks in advance.

    Hello joemc22..
    this is what i got..

    i1=5sin(50t+pi/3)
    =5sin(50t+60)----- where pi/3=60 degrees

    i2=6cos50t

    we need to express i1 in cosine form. the rule for converting sine to cosine is to subtract 90degrees. hence,
    i1=5cos(50t+60-90)=5cos(50t-30)

    then we need to convert in phasor form, using scientific calculator..
    i1=5/_-30------>> i'm not sure how to write it here, i hope u understand
    i1=6/_0

    i(total)=i1 + i2
    =5/_-30 + 6/_0
    =10.33 - j2.5-------->>you will get the answer in rectangular form
    =10.628/_-13.6------>>in phasor form


    i hope this will help you.
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  3. #3
    Member u2_wa's Avatar
    Joined
    Nov 2008
    Posts
    119

    answer

    Solving the equations:
    5(sin (50t).cos( \pi/3) + cos(50t).sin ( \pi/3) + 6cos(50t)
    5(1/2 sin(50t) + √3 /2 cos(50t)) + 6cos(50t)
    5/2 sin (50t) + 5√3/2 cos(50t) + 6cos(50t)
    5/2 sin(50t) + (5√3 + 12)/2 cos (50t)
    R= √((5/2)^2+ ((5√3 + 12)/2)^2), R=10.63
    arctan (θ) = (5/2)/((5√3 + 12)/2) , so θ = 0.24 rad
    now equation is one function is:
    R cos(x-θ)
    10.63 cos(50t-0.24)
    max. value of cos(50t-0.24) = 1 so 10.63(1) = 10.63 is the max. value
    cos(0)= 1 i.e max. value of cos function
    so: 50t-0.24=0, hence t= 4.8*10^(-3)
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