Not sure why cos(phi) = sin(phi) = 1

Discussion in 'Homework Help' started by raddian, Jun 25, 2015.

  1. raddian

    Thread Starter New Member

    Jun 20, 2015
    6
    0
    Screenshot - 06252015 - 07:38:57 AM.png
    [Taken from Basic Engineering Circuit Analysis, 10th Ed. Irwin.]

    <br />
\begin{align}<br />
i(t) &= A \cos(\omega t+ \phi)<br />
i(t) &=A \cos(\phi)\cos(\omega t) - A \sin(\phi) \sin(\omega t)<br />
\end{align}<br />
i(t) =A \cos(\omega t) + A \sin(\omega t)<br />
    And they decided to just make cos(phi) = 1 = sin(phi), Why did this happen?

    (here is the rest of the example, which I dont think provides explanation)
    Screenshot - 06252015 - 07:46:50 AM.png
     
    Last edited: Jun 25, 2015
  2. raddian

    Thread Starter New Member

    Jun 20, 2015
    6
    0
    I now think cos(phi) and sin(phi) are constants, given that the A turned into A_1 and A_2 :confused:
     
  3. Papabravo

    Expert

    Feb 24, 2006
    10,137
    1,786
    The only value of φ for which they are equal is π/4 or 45°. They can never be simultaneously equal to 1 -- EVER!
    It is true that φ is a constant, and it is true that A is a constant, and it is also true that
    A_1 and A_2
    are constants, but they don't have to be equal.
     
    Last edited: Jun 25, 2015
  4. WBahn

    Moderator

    Mar 31, 2012
    17,716
    4,788
    Attention to detail.

    <br />
A_1 \; = \; A \cos (\phi)<br />
A_2 \; = \; A \sin (\phi)<br />
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
    2,423
    490
    Hi,

    Yes that is the perfect reason for the misunderstanding Wbahn.

    This is a trigonometric identity that is often used in electronics but written sometimes in a slightly more apparent form is:
    A*cos(wt)+B*sin(wt)=C*cos(wt+ph)

    or alternately:
    C*cos(wt+ph)=A*cos(wt)+B*sin(wt)

    You're goal is either to find A and B, or to find C and ph, depending on which way you are going, in order to either get rid of the phase shift or to include a phase shift in the final form.
    Writing it this way makes it a little more apparent, but then we just have C equal to A in the previous form, and A1 and A2 equal to A and B in the second form. So it's the same thing just written using A, B, and C rather than A1, A2, and A.

    So the basic idea is to either get rid of the phase shift or to convert to a form with a phase shift. The phase shift can also be calculated, given the two term form, as well as the constant C.
     
    Last edited: Jun 25, 2015
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