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How to go from a*cosx + b*sinx TO A*cos(wt - phi)?

  1. Mar 10, 2016 #1
    Moved from non-homework forum section, so homework template is not present.
    Express Cos(t - pi/8) + Sin(t - pi/8) in the form A*Cos(wt - phi).

    I got sqrt(2)*Cos(t-3pi/4).

    Not sure if thats right though
     
    Last edited by a moderator: Mar 13, 2016
  2. jcsd
  3. Mar 10, 2016 #2

    Charles Link

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    The simplest way to see this is to use ## Acos(\omega*t)+Bsin(\omega*t)=sqrt(A^2+B^2)*(A/srqrt(A^2+B^2))cos(\omega*t)+(B/sqrt(A^2+B^2))sin(\omega*t)##
    Let ## A/sqrt(A^2+B^2)=cos(\phi) ## and ## B/sqrt(A^2+B^2)=sin(\phi) ## You should recognize the expanded terms as ## cos(\omega*t-\phi) ## with a sqrt(A^2+B^2) in front of it and ## \phi ## is the angle that has ## tan(\phi)=B/A ## This actually comes up quite a lot in different physics calculations and is something worth looking over a couple of times until you have it completely memorized. I get ## 3\pi/8 ## though. Please check the arithmetic.
     
  4. Mar 11, 2016 #3
    Please post your working, the next time you ask a question. It saves everybody's time.
    Thanks.
     
  5. Mar 13, 2016 #4

    SammyS

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    You can check your answer as follows:

    −3π/4 = −π/8 − 5π/8

    Therefore, use the angle addition identity for cosine on:

    cos( (t−π/8) − 5π/8 )
     
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