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Help with trig proof

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data
    prove that Acosx+Bsinx=sqrt(A^2+B^2)sin(x+alpha) where tan(alpha)=A/B


    3. The attempt at a solution
    none so far except that sin(x+alpha)=sinx cos(alpha)+cosx sin(alpha)

    Any help is appreciated. This is due tomorrow (It was just assigned yesterday).
    I've taken a year off school so I'm kind of out of the swing of stuff like this.

    Thanks
     
  2. jcsd
  3. Aug 31, 2010 #2

    Mark44

    Staff: Mentor

    I would start on the right side and see if you can show that it's equal to the left side. It might be useful to note that tan(alpha) = sin(alpha)/cos(alpha) = A/B = (A/sqrt(A^2 + B^2))/(B/sqrt(A^2 + B^2)).

    I haven't worked this out, but that's where I would start.
     
  4. Aug 31, 2010 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Denoting sqrt(A^2+B^2)=C, you have the equation

    Acosx + Bsinx =(C*sin(alpha))cos x +(C*cos(alpha))sinx


    This has to be an identity, valid for any x. If x=0, sinx=0, cosx=1,
    A=C*sin(alpha). If x=pi/2, sinx=1, cosx=0, and B=C*cos(alpha).

    So you have two equations for C and alpha:

    A=C*sin(alpha)
    B=C*cos(alpha)

    Square both of equations and add the together. Use that cos2x+sin2x =1 what do you get for C?

    Divide the first equation by the second one, what do you get for alpha?

    ehild
     
  5. Aug 31, 2010 #4

    Mark44

    Staff: Mentor

    But that's what the OP needs to show. He/she can't just assume that it is true for all x.
     
  6. Aug 31, 2010 #5

    ehild

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    Homework Helper
    Gold Member

    Mark, you start with the right side, I would start with the left one, and show that it can be written in the form of C*sin(x+alpha) (that means it is identical to C*sin(x+alpha)) for a certain C and alpha. If it comes out that C^2=A^2+B^2 and tan(alpha)=A/B it means that the left side is identical to the right side of the original equation that had to be proved.

    ehild
     
  7. Aug 31, 2010 #6

    Mark44

    Staff: Mentor

    OK, I misunderstood what you were saying.
     
  8. Aug 31, 2010 #7
    thank you for your suggestions but I'm still a bit confused. It has been a long time since I've done stuff like this so its a bit harder for me to understand than usual. I'd appreciate if someone spelled out a few steps at least to get me started (if you don't mind) and hopefully I can figure out the rest. Thanks again.
     
  9. Sep 1, 2010 #8

    Mark44

    Staff: Mentor

    Basically what you should do is to start from the left side, as ehild suggests, and work with Acosx + Bsinx to make it look like C(A/C *cosx + B/C*sin x). Then you want to write A/C as sin(alpha) and B/C as cos(alpha) so that you end up with Csin(x + alpha).
     
  10. Sep 1, 2010 #9
    I GOT IT!!! Thank you so much....you were extraordinarily helpful :)
     
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