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When does cos(A+B)=cosA +cosB

  1. Nov 23, 2009 #1
    I recently read an article in a math magazine where some guy found anlges (A and B) when sin(A+B)=sinA + sinB. I have been trying to work with cos(A+B) and see if there are instances when cos(A+B) would equal cosA+cosB. Any ideas?
     
  2. jcsd
  3. Nov 23, 2009 #2
    assuming he didn't do it by brute force I suppose he may have used the formulae

    [tex]\sin(A+B)=\sin A \cos B + \cos A \sin B[/tex]

    [tex]\sin A + \sin B =2\sin\frac{A+B}{2}\cos\frac{A-B}{2}[/tex]

    so he possibly assumed

    [tex]\sin A \cos B + \cos A \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}[/tex]

    and played around with it or other formulae?

    but honestly I have no idea how he did it, I didn't really "think"/"work" this over.:rolleyes:
    btw, can you post a link or something to that article?
     
    Last edited: Nov 23, 2009
  4. Nov 23, 2009 #3

    Mark44

    Staff: Mentor

    If you set sinAcosB + sinBcosA = sinA + sinB, this will be true if cosB = 1 and cosA = 1, which means that A and B can be 0, pi, 2pi, etc. Any integer multiple of pi works.

    It's much harder to find a solution for cos(A + B) = cosA + cosB, since cos(A + B) = cosAcosB - sinAsinB.
     
  5. Nov 23, 2009 #4
    I think he means angles besides those (which are trivial cases... I don't think such a "discovery" would be that interesting to be published in a mathematics magazine like the OP said)... 0=0 isn't that fascinating...:wink:

    (btw, [tex]\cos \pi = -1[/tex] so only angles of the form [tex]2k\pi , \ k \in \mathbb{Z}[/tex] work for this solution)
     
  6. Nov 23, 2009 #5

    Mark44

    Staff: Mentor

    [itex]k\pi[/itex], [itex]2k\pi[/itex], what's the difference?:smile:
     
  7. Nov 23, 2009 #6

    Borek

    User Avatar

    Staff: Mentor

    π/4 & 3π/2
     
  8. Nov 23, 2009 #7
    you said we must have cosA=cosB=1 , I was replying to that...

    but anyway in this case yeah, it doesn't matter since it's 0=0 anyway... :approve:

    anyway @OP:

    if this is not a farce/joke (which I suspect to be the case) I'd like to know about that "result", thanks. :smile:

    lol :rofl:
    nice one.

    edit: I just noticed you have a "best humour" badge. indeed, you're good. :surprised
     
  9. Nov 23, 2009 #8
    Giving one particular solution is useless from a mathematicians point of view. It's important to find all solutions.

    I'm not sure what that guy did, but one easily gets
    cos(a+b)=cos(a)+cos(b)
    [tex]a=\arccos t+\arccos\left(t-\frac{1}{2t}\right)[/tex]
    [tex]b=\arccos t-\arccos\left(t-\frac{1}{2t}\right)[/tex]
    [tex]\frac{\sqrt{3}-1}{2}<|t|<1[/tex]
    which is a continous set of solutions.

    And btw, sin(a+b)=sin(a)+sin(b) is solved with
    [tex]a=\pi+t, b=\pi-t[/tex]
    or
    [tex]a=t, b=0[/tex]
    or
    [tex]a=0, b=t[/tex]
     
    Last edited: Nov 23, 2009
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